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stringclasses 106
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17.9k
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stringclasses 5
values | level
float64 1
3
⌀ | solution
stringlengths 0
29.4k
| options
stringlengths 2
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2a9e96fbe37059749619ba6565a451b2
|
BITSAT_MATH
|
Vector Algebra
|
If in a $\Delta A B C, \mathbf{O}$ and $\mathbf{O}^{\prime}$ are the incentre and orthocentre respectively, then $\left(\boldsymbol{O}^{\prime} \mathbf{A}+\boldsymbol{O}^{\prime} \mathbf{B}\right.$
$\left.+\mathbf{O}^{\prime} \mathbf{C}\right)$ is equal to
|
singleCorrect
| 2 |
O' $\mathbf{A}=\mathbf{O}^{\prime} \mathbf{O}+\mathbf{O A}$
$\mathbf{O}^{\prime} \mathbf{B}=\mathbf{O}^{\prime} \mathbf{O}+\mathbf{O B}$
$\mathbf{O}^{\prime} \mathbf{C}=\mathbf{O}^{\prime} \mathbf{O}+\mathbf{O C}$
<img src="https://cdn-question-pool.getmarks.app/pyq/mht_cet/IK-nfwTpfNQpBGiI03ECR8PPzo9KVp05IFGWmJKiOiI.original.fullsize.png"><br>
$\Rightarrow \mathbf{O}^{\prime} \mathbf{A}+\mathbf{O}^{\prime} \mathbf{B}+\mathbf{O}^{\prime} \mathbf{C}$
$=3 \mathbf{0} \mathbf{0}+(\mathbf{O} \mathbf{A}+\mathbf{O B}+\mathbf{O C}) \ldots(\mathrm{i})$
$\because \quad \mathbf{O A}+\mathbf{O B}+\mathbf{O C}=\mathbf{O O}^{\prime}=-\mathbf{O}^{\prime} \mathbf{O}$
$\therefore \quad \mathbf{O}^{\prime} \mathbf{A}+\mathbf{O}^{\prime} \mathbf{B}+\mathbf{O}^{\prime} \mathbf{C}=3 \mathbf{O}^{\prime} \mathbf{O}-\mathbf{O}^{\prime} \mathbf{O}$
[from Eq. (i)]
$\mathbf{O}^{\prime} \mathbf{A}+\mathbf{O}^{\prime} \mathbf{B}+\mathbf{O}^{\prime} \mathbf{C}=2 \mathbf{O}^{\prime} \mathbf{O}$
|
["$2 \\mathbf{O}^{\\prime} \\mathbf{O}$", "$0^{\\prime} 0$", "$00^{\\prime}$", "2 OO"]
|
[0]
| null |
Practise Problem
|
f1ea2b41b302feed7e0212209d4c32d4
|
BITSAT_MATH
|
Vector Algebra
|
If the sum of two unit vectors is a unit vector, then the magnitude of their difference is
|
singleCorrect
| 1 |
$\begin{aligned} &\text { Let a and } \mathbf{b} \text { are two unit vectors. }\\ &\text { Then, } \quad|\mathbf{a}+\mathbf{b}|=1\\ &\Rightarrow \quad|\mathbf{a}+\mathbf{b}|^{2}=1 \end{aligned}$
$\Rightarrow \quad(a+b) \cdot(a+b)=1$
$\Rightarrow|a|^{2}+|b|^{2}+2 a \cdot b=1$
$\Rightarrow \quad 1+1+2 \mathbf{a} \cdot \mathbf{b}=1$
$\Rightarrow \quad 2 \mathbf{a} \cdot \mathbf{b}=-1$
Again, $|\hat{\mathbf{a}}-\mathbf{b}|^{2}=(\mathbf{a}-\hat{\mathbf{b}}) \cdot(\mathbf{a}-\hat{\mathbf{b}})$
$=|\mathbf{a}|^{2}+|\mathbf{b}|^{2}-2 \mathbf{a} \cdot \mathbf{b}$
$=1+1-(-1)$
$=1+1+1=3$
$|\hat{a}-\hat{b}|=\sqrt{3}$
|
["$\\sqrt{2}$ units", "2 units", "$\\sqrt{3}$ units", "$\\sqrt{5}$ units"]
|
[2]
| null |
Practise Problem
|
1f7c0c034bd73eb360404e9f20edf2b0
|
BITSAT_MATH
|
Vector Algebra
|
The vector $\mathbf{c} .(\mathbf{b}+\mathbf{c}) \times(\mathbf{a}+\mathbf{b}+\mathbf{c})$ is equal to
|
singleCorrect
| 1 |
<img src="https://cdn-question-pool.getmarks.app/pyq/ap_eamcet/sYUwJoDiiRfQ1uIogcq5SbEFwPyr8DkbiykuLMsS69Y.original.fullsize.png"><br>
|
["c. $\\mathbf{b} \\times \\mathbf{a}$", "$0$", "c. $\\mathbf{a} \\times \\mathbf{b}$", "c. $\\mathbf{a} \\times \\mathbf{b}$"]
|
[0]
| null |
Practise Problem
|
de7516949803bd04d7cbac3dfe499e13
|
BITSAT_MATH
|
Vector Algebra
|
If $3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\sqrt{3} \hat{\mathbf{k}}, \hat{\mathbf{i}}+\hat{\mathbf{k}}, \sqrt{3} \hat{\mathbf{i}}+\sqrt{3} \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$ are coplanar, then $\lambda$ is equal to
|
singleCorrect
| 1 |
Let $\mathbf{a}=3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\sqrt{3} \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{k}}$ and
$$
\mathbf{c}=\sqrt{3} \hat{\mathbf{i}}+\sqrt{3} \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}
$$
Since, these vectors are coplanar
$$
\begin{aligned}
& \therefore \quad \mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})=0 \\
& \Rightarrow \quad\left|\begin{array}{ccc}
3 & 3 & \sqrt{3} \\
1 & 0 & 1 \\
\sqrt{3} & \sqrt{3} & \lambda
\end{array}\right|=0 \\
& \Rightarrow \quad 3(0-\sqrt{3})-3(\lambda-\sqrt{3})+\sqrt{3}(\sqrt{3}-0)=0 \\
& \Rightarrow \quad-3 \sqrt{3}-3 \lambda+3 \sqrt{3}+3=0 \\
& \Rightarrow \quad \lambda=1 \\
&
\end{aligned}
$$
|
["1", "2", "3", "4"]
|
[0]
| null |
Practise Problem
|
0f3d68eb953e44c7a6cfae1a763b81e7
|
BITSAT_MATH
|
Vector Algebra
|
If the vector $\mathbf{a}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$ and $\mathbf{b}$ are collinear and $|\mathbf{b}|=21$, then $\mathbf{b}$ equal to:
|
singleCorrect
| 1 |
Given that $\mathbf{a}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$
and
$|\mathbf{b}|=21$
Now, taking option (b)
Let
$\begin{aligned}
\mathbf{b} & = \pm 3(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \\
|\mathbf{b}| & =3 \sqrt{4+9+36}=21 \\
\mathbf{b} & = \pm 3 \mathbf{a}
\end{aligned}$
and
$\therefore \mathbf{a}$ and $\mathbf{b}$ are collinear and magnitude of $\mathbf{b}$ is 21 .
|
["$\\pm(2 \\hat{\\mathbf{i}}+3 \\hat{\\mathbf{j}}+6 \\hat{\\mathbf{k}})$", "$\\pm 3(2 \\hat{\\mathbf{i}}+3 \\hat{\\mathbf{j}}+6 \\hat{\\mathbf{k}})$", "$(\\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}+\\hat{\\mathbf{k}})$", "$\\pm 21(2 \\hat{\\mathbf{i}}+3 \\hat{\\mathbf{j}}+6 \\hat{\\mathbf{k}})$"]
|
[1]
| null |
Practise Problem
|
1f45ec56c11d3dbe466fec085019d77f
|
BITSAT_MATH
|
Vector Algebra
|
If $\bar{a}=3 \hat{\imath}+\hat{\jmath}-\hat{k}, \bar{b}=2 \hat{\imath}-\hat{\jmath}+7 \hat{k}$ and $\bar{c}=7 \hat{\imath}-\hat{\jmath}+23 \hat{k}$ are three vectors,
then which of the following statement is true.
|
singleCorrect
| 2 |
$\bar{a}=3 \hat{i}+\hat{j}-\hat{k}, \bar{b}=2 \hat{i}-\hat{j}+7 \hat{k}, \bar{c}=7 \hat{i}-\hat{j}+23 \hat{k}$
$\begin{aligned}\left[\begin{array}{ccc}\bar{a} & \bar{b} & \bar{c}\end{array}\right] &=\left|\begin{array}{ccc}3 & 1 & -1 \\ 2 & -1 & 7 \\ 7 & -1 & 23\end{array}\right| \\ &=3(-23+7)-1(46-49)-1(-2+7) \\ &=3(-16)-(-3)-(5)=-50 \neq 0 \end{aligned}$
$\therefore \overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are non coplanar.
|
["$\\bar{a}, \\bar{b}$ and $\\bar{c}$ are non-coplanar.", "$\\bar{a}, \\bar{b}$ and $\\bar{c}$ are coplanar.", "$\\bar{a}, \\bar{b}, \\bar{c}$ are mutually perpendicular.", "$\\bar{a}$ and $\\bar{b}$ are collinear."]
|
[0]
| null |
Practise Problem
|
0b72a779a3542c369db9c773523ab499
|
BITSAT_MATH
|
Vector Algebra
|
If $\bar{a}, \bar{b}, \bar{c}$ are nonzero vectors along the coterminus edges of a parallelopiped with
volume 7 cubic units, then the volume of a parallelopiped with $\overline{\mathrm{a}}+\overline{\mathrm{b}}, \overline{\mathrm{b}}+\overline{\mathrm{c}}, \overline{\mathrm{c}}+\overline{\mathrm{a}}$ as
the coterminus edges is
|
singleCorrect
| 2 |
We have $[\bar{a} \cdot(\bar{b} \times \bar{c})]=7$
$[\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad \bar{c}+\bar{a}]$
$=(\bar{a}+\bar{b}) \cdot[(\bar{b}+\bar{c}) \times(\bar{c}+\bar{a})]$
$=(\bar{a}+\bar{b}) \cdot[(\bar{b} \times \bar{c})+(\bar{b} \times \bar{a})+(\bar{c} \times \bar{c})+(\bar{c} \times \bar{a})]$
$=[\bar{a} \cdot(\bar{b} \times \bar{c})]+[\bar{a} \cdot(\bar{b} \times \bar{a})]+[\bar{a}(\bar{c} \times \bar{a})]+[\bar{b} \cdot(\bar{b} \times \bar{c})]+[\bar{b} \cdot(\bar{b} \times \bar{a})]+[\bar{b} \cdot(\bar{c} \times \bar{a})]$
$=[\bar{a} \cdot(\bar{b} \times \bar{c})]+0+[\bar{b} \cdot(\bar{c} \times \bar{a})]$
$=[\bar{a} \cdot(\bar{b} \times \bar{c})]+[\bar{a} \cdot(\bar{b} \times \bar{c})]=2[\bar{a} \cdot(\bar{b} \times \bar{c})]$
$=2(7)=14$
|
["49 cubic units", "2 cubic units", "14 cubic units", "7 cubic units"]
|
[2]
| null |
Practise Problem
|
3cd6e08a8eb6083bfc18541633fc498a
|
BITSAT_MATH
|
Vector Algebra
|
The direction ratios of the line perpendicular to the lines having direction ratios $2,3,1$
and $1,2,1$ are
|
singleCorrect
| 2 |
Let $\bar{a}$ and $\bar{b}$ be the vectors along the lines whose direction ratios are $2,3,1$ and $1,2,1$ respectively.
$\therefore \overline{\mathrm{a}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}} \quad \text { and } \quad \overline{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
A vector perpendicular to both $\bar{a}$ and $\bar{b}$ is given by,
$\bar{a} \times \bar{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 1 \\
1 & 2 & 1
\end{array}\right|=\hat{i}(3-2)-\hat{j}(2-1)+\hat{k}(4-3)=\hat{i}-\hat{j}+\hat{k}$
Hence d.r.s are $1,-1,1$
|
["$-2,1,1$", "$1,1,1$", "$1,-1,1$", "$2,2,-2$"]
|
[2]
| null |
Practise Problem
|
82964d7508cf8f97f1ce304f2305f9a5
|
BITSAT_MATH
|
Vector Algebra
|
If <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>a</mi><mo>→</mo></mover><mo>,</mo><mover><mi>b</mi><mo>→</mo></mover><mo>,</mo><mover><mi>c</mi><mo>→</mo></mover></math> are non-coplaner vectors such that <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi mathvariant="italic">b</mi><mo mathvariant="italic">→</mo></mover><mo mathvariant="italic">×</mo><mover><mi mathvariant="italic">c</mi><mo mathvariant="italic">→</mo></mover><mo mathvariant="italic">=</mo><mover><mi mathvariant="italic">a</mi><mo mathvariant="italic">→</mo></mover><mo mathvariant="italic">;</mo><mo mathvariant="italic"> </mo><mover><mi mathvariant="italic">c</mi><mo mathvariant="italic">→</mo></mover><mo mathvariant="italic">×</mo><mover><mi mathvariant="italic">a</mi><mo mathvariant="italic">→</mo></mover><mo mathvariant="italic">=</mo><mover><mi mathvariant="italic">b</mi><mo mathvariant="italic">→</mo></mover><mo mathvariant="italic">;</mo><mo mathvariant="italic"> </mo><mover><mi mathvariant="italic">a</mi><mo mathvariant="italic">→</mo></mover><mo mathvariant="italic">×</mo><mover><mi mathvariant="italic">b</mi><mo mathvariant="italic">→</mo></mover><mo mathvariant="italic">=</mo><mover><mi mathvariant="italic">c</mi><mo mathvariant="italic">→</mo></mover></math>, then which of the following is not TRUE?
|
singleCorrect
| 1 |
<p>Given, </p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>a</mi><mo>→</mo></mover><mo>×</mo><mover><mi>b</mi><mo>→</mo></mover><mo>=</mo><mover><mi>c</mi><mo>→</mo></mover></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>b</mi><mo>→</mo></mover><mo>×</mo><mover><mi>c</mi><mo>→</mo></mover><mo>=</mo><mover><mi>a</mi><mo>→</mo></mover></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>c</mi><mo>→</mo></mover><mo>×</mo><mover><mi>a</mi><mo>→</mo></mover><mo>=</mo><mover><mi>b</mi><mo>→</mo></mover></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>(</mo><mover><mi>b</mi><mo>→</mo></mover><mo>×</mo><mover><mi>c</mi><mo>→</mo></mover><mo>)</mo><mo>·</mo><mover><mi>a</mi><mo>→</mo></mover><mo>=</mo><mover><mi>a</mi><mo>→</mo></mover><mo>·</mo><mover><mi>a</mi><mo>→</mo></mover></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>(</mo><mover><mi>c</mi><mo>→</mo></mover><mo>×</mo><mover><mi>a</mi><mo>→</mo></mover><mo>)</mo><mo>·</mo><mover><mi>b</mi><mo>→</mo></mover><mo>=</mo><mover><mi>b</mi><mo>→</mo></mover><mo>·</mo><mover><mi>b</mi><mo>→</mo></mover></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mover><mi>a</mi><mo>→</mo></mover><mn>2</mn></msup><mo>-</mo><msup><mover><mi>b</mi><mo>→</mo></mover><mn>2</mn></msup><mo>=</mo><mo>(</mo><mover><mi>b</mi><mo>→</mo></mover><mo>×</mo><mover><mi>c</mi><mo>→</mo></mover><mo>)</mo><mo>·</mo><mover><mi>a</mi><mo>→</mo></mover><mo>-</mo><mo>(</mo><mover><mi>c</mi><mo>→</mo></mover><mo>×</mo><mover><mi>a</mi><mo>→</mo></mover><mo>)</mo><mo>·</mo><mover><mi>b</mi><mo>→</mo></mover></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mo>[</mo><mtable><mtr><mtd><mover><mi>c</mi><mo>→</mo></mover></mtd><mtd><mover><mi>a</mi><mo>→</mo></mover></mtd><mtd><mover><mi>b</mi><mo>→</mo></mover></mtd></mtr></mtable><mo>]</mo><mo>-</mo><mo>[</mo><mtable><mtr><mtd><mover><mi>b</mi><mo>→</mo></mover></mtd><mtd><mover><mi>c</mi><mo>→</mo></mover></mtd><mtd><mover><mi>a</mi><mo>→</mo></mover></mtd></mtr></mtable><mo>]</mo><mo>=</mo><mn>0</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mo>|</mo><mover><mi>a</mi><mo>→</mo></mover><msup><mo>|</mo><mn>2</mn></msup><mo>-</mo><mo>|</mo><mover><mi>b</mi><mo>→</mo></mover><msup><mo>|</mo><mn>2</mn></msup><mo>=</mo><mn>0</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>|</mo><mover><mi>a</mi><mo>→</mo></mover><mo>|</mo><mo>-</mo><mo>|</mo><mover><mi>b</mi><mo>→</mo></mover><mo>|</mo><mo>=</mo><mn>0</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo>|</mo><mover><mi>a</mi><mo>→</mo></mover><mo>|</mo><mo>=</mo><mo>|</mo><mover><mi>b</mi><mo>→</mo></mover><mo>|</mo></math></p><p>We know that, </p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="[" close="]"><mrow><mover><mi>a</mi><mo>→</mo></mover><mo>×</mo><mover><mi>b</mi><mo>→</mo></mover><mo> </mo><mo> </mo><mover><mi>b</mi><mo>→</mo></mover><mo>×</mo><mover><mi>c</mi><mo>→</mo></mover><mo> </mo><mo> </mo><mover><mi>c</mi><mo>→</mo></mover><mo>×</mo><mover><mi>a</mi><mo>→</mo></mover></mrow></mfenced><mo>=</mo><mo>[</mo><mover><mi>a</mi><mo>→</mo></mover><mover><mi>b</mi><mo>→</mo></mover><mover><mi>c</mi><mo>→</mo></mover><msup><mo>]</mo><mn>2</mn></msup></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="[" close="]"><mtable columnalign="left"><mtr><mtd><mtable><mtr><mtd><mover><mi>a</mi><mo>→</mo></mover></mtd><mtd><mover><mi>b</mi><mo>→</mo></mover></mtd><mtd><mover><mi>c</mi><mo>→</mo></mover></mtd></mtr></mtable></mtd></mtr></mtable></mfenced><mo>=</mo><msup><mfenced open="[" close="]"><mtable columnalign="left"><mtr><mtd><mover><mi>a</mi><mo>→</mo></mover></mtd><mtd><mover><mi>b</mi><mo>→</mo></mover></mtd><mtd><mover><mi>c</mi><mo>→</mo></mover></mtd></mtr></mtable></mfenced><mn>2</mn></msup></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mfenced open="[" close="]"><mtable><mtr><mtd><mover><mi>a</mi><mo>→</mo></mover></mtd><mtd><mover><mi>b</mi><mo>→</mo></mover></mtd><mtd><mover><mi>c</mi><mo>→</mo></mover></mtd></mtr></mtable></mfenced><mo>=</mo><mn>1</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>|</mo><mover><mi>a</mi><mo>→</mo></mover><mo>|</mo><mo>|</mo><mover><mi>b</mi><mo>→</mo></mover><mo>|</mo><mo>|</mo><mover><mi>c</mi><mo>→</mo></mover><mo>|</mo><mo>=</mo><mn>1</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mo>|</mo><mover><mi>a</mi><mo>→</mo></mover><mo>|</mo><mo>=</mo><mo>|</mo><mover><mi>b</mi><mo>→</mo></mover><mo>|</mo><mo>=</mo><mo>|</mo><mover><mi>c</mi><mo>→</mo></mover><mo>|</mo><mo>≠</mo><mn>2</mn></math></p>
|
["<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mo>|</mo><mover><mi>a</mi><mo>→</mo></mover><mo>|</mo><mo>-</mo><mo>|</mo><mover><mi>b</mi><mo>→</mo></mover><mo>|</mo><mo>=</mo><mn>0</mn></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mo>|</mo><mover><mi>a</mi><mo>→</mo></mover><mo>|</mo><mo>=</mo><mo>|</mo><mover><mi>b</mi><mo>→</mo></mover><mo>|</mo><mo>=</mo><mo>|</mo><mover><mi>c</mi><mo>→</mo></mover><mo>|</mo><mo>=</mo><mn>2</mn></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mfenced open=\"[\" close=\"]\"><mtable><mtr><mtd><mover><mi>a</mi><mo>→</mo></mover></mtd><mtd><mover><mi>b</mi><mo>→</mo></mover></mtd><mtd><mover><mi>c</mi><mo>→</mo></mover></mtd></mtr></mtable></mfenced><mo>=</mo><mn>1</mn></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mo>|</mo><mover><mi>a</mi><mo>→</mo></mover><mo>|</mo><mo>|</mo><mover><mi>b</mi><mo>→</mo></mover><mo>|</mo><mo>|</mo><mover><mi>c</mi><mo>→</mo></mover><mo>|</mo><mo>=</mo><mn>1</mn></math>"]
|
[1]
| null |
Practise Problem
|
0411cd3a6bec3f9e7fdc1f18d3b0e6cb
|
BITSAT_MATH
|
Vector Algebra
|
If <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi mathvariant="normal">a</mi><mo>→</mo></mover><mo>=</mo><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover><mo>+</mo><mn>2</mn><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover><mo>+</mo><mn>3</mn><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover><mo>,</mo><mo> </mo><mover><mi mathvariant="normal">b</mi><mo>→</mo></mover><mo>=</mo><mo>-</mo><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover><mo>+</mo><mn>2</mn><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover><mo>+</mo><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover><mo>,</mo><mo> </mo><mover><mi mathvariant="normal">c</mi><mo>→</mo></mover><mo>=</mo><mn>3</mn><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover><mo>+</mo><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi mathvariant="normal">a</mi><mo>→</mo></mover><mo>+</mo><mi>p</mi><mover><mi mathvariant="normal">b</mi><mo>→</mo></mover></math> is normal to <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi mathvariant="normal">c</mi><mo>→</mo></mover></math>, then <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>p</mi></math> is equal to
|
singleCorrect
| 1 |
<p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mrow><mover><mi mathvariant="normal">a</mi><mo>→</mo></mover><mo>+</mo><mi>p</mi><mover><mi mathvariant="normal">b</mi><mo>→</mo></mover></mrow></mfenced></math> is normal to <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi mathvariant="normal">c</mi><mo>→</mo></mover></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mfenced><mrow><mover><mi mathvariant="normal">a</mi><mo>→</mo></mover><mo>+</mo><mi>p</mi><mover><mi mathvariant="normal">b</mi><mo>→</mo></mover></mrow></mfenced><mo>·</mo><mover><mi mathvariant="normal">c</mi><mo>→</mo></mover><mo>=</mo><mn>0</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mrow><mfenced><mrow><mn>1</mn><mo>-</mo><mi>p</mi></mrow></mfenced><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover><mo>+</mo><mfenced><mrow><mn>2</mn><mo>+</mo><mn>2</mn><mi>p</mi></mrow></mfenced><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover><mo>+</mo><mfenced><mrow><mn>3</mn><mo>+</mo><mi>p</mi></mrow></mfenced><mover><mi mathvariant="normal">k</mi><mo>^</mo></mover></mrow></mfenced><mo>·</mo><mfenced><mrow><mn>3</mn><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover><mo>+</mo><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></mrow></mfenced><mo>=</mo><mn>0</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>3</mn><mo>-</mo><mn>3</mn><mi>p</mi><mo>+</mo><mn>2</mn><mo>+</mo><mn>2</mn><mi>p</mi><mo>=</mo><mn>0</mn><mo>⇒</mo><mi>p</mi><mo>=</mo><mn>5</mn></math></p>
|
["<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>0</mn></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>1</mn></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>5</mn></math>", "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mn>3</mn></math>"]
|
[2]
| null |
Practise Problem
|
10a77096e160773b9675f5e0d9101275
|
BITSAT_MATH
|
Vector Algebra
|
If $\mathbf{p}=\hat{\mathbf{i}}+\hat{\mathbf{j}}, \mathbf{q}=4 \hat{\mathbf{k}}-\hat{\mathbf{j}}$ and $\mathbf{r}=\hat{\mathbf{i}}+\hat{\mathbf{k}}$, then the unit vector in tile direction of $3 p+q-2 r$ is
|
singleCorrect
| 1 |
We have, $\mathbf{p}=\hat{\mathbf{i}}+\hat{\mathbf{j}}, \mathbf{q}=4 \hat{\mathbf{k}}-\hat{\mathbf{j}}$ and $\mathbf{r}=\hat{\mathbf{i}}+\hat{\mathbf{k}}$
$\begin{aligned}
\text { So, } 3 \mathbf{p}+\mathbf{q}-2 \mathbf{i} &=3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}-\hat{\mathfrak{j}}-2 \hat{\mathbf{i}}-2 \hat{\mathbf{k}} \\
&=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}
\end{aligned}$
Now, required unit vector $=\frac{\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{\sqrt{1+4+4}}$
$=\frac{1}{3}(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})$
|
["$\\hat{\\mathbf{i}}+2 \\hat{\\mathbf{j}}+2 \\hat{\\mathbf{k}}$", "$\\frac{1}{3}(\\hat{\\mathbf{i}}-2 \\hat{\\mathbf{j}}+2 \\hat{\\mathbf{k}})$", "$\\frac{1}{3}(\\hat{\\mathbf{i}}-2 \\hat{\\mathbf{j}}-2 \\hat{\\mathbf{k}})$", "$\\frac{1}{3}(\\hat{\\mathbf{i}}+2 \\hat{\\mathbf{j}}+2 \\hat{\\mathbf{k}})$"]
|
[3]
| null |
Practise Problem
|
d45624f4b5b709f997a647919a6f74bb
|
BITSAT_MATH
|
Vector Algebra
|
If $\mathbf{a}$ and $\mathbf{b}$ are the two vectors such that $|\mathrm{a}|=3 \sqrt{3},|b|=4$ and $|\mathrm{a}+\mathrm{b}|=\sqrt{7}$, the angle between $\mathbf{a}$ and $\mathbf{b}$ is
|
singleCorrect
| 1 |
We have, $|a|=3 \sqrt{3},|b|=4$ and $|\mathbf{a}+b|=\sqrt{7}$
Now, $|\mathbf{a}+\mathbf{b}|^{2}=|\mathbf{a}|^{2}+|\mathbf{b}|^{2}+2|\mathbf{a}||\mathbf{b}| \cos \theta$
$\Rightarrow \quad(\sqrt{7})^{2}=(3 \sqrt{3})^{2}+16+2(3 \sqrt{3})(4) \cos \theta$
$\Rightarrow \quad 7=27+16+24 \sqrt{3} \cos \theta$
$\Rightarrow \quad 24 \sqrt{3} \cos \theta=-36$
$\Rightarrow \quad \cos \theta=-\frac{36}{24 \sqrt{3}}$
$\Rightarrow \quad \cos \theta=-\frac{\sqrt{3}}{2}$
$\Rightarrow \quad \theta=150^{\circ}$
|
["$150^{\\circ}$", "$30^{\\circ}$", "$60^{\\circ}$", "$120^{\\circ}$"]
|
[0]
| null |
Practise Problem
|
c2790ab910c3052b29d54a91298c8985
|
BITSAT_MATH
|
Vector Algebra
|
If $a$ is vector perpendicular to both $b$ and $c$, then
|
singleCorrect
| 1 |
Given, $a$ is perpendicular to $b$ and $c$. Thus, $\mathbf{a}$ is perpendicular to the plane of $\mathbf{b}$ and $\mathbf{c}$. Now, cross product of $b$ and $c$ will give a vector perpendicular to plane of $b$ and $\boldsymbol{c}$. This vector will be parallel to a.
Now, cross product of two parallel is zero vector.
Thus, $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=0$
|
["$\\mathrm{a} \\cdot(\\mathrm{b} \\times \\mathrm{c})=0$", "$a \\times(b \\times c)=0$", "$a \\times(b+c)=0$", "$a+(b+c)=0$"]
|
[1]
| null |
Practise Problem
|
a61b7dd088c3cd6449d8486246ff98fa
|
BITSAT_MATH
|
Vector Algebra
|
If $\mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}$ and $|\mathbf{a}|=5,|\mathbf{b}|=3$ and $|\mathbf{c}|=7$
then angle between $\mathbf{a}$ and $\mathbf{b}$ is
|
singleCorrect
| 1 |
Given, $\mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}$
and $|\mathbf{a}|=5,|\mathbf{b}|=3,|\mathbf{c}|=7$
$\Rightarrow \quad \mathbf{a}+\mathbf{b}=-\mathbf{c}$
On squaring both sides, we get $(\mathbf{a}+\mathbf{b})^{2}=(-\mathbf{c})^{2}$
$\Rightarrow$
$|\mathbf{a}+\mathbf{b}|^{2}=|\mathbf{c}|^{2}$
$\Rightarrow|\mathbf{a}|^{2}+|\mathbf{b}|^{2}+2 \mathbf{a} \cdot \mathbf{b}=|\mathbf{c}|^{2}$
$(\because \theta$ be the angle between $\mathbf{a}$ and $\mathbf{b}$ )
$\Rightarrow(5)^{2}+(3)^{2}+2|\mathbf{a}||\mathbf{b}| \cos \theta=(7)^{2}$
$\Rightarrow \quad 25+9+2 \cdot 5 \cdot 3 \cdot \cos \theta=49$
$\Rightarrow \quad 30 \cos \theta=15$
$\Rightarrow \quad \cos \theta=\frac{1}{2}=\cos 60^{\circ}$
$\Rightarrow \quad \theta=\frac{\pi}{3}$
|
["$\\frac{\\pi}{2}$", "$\\frac{\\pi}{3}$", "$\\frac{\\pi}{4}$", "$\\frac{\\pi}{6}$"]
|
[1]
| null |
Practise Problem
|
13600d118145da1458d43462bde36207
|
BITSAT_MATH
|
Vector Algebra
|
$\mathbf{i} \cdot(\mathbf{j} \times \mathbf{k})+\mathbf{j} \cdot(\mathbf{k} \times \mathbf{i})+\mathbf{k} \cdot(\mathbf{j} \times \mathbf{i})$ is equal to
|
singleCorrect
| 1 |
Given,
$\begin{aligned} \mathbf{i} \cdot(\mathbf{j} \times \mathbf{k}) &+\mathbf{j} \cdot(\mathbf{k} \times \mathbf{i})+\mathbf{k} \cdot(\mathbf{j} \times \mathbf{i}) \\=& \mathbf{i} \cdot(\mathbf{i})+\mathbf{j} \cdot(\mathbf{j})+\mathbf{k} \cdot(-\mathbf{k}) \\=&(\mathbf{i} \cdot \mathbf{i})+(\mathbf{j} \cdot \mathbf{j})-(\mathbf{k} \cdot \mathbf{k}) \\=& 1+1-1=1 \end{aligned}$
|
["3", "2", "1", "0"]
|
[2]
| null |
Practise Problem
|
d89519c2c4768992a594065cdce58b0b
|
BITSAT_MATH
|
Vector Algebra
|
If <math> <mrow> <mover accent="true"> <mi>a</mi> <mo stretchy="true">→</mo> </mover> <mo>=</mo><mover accent="true"> <mi>i</mi> <mo stretchy="true">^</mo> </mover> <mo>+</mo><mover accent="true"> <mi>j</mi> <mo stretchy="true">^</mo> </mover> <mo>−</mo><mn>2</mn><mover accent="true"> <mi>k</mi> <mo stretchy="true">^</mo> </mover> <mo>,</mo><mover accent="true"> <mi>b</mi> <mo stretchy="true">→</mo> </mover> <mo>=</mo><mn>2</mn><mover accent="true"> <mi>i</mi> <mo stretchy="true">^</mo> </mover> <mo>−</mo><mover accent="true"> <mi>j</mi> <mo stretchy="true">^</mo> </mover> <mo>+</mo><mover accent="true"> <mi>k</mi> <mo stretchy="true">^</mo> </mover> </mrow> </math> and <math> <mrow> <mover accent="true"> <mi>c</mi> <mo stretchy="true">→</mo> </mover> <mo>=</mo><mn>3</mn><mover accent="true"> <mi>i</mi> <mo stretchy="true">^</mo> </mover> <mo>−</mo><mover accent="true"> <mi>k</mi> <mo stretchy="true">^</mo> </mover> </mrow> </math>. If <math> <mrow> <mover accent="true"> <mi>c</mi> <mo stretchy="true">→</mo> </mover> <mo>=</mo><mi>m</mi><mover accent="true"> <mi>a</mi> <mo stretchy="true">→</mo> </mover> <mo>+</mo><mi>n</mi><mover accent="true"> <mi>b</mi> <mo stretchy="true">→</mo> </mover> </mrow> </math> then <math><mi>m</mi><mo>+</mo><mi>n</mi><mo>=</mo></math>
|
singleCorrect
| 1 |
<math> <mrow> <mover accent="true"> <mtext>c</mtext> <mo>¯</mo> </mover> <mo>=</mo> <mtext>m</mtext> <mover accent="true"> <mtext>a</mtext> <mo>¯</mo> </mover> <mo>+</mo> <mtext>n</mtext> <mover accent="true"> <mtext>b</mtext> <mo>¯</mo> </mover> </mrow> </math><br /><math><mo>⇒</mo><mn>3</mn><mover accent="false"><mrow><mi>i</mi></mrow><mo>¯</mo></mover><mo>-</mo><mover accent="false"><mrow><mi>k</mi></mrow><mo>¯</mo></mover><mo>=</mo><mi>m</mi><mover accent="false"><mrow><mi>i</mi></mrow><mo>¯</mo></mover><mo>+</mo><mi>m</mi><mover accent="false"><mrow><mi>j</mi></mrow><mo>¯</mo></mover><mo>-</mo><mn>2</mn><mi>m</mi><mover accent="false"><mrow><mi>k</mi></mrow><mo>¯</mo></mover><mo>+</mo><mn>2</mn><mi>n</mi><mover accent="false"><mrow><mi>i</mi></mrow><mo>¯</mo></mover><mo>-</mo><mi>n</mi><mover accent="false"><mrow><mi>j</mi></mrow><mo>¯</mo></mover><mo>+</mo><mi>n</mi><mover accent="false"><mrow><mi>k</mi></mrow><mo>¯</mo></mover></math><br /><math><mo>⇒</mo><mi> </mi><mi> </mi><mn>3</mn><mover accent="false"><mrow><mi>i</mi></mrow><mo>¯</mo></mover><mo>-</mo><mover accent="false"><mrow><mi>k</mi></mrow><mo>¯</mo></mover><mo>=</mo><mfenced separators="|"><mrow><mi>m</mi><mo>+</mo><mn>2</mn><mi>n</mi></mrow></mfenced><mover accent="false"><mrow><mi>i</mi></mrow><mo>¯</mo></mover><mo>+</mo><mfenced separators="|"><mrow><mi>m</mi><mo>-</mo><mi>n</mi></mrow></mfenced><mover accent="false"><mrow><mi>i</mi></mrow><mo>¯</mo></mover><mo>+</mo><mfenced separators="|"><mrow><mi>n</mi><mo>-</mo><mn>2</mn><mi>m</mi></mrow></mfenced><mover accent="false"><mrow><mi>k</mi></mrow><mo>¯</mo></mover></math><br /><math><mo>∴</mo><mi> </mi><mi> </mi><mi>m</mi><mo>+</mo><mn>2</mn><mi>n</mi><mo>=</mo><mn>3</mn></math> ...... (i)<br /><math><mi>m</mi><mo>-</mo><mi>n</mi><mo>=</mo><mn>0</mn></math> ...... (ii)<br /><math><mi>n</mi><mo>-</mo><mn>2</mn><mi>m</mi><mo>=</mo><mo>-</mo><mn>1</mn></math> .... (iii)<br />As <math><mi>m</mi><mo>-</mo><mi>n</mi><mo>=</mo><mn>0</mn></math><br />from (ii)<br /><math><mo>∴</mo><mi> </mi><mi> </mi><mi>m</mi><mo>=</mo><mi>n</mi></math><br /><math><mo>∴</mo><mi> </mi><mi> </mi><mn>3</mn><mi>m</mi><mo>=</mo><mn>3</mn></math><br /><math><mo>∴</mo><mi> </mi><mi> </mi><mi>m</mi><mo>=</mo><mn>1</mn><mi> </mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal">d</mi><mi> </mi><mi>n</mi><mo>=</mo><mn>1</mn></math><br /><math><mo>∴</mo><mi> </mi><mi> </mi><mi>m</mi><mo>+</mo><mi>n</mi><mo>=</mo><mn>2</mn></math>
|
["<math > <mn>0</mn></math>", "<math > <mn>1</mn></math>", "<math > <mn>2</mn></math>", "<math > <mrow> <mo>−</mo><mn>1</mn> </mrow></math>"]
|
[2]
| null |
Practise Problem
|
cd05cc5629acdec3b0cbdb7ebc319f82
|
BITSAT_MATH
|
Vector Algebra
|
Let $\hat{\alpha}, \hat{\beta}, \hat{\gamma}$ be three unit vectors such that $\hat{\alpha} \times (\hat{\beta} \times \hat{\gamma})=\frac{1}{2}(\hat{\beta}+\hat{\gamma})$ where $\hat{\alpha} \times(\hat{\beta} \times \hat{\gamma})=(\hat{\alpha} \cdot \hat{\gamma}) \hat{\beta}-(\hat{\alpha} \cdot \hat{\beta}) \gamma \cdot$ If $\hat{\beta}$ is not parallel to $\hat{\gamma},$ then the angle between $\hat{\alpha}$ and $\hat{\beta}$ is
|
singleCorrect
| 2 |
Given,
$|\hat{\alpha}|=|\hat{\beta}|=|\hat{\gamma}|=1$
and $\quad \hat{\alpha} \times(\hat{\beta} \times \hat{\gamma})=\frac{1}{2}(\hat{\beta}+\hat{\gamma})$
$\left.\operatorname{Now}, \hat{\alpha} \times(\hat{\beta} \times \hat{\gamma})=\frac{1}{2} \hat{(\hat{\beta}}+\hat{\gamma}\right)$
$\Rightarrow \quad(\hat{\alpha} \cdot \hat{\gamma}) \hat{\beta}-(\hat{\alpha} \cdot \hat{\beta}) \hat{\gamma}=\frac{1}{2} \hat{\beta}+\frac{1}{2} \hat{\gamma}$
On comparing we get, $-(\hat{\alpha} \cdot \hat{\beta})=\frac{1}{2}$
$\Rightarrow|\hat{\alpha}||\hat{\beta}| \cos \theta=-\frac{1}{2}$
$\Rightarrow \quad \cos \theta=-\frac{1}{2} \quad[\because|\hat{\alpha}|=|\hat{\beta}|=1]$
$\Rightarrow \quad \theta=120^{\circ}$ or $\frac{2 \pi}{3}$
|
["$\\frac{5 \\pi}{6}$", "$\\frac{\\pi}{6}$", "$\\frac{\\pi}{3}$", "$\\frac{2 \\pi}{3}$"]
|
[3]
| null |
Practise Problem
|
210147b8237dce8120d6f917c784007f
|
BITSAT_MATH
|
Vector Algebra
|
I. Two non-zero, non-collinear vectors are linearly independent.
II. Any three coplanar vectors are linearly dependent.
Which of the above statements is/are true?
|
singleCorrect
| 1 |
I : It is true that non-zero, non-collinear vectors are linearly independent.
II : It is also true that any three coplanar vectors are linearly dependent.
$\therefore$ Both I and II are true.
|
["Only I", "Only II", "Both I and II", "Neither I nor II"]
|
[2]
| null |
Practise Problem
|
e4330ed76855ae45e5cd6afb51730da1
|
BITSAT_MATH
|
Vector Algebra
|
Observe the following lists
<img src="https://cdn-question-pool.getmarks.app/pyq/ap_eamcet/zyzdni0rtOsTTDEn8ihI4ei8uP4XOsYVccUZnM5gb40.original.fullsize.png"><br>
Then the correct match for List I from List II is
A B C D
|
singleCorrect
| 2 |
(A) $[\mathbf{a} \mathbf{b} \mathbf{c}]=\mathbf{a} \cdot \mathbf{b} \times \mathbf{c}$
(B) $(\mathbf{c} \times \mathbf{a}) \times \mathbf{b}=(\mathbf{b} \cdot \mathbf{c}) \mathbf{a}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}$
(C) $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}$
(D) $\mathbf{a} \cdot \mathbf{b}=|\mathbf{a}||\mathbf{b}| \cos (\mathbf{a} \cdot \mathbf{b})$
$\therefore$ Option (b) is correct.
|
["1 2 3 4", "3 5 2 1", "3 5 5 1", "3 5 5 1"]
|
[1]
| null |
Practise Problem
|
0af5958a64301cbab2a5eb3772c8de43
|
BITSAT_MATH
|
Vector Algebra
|
Observe the following statements
A. Three vectors are coplanar if one of them is expressible as a linear combination of the other two.
R. Any three coplanar vectors are linearly dependent.
Then, which of the following is true?
|
singleCorrect
| 2 |
Both Statement A and $\mathrm{R}$ are true. But $\mathrm{R}$ is not correct explanation of A.
|
["Both $A$ and $R$ are true and $R$ is the correct explanation of $A$", "Both $A$ and $R$ are true but $R$ is not the correct explanation of $A$", "$A$ is true, but $R$ is false", "$A$ is false, but $R$ is true"]
|
[1]
| null |
Practise Problem
|
5c28efb4c042e371906414e7acb318e4
|
BITSAT_MATH
|
Vector Algebra
|
The value of if \( x(\hat{i}+\hat{j}+\hat{k}) \) is a unit vector is
|
singleCorrect
| 1 |
Given that, \( |x(i+j+k)|=1 \)<br>\[
\begin{array}{l}
\Rightarrow|x||\hat{i}+\hat{j}+\hat{k}|=1 \\
\Rightarrow|x| \sqrt{1+1+1}=1 \\
\Rightarrow|x| \sqrt{3}=1 \Rightarrow x=\pm \frac{1}{\sqrt{3}}
\end{array}
\]
|
["\\( \\pm \\frac{1}{\\sqrt{3}} \\)", "\\( 0 \\pm \\sqrt{3} \\)", "\\( \\pm 3 \\)", "\\( \\pm \\frac{1}{3} \\)"]
|
[0]
| null |
Practise Problem
|
7fb0843a0725de3ae74b093578595b69
|
BITSAT_MATH
|
Vector Algebra
|
If the volume of the parallelopiped whose conterminus edges are along the vectors $\bar{a}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ is 12, then the volume of the tetrahedron whose conterminus edges are $\bar{a}+\overline{\mathrm{b}}, \overline{\mathrm{b}}+\overline{\mathrm{c}}$ and $\overline{\mathrm{c}}+\overline{\mathrm{a}}$ is
|
singleCorrect
| 2 |
Volume of parallelepiped $=[\bar{a} \bar{b} \bar{c}]=12$
$\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]=12 \ldots(1)$
volume of tetrahedrom
$=\frac{1}{6}\left[\begin{array}{lll}
\bar{a}+\bar{b} & \bar{b}+\bar{c} & \bar{c}+\bar{a}
\end{array}\right]$
$=\frac{1}{6}(\bar{a}+\bar{b}) \cdot[(\bar{b}+\bar{c}) \times(\bar{c}+\bar{a})]=\frac{1}{6}(\bar{a}+\bar{b}) \cdot[(\bar{b} \times \bar{c})+(\bar{b} \times \bar{a})+(\bar{c} \times \bar{c})+(\bar{c} \times \bar{a})]$
$=\frac{1}{6}\{\bar{a} \cdot(\bar{b} \times \bar{c})+\bar{a}(\bar{b} \times \bar{a})+\bar{a}(\bar{c} \times \bar{a})+\bar{b} \cdot(\bar{b} \times \bar{c})+\bar{b} \cdot(\bar{b} \times \bar{a})+\bar{b}(\bar{c} \times \bar{a})\} \quad \ldots[\because \bar{c} \times \bar{c}=0]$
$=\frac{1}{6}[\bar{a} \cdot(\bar{b} \times \bar{c})+0+\bar{b} \cdot(\bar{c} \times \bar{a})]=\frac{1}{6}\{[\bar{a} \bar{b} \bar{c}]+[\bar{a} \bar{b} \bar{c}]\}$
$=\frac{2}{6}[\bar{a} \bar{b} \bar{c}]=\frac{1}{3}(12)=4$
|
["4 (units) $^{3}$", "24 (units) $^{3}$", "6 (units) $^{3}$", "12 (units) $^{3}$"]
|
[0]
| null |
Practise Problem
|
d5052943e1f936f7f32252fee6d82faf
|
BITSAT_MATH
|
Vector Algebra
|
If $[\bar{a} \bar{b} \bar{c}] \neq 0$, then $\frac{[\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad \bar{c}+\bar{a}]}{[\bar{b} \bar{c} \bar{a}]}=$
|
singleCorrect
| 2 |
$\left[\begin{array}{lll}\bar{a}+\bar{b} & \bar{b}+\bar{c} & \bar{c}+\bar{a}\end{array}\right]$
$=(\bar{a}+\bar{b}) \cdot[(\bar{b}+\bar{c}) \times(\bar{c}+\bar{a})]$
$=(\bar{a}+\bar{b}) \cdot[(\bar{b} \times \bar{c})+(\bar{b} \times \bar{a})+(\bar{c} \times \bar{c})+(\bar{c} \times \bar{a})]$
$=[\bar{a} \cdot(\bar{b} \times \bar{c})]+[\bar{a} \cdot(\bar{b} \times \bar{a})]+[\bar{a} \cdot(\bar{c} \times \bar{a})]+[\bar{b} \cdot(\bar{b} \times \bar{c})]+[\bar{b} \cdot(\bar{b} \times \bar{a})]+[\bar{b} \cdot(\bar{c} \times \bar{a})]$
$=[\bar{a} \cdot(\bar{b} \times \bar{c})]+[\bar{b} \cdot(\bar{c} \times \bar{a})]$
$=2[\bar{a} \cdot(\bar{b} \times \bar{c})] \quad=2\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$
Hence give expression $=\frac{2[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]}=2$
|
["0", "1", "2", "4"]
|
[2]
| null |
Practise Problem
|
6652ba48b915692ef0a610984abe1cdb
|
BITSAT_MATH
|
Vector Algebra
|
If $\vec{r}$ and $\vec{s}$ arenon-zero constant vectors and the scalar $b$ is chosen such that $|\vec{r}+b \vec{s}|$ is minimum, then the value of $|b \vec{s}|^{2}+|\vec{r}+b \vec{s}|^{2}$ is equal to
|
singleCorrect
| 1 |
For minimum value $|\vec{r}+b \vec{s}|=0$. Let $\vec{r}$ and $\vec{s}$ are anti-parallel so $b \vec{s}=-\vec{r}$
$\therefore|b \vec{s}|^{2}+|\vec{r}+b \vec{s}|^{2}=|-\vec{r}|^{2}+|\vec{r}-\vec{r}|^{2}=|\vec{r}|^{2}$
|
["$2|\\vec{r}|^{2}$", "$|\\vec{r}|^{2} / 2$", "$3|\\vec{r}|^{2}$", "$|\\vec{r}|^{2}$"]
|
[3]
| null |
Practise Problem
|
2e6aca22fd3809f00425f15db568f073
|
BITSAT_MATH
|
Vector Algebra
|
If $\vec{a}$ is a vector of magnitude 50 , collinear with the vector $\vec{b}=6 \hat{i}-8 \hat{j}-\frac{15}{2} \hat{k}$ and makes an acute angle with the positive direction of $Z$ - axis, then $\vec{a}$ is equal to
|
singleCorrect
| 1 |
Since $\vec{a}=m \vec{b}$ for some scalar $m$ i.e.,
$\begin{array}{l}
\vec{a}=m\left(6 \hat{i}-8 \hat{j}-\frac{15}{2} \hat{k}\right) \\
\Rightarrow|a|=|m| \sqrt{36+64+\frac{225}{4}} \\
\Rightarrow 50=\frac{25}{2}|m| \\
\Rightarrow|m|=4 \\
\Rightarrow m=\pm 4
\end{array}$
Since, a makes an acute angle with the positive direction of $Z$ - axis, so its $z$ component must be positive and hence, $m$ must be $-4$.
$\therefore a=-4\left(6 \hat{i}+8 \hat{j}-\frac{15}{2} \hat{k}\right)=-24 \hat{i}+32 \hat{j}+30 \hat{k}$
|
["$-24 \\hat{i}+32 \\hat{j}+30 \\hat{k}$", "$24 \\hat{i}-32 \\hat{j}-30 \\hat{k}$", "$-12 \\hat{i}+16 \\hat{j}-15 \\hat{k}$", "$12 \\hat{i}-16 \\hat{j}-15 \\hat{k}$"]
|
[0]
| null |
Practise Problem
|
fa45d751b8a90b246a870ad202ce6cb6
|
BITSAT_MATH
|
Vector Algebra
|
If the area of the parallelogram with $\mathbf{a}$ and $\mathbf{b}$ as two adjacent sides is 15 sq units, then the area of the parallelogram having $3 \mathbf{a}+2 \mathbf{b}$ and $\mathbf{a}+3 \mathbf{b}$ as two adjacent sides in sq units is
|
singleCorrect
| 1 |
We know that, if $\mathbf{a}$ and $\mathbf{b}$ are two adjacent sides of a parallelogram, then
Area $=|a \times b|=15$ (given) ...(i)
If the sides are $(3 \mathbf{a}+2 \mathbf{b})$ and $(\mathbf{a}+3 \mathbf{b})$
Then, area of parallelogram
$\begin{aligned}
&=3|(3 \mathbf{a}+2 \mathbf{b}) \times(\mathbf{a}+3 \mathbf{b})| \\
&=|3 \mathbf{a} \times \mathbf{a}+9 \mathbf{a} \times \mathbf{b}+2 \mathbf{b} \times \mathbf{a}+6 \mathbf{b} \times \mathbf{a}| \\
&=|0+9 \mathbf{a} \times \mathbf{b}-2 \mathbf{a} \times \mathbf{b}+0| \\
&=|7(\mathbf{a} \times \mathbf{b})| \\
&=7|\mathbf{a} \times \mathbf{b}| \\
&=7 \times 15=105 \text { sq units }
\end{aligned}$
|
["45", "75", "105", "120"]
|
[2]
| null |
Practise Problem
|
819f87487b2f9b22320b93bcd9027007
|
BITSAT_MATH
|
Vector Algebra
|
If $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{b}=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$, then find $(a+3 b) \cdot(2 a-b) .$
|
singleCorrect
| 1 |
Here, $\quad \mathbf{a}+3 \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}+3(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})$
$=10 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
and $2 \mathbf{a}-\mathbf{b}=2(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})-(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})=-\hat{\mathbf{i}}+5 \hat{\mathbf{k}}$
$(\mathbf{a}+3 \mathbf{b}) \cdot(2 \mathbf{a}-\mathbf{b})=(10 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \cdot(-\hat{\mathbf{i}}+5 \hat{\mathbf{k}})$
$=10 \times(-1)+7 \times 0+-1 \times 5=-15$
|
["$-15$", "12", "13", "$-10$"]
|
[0]
| null |
Practise Problem
|
89773562a0ddf4cc550c169f905bfe24
|
BITSAT_MATH
|
Vector Algebra
|
The value of $[a-b \quad b-c \quad c-a]$, where $|a|=1$, $|\mathrm{b}|=5|\mathrm{c}|=3$, is
|
singleCorrect
| 2 |
$[\mathbf{a}-\mathbf{b} \mathbf{b}-\mathbf{c} \mathbf{c}-\mathbf{a}]=(\mathbf{a}-\mathbf{b}) \cdot[(\mathbf{b}-\mathbf{c}) \times(\mathbf{c}-\mathbf{a})]$
$=(\mathbf{a}-\mathbf{b}) \cdot[\mathbf{b} \times \mathbf{c}-\mathbf{b} \times \mathbf{a}-\mathbf{c} \times \mathbf{c}+\mathbf{c} \times \mathbf{a})]$
$=(\mathbf{a}-\mathbf{b}) .[\mathbf{b} \times \mathbf{c}-\mathbf{b} \times \mathbf{a}+\mathbf{c} \times \mathbf{a}]$
$=\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})-\mathbf{a} \cdot(\mathbf{b} \times \mathbf{a})+\mathbf{a} \cdot(\mathbf{c} \times \mathbf{a})-\mathbf{b}(\mathbf{b} \times \mathbf{c})$
$+\mathbf{b} \cdot(\mathbf{b} \times \mathbf{a})-\mathbf{b}(\mathbf{c} \times \mathbf{a})$
$=[\mathbf{a b c}]-[\mathbf{a b a}]+[\mathbf{a c a}]-[\mathbf{b b c}]+[\mathbf{b b a}]-[$ bca $]$
$=[\mathbf{a b c}]-[\mathbf{b c a}]=0 \quad\{\because[\mathbf{a b c}]=[\mathbf{b c a}]=[\mathbf{c a b}]\}$
|
["0", "1", "6", "None of these"]
|
[0]
| null |
Practise Problem
|
a0a62d9da02d27b963423a41f1462281
|
BITSAT_MATH
|
Vector Algebra
|
If <math> <mi>G</mi><mrow><mo>(</mo> <mrow> <mover accent="true"> <mi>g</mi> <mo stretchy="true">→</mo> </mover> </mrow> <mo>)</mo></mrow><mo>,</mo><mi>H</mi><mrow><mo>(</mo> <mrow> <mover accent="true"> <mi>h</mi> <mo stretchy="true">→</mo> </mover> </mrow> <mo>)</mo></mrow> </math> and <math> <mrow> <mi>P</mi><mrow><mo>(</mo> <mrow> <mover accent="true"> <mi>p</mi> <mo stretchy="true">→</mo> </mover> </mrow> <mo>)</mo></mrow> </mrow> </math> are centroid, orthocenter and circumcenter of a triangle and <math> <mi>x</mi><mover accent="true"> <mi>p</mi> <mo stretchy="true">→</mo> </mover> <mo>+</mo><mi>y</mi><mover accent="true"> <mi>h</mi> <mo stretchy="true">→</mo> </mover> <mo>+</mo><mi>z</mi><mover accent="true"> <mi>g</mi> <mo stretchy="true">→</mo> </mover> <mo>=</mo><mn>0</mn> </math> then <math><mfenced separators="|"><mrow><mi>x</mi><mo>,</mo><mi> </mi><mi>y</mi><mo>,</mo><mi> </mi><mi>z</mi></mrow></mfenced><mo>= </mo></math>
|
singleCorrect
| 2 |
We know that, in a triangle centroid (G), orthocenter (H) and circumcenter (P) are always collinear and G divides the line segment PH in ratio 1:2<br />So using section formula, <math> <mrow> <mover accent="true"> <mi>g</mi> <mo stretchy="true">¯</mo> </mover> <mo>=</mo><mfrac> <mrow> <mn>2</mn><mover accent="true"> <mi>p</mi> <mo stretchy="true">¯</mo> </mover> <mo>+</mo><mover accent="true"> <mi>h</mi> <mo stretchy="true">¯</mo> </mover> </mrow> <mn>3</mn> </mfrac> </mrow> </math><br /><math> <mrow> <mo>⇒</mo><mn>3</mn><mover accent="true"> <mi>g</mi> <mo stretchy="true">¯</mo> </mover> <mo>=</mo><mn>2</mn><mover accent="true"> <mi>p</mi> <mo stretchy="true">¯</mo> </mover> <mo>+</mo><mover accent="true"> <mi>h</mi> <mo stretchy="true">¯</mo> </mover> </mrow> </math><br /><math> <mrow> <mo>⇒</mo><mn>2</mn><mover accent="true"> <mi>p</mi> <mo stretchy="true">¯</mo> </mover> <mo>+</mo><mover accent="true"> <mi>h</mi> <mo stretchy="true">¯</mo> </mover> <mo>−</mo><mn>3</mn><mover accent="true"> <mi>g</mi> <mo stretchy="true">¯</mo> </mover> <mo>=</mo><mn>0</mn></mrow> </math>
|
["<math><mo>(</mo><mn>1</mn><mo>,</mo><mi> </mi><mn>1</mn><mo>,</mo><mo>-</mo><mn>2</mn><mo>)</mo></math>", "<math><mo>(</mo><mn>2</mn><mo>,</mo><mn>1</mn><mo>,</mo><mo>-</mo><mn>3</mn><mo>)</mo></math>", "<math><mo>(</mo><mn>1</mn><mo>,</mo><mn>3</mn><mo>,</mo><mo>-</mo><mn>4</mn><mo>)</mo></math>", "<math><mo>(</mo><mn>2</mn><mo>,</mo><mn>3</mn><mo>,</mo><mo>-</mo><mn>5</mn><mo>)</mo></math>"]
|
[1]
| null |
Practise Problem
|
f53fcc7c1623f5fda571320d442c61e2
|
BITSAT_MATH
|
Vector Algebra
|
If <math> <mrow> <mover accent="true"> <mi>a</mi> <mo stretchy="true">→</mo> </mover> <mo>=</mo><mover accent="true"> <mi>i</mi> <mo stretchy="true">^</mo> </mover> <mo>+</mo><mover accent="true"> <mi>j</mi> <mo stretchy="true">^</mo> </mover> <mo>+</mo><mover accent="true"> <mi>k</mi> <mo stretchy="true">^</mo> </mover> <mo>,</mo><mover accent="true"> <mi>b</mi> <mo stretchy="true">→</mo> </mover> <mo>=</mo><mn>2</mn><mover accent="true"> <mi>i</mi> <mo stretchy="true">^</mo> </mover> <mo>+</mo><mi>λ</mi><mo>+</mo><mover accent="true"> <mi>k</mi> <mo stretchy="true">^</mo> </mover> <mo>,</mo><mover accent="true"> <mi>c</mi> <mo stretchy="true">→</mo> </mover> <mo>=</mo><mover accent="true"> <mi>i</mi> <mo stretchy="true">^</mo> </mover> <mo>−</mo><mover accent="true"> <mi>j</mi> <mo stretchy="true">^</mo> </mover> <mo>+</mo><mn>4</mn><mover accent="true"> <mi>k</mi> <mo stretchy="true">^</mo> </mover> </mrow> </math> and <math> <mrow> <mover accent="true"> <mi>a</mi> <mo stretchy="true">→</mo> </mover> <mo>.</mo><mrow><mo>(</mo> <mrow> <mover accent="true"> <mi>b</mi> <mo stretchy="true">→</mo> </mover> <mo>×</mo><mover accent="true"> <mi>c</mi> <mo stretchy="true">→</mo> </mover> </mrow> <mo>)</mo></mrow><mo>=</mo><mn>10</mn><mo>,</mo> </mrow> </math> then <math><mi>λ</mi></math> is equal to
|
singleCorrect
| 2 |
<math><mfenced close="|" open="|" separators="|"><mrow><mtable><mtr><mtd><mn>1</mn></mtd><mtd><mn>1</mn></mtd><mtd><mn>1</mn></mtd></mtr><mtr><mtd><mn>2</mn></mtd><mtd><mi>λ</mi></mtd><mtd><mn>1</mn></mtd></mtr><mtr><mtd><mn>1</mn></mtd><mtd><mo>-</mo><mn>1</mn></mtd><mtd><mn>4</mn></mtd></mtr></mtable></mrow></mfenced><mo>=</mo><mn>10</mn></math><br /><math><mn>1</mn><mfenced separators="|"><mrow><mn>4</mn><mi>λ</mi><mo>+</mo><mn>1</mn></mrow></mfenced><mo>-</mo><mn>1</mn><mfenced separators="|"><mrow><mn>8</mn><mo>-</mo><mn>1</mn></mrow></mfenced><mo>+</mo><mn>1</mn><mfenced separators="|"><mrow><mo>-</mo><mn>2</mn><mo>-</mo><mi>λ</mi></mrow></mfenced><mo>=</mo><mn>10</mn></math><br /><math><mo>⇒</mo><mi> </mi><mi> </mi><mn>4</mn><mi>λ</mi><mo>+</mo><mn>1</mn><mo>-</mo><mn>7</mn><mo>-</mo><mn>2</mn><mo>-</mo><mi>λ</mi><mo>=</mo><mn>10</mn></math><br /><math><mo>⇒</mo><mi> </mi><mi> </mi><mn>3</mn><mi>λ</mi><mo>=</mo><mn>18</mn></math><br /><math><mo>⇒</mo><mi> </mi><mi> </mi><mi>λ</mi><mo>=</mo><mn>6</mn></math>
|
["<math> <mn>6</mn></math>", "<math> <mn>7</mn></math>", "<math> <mn>9</mn></math>", "<math> <mn>10</mn></math>"]
|
[0]
| null |
Practise Problem
|
9b9d5b95e8c971575a04904829b6dcb1
|
BITSAT_MATH
|
Vector Algebra
|
If the origin and the points <math> <mrow> <mi>P</mi><mrow><mo>(</mo> <mrow> <mn>2</mn><mo>,</mo><mn>3</mn><mo>,</mo><mn>4</mn> </mrow> <mo>)</mo></mrow><mo>,</mo><mi>Q</mi><mrow><mo>(</mo> <mrow> <mn>1</mn><mo>,</mo><mn>2</mn><mo>,</mo><mn>3</mn> </mrow> <mo>)</mo></mrow> </mrow> </math> and <math> <mrow> <mi>R</mi><mrow><mo>(</mo> <mrow> <mi>x</mi><mo>,</mo><mi>y</mi><mo>,</mo><mi>z</mi> </mrow> <mo>)</mo></mrow> </mrow> </math> are co-planar then
|
singleCorrect
| 1 |
<math > <mrow> <mi>O</mi><mo>,</mo><mi>P</mi><mo>,</mo><mi>Q</mi><mo>,</mo><mi>R</mi> </mrow></math>are co-planar<br /><math> <mo>⇒</mo> </math> <math><mfenced close="]" open="[" separators="|"><mrow><mi> </mi><mover accent="false"><mrow><mi>O</mi><mi>R</mi></mrow><mo>¯</mo></mover><mi> </mi><mi> </mi><mi> </mi><mi> </mi><mover accent="false"><mrow><mi>O</mi><mi>P</mi></mrow><mo>¯</mo></mover><mi> </mi><mi> </mi><mi> </mi><mi> </mi><mover accent="false"><mrow><mi>O</mi><mi>Q</mi></mrow><mo>¯</mo></mover><mi> </mi></mrow></mfenced><mo>=</mo><mn>0</mn><mi> </mi></math><br /><math> <mo>⇒</mo> </math> <math><mfenced close="|" open="|" separators="|"><mrow><mtable><mtr><mtd><mi>x</mi></mtd><mtd><mi>y</mi></mtd><mtd><mi>z</mi></mtd></mtr><mtr><mtd><mn>2</mn></mtd><mtd><mn>3</mn></mtd><mtd><mn>4</mn></mtd></mtr><mtr><mtd><mn>1</mn></mtd><mtd><mn>2</mn></mtd><mtd><mn>3</mn></mtd></mtr></mtable></mrow></mfenced><mo>=</mo><mn>0</mn></math><br /><math> <mo>⇒</mo> </math> <math><mi>x</mi><mfenced separators="|"><mrow><mn>9</mn><mo>-</mo><mn>8</mn></mrow></mfenced><mo>-</mo><mi>y</mi><mfenced separators="|"><mrow><mn>6</mn><mo>-</mo><mn>4</mn></mrow></mfenced><mo>+</mo><mi>z</mi><mfenced separators="|"><mrow><mn>4</mn><mo>-</mo><mn>3</mn></mrow></mfenced><mo>=</mo><mn>0</mn></math><br /><math> <mo>⇒</mo> </math> <math><mi>x</mi><mo>-</mo><mn>2</mn><mi>y</mi><mo>+</mo><mi>z</mi><mo>=</mo><mn>0</mn></math><br /><b>Alternative</b><br />Points <math > <mrow> <mi>P</mi><mo>,</mo><mi>Q</mi> </mrow></math> satisfy equations given in option.
|
["<math><mi>x</mi><mo>-</mo><mn>2</mn><mi>y</mi><mo>-</mo><mi>z</mi><mo>=</mo><mn>0</mn></math>", "<math><mi>x</mi><mo>+</mo><mn>2</mn><mi>y</mi><mo>+</mo><mi>z</mi><mo>=</mo><mn>0</mn></math>", "<math><mi>x</mi><mo>-</mo><mn>2</mn><mi>y</mi><mo>+</mo><mi>z</mi><mo>=</mo><mn>0</mn></math>", "<math><mn>2</mn><mi>x</mi><mo>-</mo><mn>2</mn><mi>y</mi><mo>+</mo><mi>z</mi><mo>=</mo><mn>0</mn></math>"]
|
[2]
| null |
Practise Problem
|
eb01d1e89d870caa41e223898bd51b4f
|
BITSAT_MATH
|
Vector Algebra
|
Let $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ be three non-zero vectors which are pairwise non-collinear. If $\vec{\alpha}+3 \vec{\beta}$ is collinear with $\vec{\gamma}$ and $\vec{\beta}+2 \vec{\gamma}$ is collinear with $\vec{\alpha}$, then $\vec{\alpha}+3 \vec{\beta}+6 \vec{\gamma}$ is
|
singleCorrect
| 1 |
$\vec{\alpha}+3 \vec{\beta}=\mathrm{k}_{1} \vec{\gamma} \Rightarrow \vec{\beta}=\frac{\mathrm{k}_{1}}{3} \vec{\gamma}-\frac{\vec{\alpha}}{3}$
$\vec{\beta}+2 \vec{\gamma}=\mathrm{k}_{2} \vec{\alpha} \Rightarrow \vec{\beta}=\mathrm{k}_{2} \vec{\alpha}-2 \vec{\gamma}$
$\Rightarrow \frac{\mathrm{k}_{1} \vec{\gamma}}{3}-\frac{\vec{\alpha}}{3}=\mathrm{k}_{2} \vec{\alpha}-2 \vec{\gamma} \Rightarrow \vec{\alpha}\left(\mathrm{k}_{2}+\frac{1}{3}\right)=\vec{\gamma}\left(\frac{\mathrm{k}_{1}}{3}+2\right) \Rightarrow \mathrm{k}_{2}=-\frac{1}{3}$ and $\frac{\mathrm{k}_{1}}{3}=-2 \Rightarrow \mathrm{k}_{1}=-6$
$\Rightarrow \vec{\alpha}+3 \vec{\beta}+6 \vec{\gamma}=\overrightarrow{0}$
|
["$\\vec{\\gamma}$", "$\\overrightarrow{0}$", "$\\vec{\\alpha}+\\vec{\\gamma}$", "$\\vec{\\alpha}$"]
|
[1]
| null |
Practise Problem
|
182643a36d875ba163f69b9e591358de
|
BITSAT_MATH
|
Vector Algebra
|
If $\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{0}}$ and $|\overrightarrow{\mathbf{a}}|=3,|\overrightarrow{\mathbf{b}}|=4$ and $|\overrightarrow{\mathbf{c}}|=\sqrt{37}$, then the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ is :
|
singleCorrect
| 1 |
$\because \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}=0$
$\Rightarrow \quad \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}=-\overrightarrow{\mathbf{c}}$
$\Rightarrow \quad(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})^2=(-\overrightarrow{\mathbf{c}})^2$
$\Rightarrow \quad|\overrightarrow{\mathbf{a}}|^2+|\overrightarrow{\mathbf{b}}|^2+2|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}| \cos \theta=|\overrightarrow{\mathbf{c}}|^2$
$\Rightarrow \quad 9+16+2 \cdot 3 \cdot 4 \cos \theta=37$
$\Rightarrow \quad 24 \cos \theta=37-25$
$\Rightarrow \quad \cos \theta=\frac{1}{2}=\cos \frac{\pi}{3}$
$\Rightarrow \quad \theta=\frac{\pi}{3}$
|
["$\\frac{\\pi}{4}$", "$\\frac{\\pi}{2}$", "$\\frac{\\pi}{6}$", "$\\frac{\\pi}{3}$"]
|
[3]
| null |
Practise Problem
|
3dc71e0a38b1341ce9339bc865bf2b5e
|
BITSAT_MATH
|
Vector Algebra
|
The position vector of a point lying on the line joining the points whose positions vectors are $\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ is :
|
singleCorrect
| 1 |
The position vector of mid point of line joining the points whose position vector are $\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$.
$=\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}+\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}}{2}=\frac{2 \hat{\mathbf{i}}}{2}=\hat{\mathbf{i}}$
|
["$\\hat{\\mathbf{j}}$", "$\\hat{\\mathbf{i}}$", "$\\hat{\\mathbf{k}}$", "$\\overrightarrow{\\mathbf{0}}$"]
|
[1]
| null |
Practise Problem
|
9708a8c3165a1e9d60fa6a42b6be49d2
|
BITSAT_MATH
|
Vector Algebra
|
If $\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\lambda \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$ are coplanar, then $\lambda$ is equal to :
|
singleCorrect
| 1 |
Since $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}, 3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\lambda \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$ are coplanar, then
$\left|\begin{array}{ccc}1 & -2 & 0 \\ 0 & 3 & 1 \\ \lambda & 3 & 0\end{array}\right|=0$
$\Rightarrow \quad 1\left|\begin{array}{ll}3 & 1 \\ 3 & 0\end{array}\right|+2\left|\begin{array}{ll}0 & 1 \\ \lambda & 0\end{array}\right|=0$
$\Rightarrow \quad-3+2(-\lambda)=0$
$\Rightarrow \quad \lambda=-\frac{3}{2}$
|
["-1", "$1 / 2$", "$-3 / 2$", "2"]
|
[2]
| null |
Practise Problem
|
362d1279f898349a114baf90d414897f
|
BITSAT_MATH
|
Vector Algebra
|
If \( |\bar{a}| \) and \( |\bar{b}| \) are mutually perpendicular unit vectors, then<br>\( (3|\bar{a}|+2|\bar{b}|) \cdot(5|\bar{a}|-6|\bar{b}|)= \)
|
singleCorrect
| 2 |
Given that, $(3 \vec{a}+2 \vec{b}) \cdot(5 \vec{a}-6 \vec{b})$
Since, $\vec{a} \cdot \vec{b}=0$
and $|\vec{a}|=|\vec{b}|=3$
Then $(3 \vec{a}+2 \vec{b}) \cdot(5 \vec{a}-6 \vec{b})$
$=15|\vec{a}|^{2}-18(0)+10(0)-12|\vec{b}|^{2}=3$
|
["5", "\\( 03 \\)", "\\( 06 \\)", "\\( 12 \\)"]
|
[1]
| null |
Practise Problem
|
99563b6d9ac9ddc369137c347a93da5b
|
BITSAT_MATH
|
Vector Algebra
|
If the vector \( a \hat{i}+\hat{j}+\hat{k}, \hat{i}+b \hat{j}+\hat{k} \) and \( \hat{i}+\hat{j}+c \hat{k} \) are colpanar \( (a \neq b \neq c \neq 1) \), then<br>the value of \( a b c-(a+b+c)= \)
|
singleCorrect
| 1 |
Given vectors, \( a \hat{i}+\hat{j}+\hat{k}, \hat{i}+b \hat{j}+\hat{k} \) and \( \hat{i}+\hat{j}+c \hat{k} \)<br>For vectors to be coplanar<br>\[
\begin{array}{l}
\left|\begin{array}{lll}
a & 1 & 1 \\
1 & b & 1 \\
1 & 1 & c
\end{array}\right|=0 \\
\Rightarrow a(b c-1)-1(c-1)+1(1-b)=0 \\
\Rightarrow a b c-a-c+2+1-b=0 \\
\Rightarrow a b c-(a+b+c)=-2
\end{array}
\]
|
["\\( 12 \\)", "\\( -2 \\)", "\\( 00 \\)", "\\( -1 \\)"]
|
[1]
| null |
Practise Problem
|
1e344f0fce36cb573426a70a18a8b48d
|
BITSAT_MATH
|
Vector Algebra
|
If the angle between \( \vec{a} \) and \( \vec{b} \) is \( \frac{2 \Pi}{3} \) and the projection of \( \vec{a} \) in the direction of \( \vec{b} \) is \( -2 \), then \( \mid \) a<br>\( 1= \)
|
singleCorrect
| 1 |
(C)<br>\[
\begin{array}{l}
\theta=\frac{2 \pi}{3} \\
\Rightarrow \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=-2 \\
\Rightarrow \frac{|\vec{a}||\vec{b}| \cos \theta}{|\vec{b}|}=-2 \Rightarrow|\vec{a}| \cos \frac{2 \Pi}{3}=-2 \Rightarrow|\vec{a}| \times-\frac{1}{2}=-2 \Rightarrow|\vec{a}|=4
\end{array}
\]
|
["\\( 03 \\)", "\\( 11 \\)", "\\( 04 \\)", "\\( 12 \\)"]
|
[2]
| null |
Practise Problem
|
7e8f8c523a138ac2809a4ae467498ae6
|
BITSAT_MATH
|
Vector Algebra
|
A vector perpendicular to the plane containing the vectors $\hat{i}-2 \hat{j}-\hat{k}$ and $3 \hat{i}-2 \hat{j}-\hat{k}$ is inclined to the vector $\hat{i}+\hat{j}+\hat{k}$ at an angle
|
singleCorrect
| 2 |
A vector perpendicular to the plane is<br />$\begin{aligned}<br />& (\hat{i}-2 \hat{j}-\hat{k}) \times(3 \hat{i}-2 \hat{j}-\hat{k}) \\<br />& =\left|\begin{array}{ccc}<br />\hat{i} & \hat{j} & \hat{k} \\<br />1 & -2 & -1 \\<br />3 & -2 & -1<br />\end{array}\right|=-2 \hat{j}+4 \hat{k} \\<br />& \Rightarrow \text { unit vector } \hat{a}=\frac{-2 \hat{j}+4 \hat{k}}{\sqrt{4+16}}=\frac{-2 \hat{j}+4 \hat{k}}{2 \sqrt{5}}<br />\end{aligned}$<br />Angle between the unit vector and $\vec{r}=\hat{i}+\hat{j}+\hat{k}$<br />$=\cos ^{-1} \frac{\vec{r} \cdot \hat{a}}{|\vec{r}| \cdot|\hat{a}|}=\cos ^{-1} \frac{1}{\sqrt{15}}=\tan ^{-1} \sqrt{14}$
|
["$\\tan ^{-1} \\sqrt{14}$", "$\\sec ^{-1} \\sqrt{14}$", "$\\tan ^{-1} \\sqrt{15}$", "<br />None of these"]
|
[0]
| null |
Practise Problem
|
75e400c2ce357b3cafc23fdd770bcd48
|
BITSAT_MATH
|
Vector Algebra
|
If $\mathbf{u}=\mathbf{a}-\mathbf{b}, \mathbf{v}=\mathbf{a}+\mathbf{b}$, and $|\mathbf{a}|=|\mathbf{b}|=2$, then $|\mathbf{u} \times \mathbf{v}|$ is
|
singleCorrect
| 1 |
Given that, $\mathbf{u}=\mathbf{a}-\mathbf{b}, \mathbf{v}=\mathbf{a}+\mathbf{b}$ and $|\mathbf{a}|=|\mathbf{b}|=2$
Now, $|\mathbf{u} \times \mathbf{v}|=|\mathbf{a}-\mathbf{b} \times \mathbf{a}+\mathbf{b}|$ $=2(\mathbf{a} \times \mathbf{b})=2|\mathbf{a}||\mathbf{b}| \sin \theta$ $=2 \times 2 \times 2 \sin \theta=8 \sin \theta=8 \sqrt{1-\cos ^{2} \theta}$ $=8 \sqrt{1-\left(\frac{a \cdot b}{|a||b|}\right)^{2}}$
$$
=8 \sqrt{1-\left(\frac{\mathbf{a} \cdot \mathbf{b}}{4}\right)^{2}}=8 \sqrt{1-\left(\frac{\mathbf{a} \cdot \mathbf{b}}{16}\right)^{2}}
$$
$$
=\frac{8}{4} \sqrt{16-(a \cdot b)^{2}}=2 \sqrt{16-(a \cdot b)^{2}}
$$
|
["$2 \\sqrt{16-(\\mathbf{a} \\cdot \\mathbf{b})^{2}}$", "$2 \\sqrt{4-(\\mathbf{a} \\cdot \\mathbf{b})^{2}}$", "$\\sqrt{16-(\\mathbf{a} \\cdot \\mathbf{b})^{2}}$", "$\\sqrt{4-(\\mathbf{a} \\cdot \\mathbf{b})^{2}}$"]
|
[0]
| null |
Practise Problem
|
52b6be929923e74be9d95c8696748ee8
|
BITSAT_MATH
|
Vector Algebra
|
The volume of the tetrahedron formed by the points $(1,1,1),(2,1,3)(3,2,2)$ and $(3,3,4)$ in cubic units is
|
singleCorrect
| 2 |
Let $A(1,1,1), B(2,1,3), C(3,2,2)$ and $D(3,3,4)$. So, $\mathbf{A B}=\hat{\mathbf{i}}+2 \hat{\mathbf{k}}, \mathbf{A C}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{A D}=2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
$$
\begin{aligned}
&\text { Volume of tetrahedron }=\frac{1}{6}\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 1 & 1 \\
2 & 2 & 3
\end{array}\right] \\
&\quad=\frac{1}{6}|1(3-2)+0(6-2)+2(4-2)| \\
&=\frac{1}{6}(1+0+4)=\frac{5}{6} \text { cubic units }
\end{aligned}
$$
|
["$\\frac{5}{6}$", "$\\frac{6}{5}$", "5", "$\\frac{2}{3}$"]
|
[0]
| null |
Practise Problem
|
9b9d213da36a17adb676d01625154012
|
BITSAT_MATH
|
Vector Algebra
|
Unit vector perpendicular to $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and lying in the plane containing $\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ is
|
singleCorrect
| 1 |
Let $\mathbf{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{c}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$
Now, $\mathbf{b} \times \mathbf{c}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & -2 \\ -1 & 2 & 1\end{array}\right|=5 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
Now, $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -2 & 2 \\ 5 & 1 & 3\end{array}\right|=-8 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}$
Now, any vector perpendicular to a and lying in the plane containing $\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ is $\pm(a \times(b \times c))$
$$
\begin{aligned}
\therefore \text { Required unit vector } &=\pm \frac{(-8 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+11 \hat{\mathbf{k}})}{\sqrt{(-8)^{2}+(7)^{2}+(11)^{2}}} \\
&=\pm \frac{(-8 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+11 \hat{\mathbf{k}})}{\sqrt{234}}
\end{aligned}
$$
|
["$8 \\hat{\\mathbf{i}}-7 \\hat{\\mathbf{j}}+11 \\hat{\\mathbf{k}}$", "$8 \\hat{\\mathbf{i}}+7 \\hat{\\mathbf{j}}-11 \\hat{\\mathbf{k}}$", "$8 \\hat{\\mathbf{i}}-7 \\hat{\\mathbf{j}}-11 \\hat{\\mathbf{k}}$", "$\\frac{1}{\\sqrt{234}}(8 \\hat{\\mathbf{i}}-7 \\hat{\\mathbf{j}}-11 \\hat{\\mathbf{k}})$"]
|
[3]
| null |
Practise Problem
|
bc408d443971a690ef193023d9ab6c89
|
BITSAT_MATH
|
Vector Algebra
|
If $\mathbf{a} \hat{\mathfrak{i}}=\mathbf{a}(\hat{\hat{i}}+\hat{\hat{j}})=\mathbf{a}(\hat{\mathbf{i}}+\hat{\hat{b}}+\hat{\mathbf{k}})=1$, then $\mathbf{a}=$
|
singleCorrect
| 1 |
Let $\mathbf{a}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$
Now, $\mathbf{a} \cdot \hat{\mathbf{i}}=(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot \hat{\mathbf{i}}=x$
$$
\mathbf{a} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}})=(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}})=x+y
$$
and $\mathbf{a} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=x+y+z$
Now, according to the question,
$$
\begin{array}{ll}
& \mathbf{a} \cdot \hat{\mathbf{i}}=\mathbf{a}(\hat{\mathbf{i}}+\hat{\mathbf{j}})=\mathbf{a} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=1 \\
\Rightarrow & x=x+y=x+y+z=1 \\
\Rightarrow & x=1, y=0, z=0 \\
\therefore & \mathbf{a}=\hat{\mathbf{i}}
\end{array}
$$
|
["$\\hat{i}+\\hat{j}$", "$\\hat{\\mathrm{i}}-\\hat{\\mathrm{k}}$", "$\\hat{\\hat{i}}$", "$\\hat{\\mathrm{i}}+\\hat{j}-\\hat{\\mathbf{i}}$"]
|
[2]
| null |
Practise Problem
|
dd63426ab379fad4006cf53b8779e79a
|
BITSAT_MATH
|
Vector Algebra
|
If <math> <mrow> <mover accent="true"> <mi>a</mi> <mo stretchy="true">→</mo> </mover> <mo>,</mo><mover accent="true"> <mi>b</mi> <mo stretchy="true">→</mo> </mover> <mo>,</mo><mover accent="true"> <mi>c</mi> <mo stretchy="true">→</mo> </mover> </mrow> </math> are mutually perpendicular vectors having magnitude <math> <mrow> <mn>1</mn><mo>,</mo><mn>2</mn><mo>,</mo><mn>3</mn> </mrow> </math> respectively, then <math> <mrow><mo>[</mo> <mrow> <mtable> <mtr> <mtd> <mrow> <mover accent="true"> <mi>a</mi> <mo stretchy="true">→</mo> </mover> <mo>+</mo><mover accent="true"> <mi>b</mi> <mo stretchy="true">→</mo> </mover> <mo>+</mo><mover accent="true"> <mi>c</mi> <mo stretchy="true">→</mo> </mover> </mrow> </mtd> <mtd> <mrow> <mover accent="true"> <mi>b</mi> <mo stretchy="true">→</mo> </mover> <mo>−</mo><mover accent="true"> <mi>a</mi> <mo stretchy="true">→</mo> </mover> </mrow> </mtd> <mtd> <mrow> <mover accent="true"> <mi>c</mi> <mo stretchy="true">→</mo> </mover> </mrow> </mtd> </mtr> </mtable> </mrow> <mo>]</mo></mrow><mo>=</mo> </math>
|
singleCorrect
| 2 |
<p>Given, <math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>|</mo><mover><mi>a</mi><mo>→</mo></mover><mo>|</mo></mrow><mo>=</mo><mn>1</mn><mo>,</mo><mtext> </mtext><mrow><mo>|</mo><mover><mi>b</mi><mo>→</mo></mover><mo>|</mo></mrow><mo>=</mo><mn>2</mn><mo>,</mo><mtext> </mtext><mrow><mo>|</mo><mover><mi>c</mi><mo>→</mo></mover><mo>|</mo></mrow><mo>=</mo><mn>3</mn></math> and <br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>a</mi><mo>→</mo></mover><mo>.</mo><mover><mi>b</mi><mo>→</mo></mover><mo>=</mo><mn>0</mn><mo>=</mo><mover><mi>b</mi><mo>→</mo></mover><mo>.</mo><mover><mi>c</mi><mo>→</mo></mover><mo>=</mo><mover><mi>c</mi><mo>→</mo></mover><mo>.</mo><mover><mi>a</mi><mo>→</mo></mover></math> (as the three vectors are mutually perpendicular)</p><p>So, <br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>[</mo><mrow><mo>(</mo><mover><mi>a</mi><mo>→</mo></mover><mo>+</mo><mover><mi>b</mi><mo>→</mo></mover><mo>+</mo><mover><mi>c</mi><mo>→</mo></mover><mo>)</mo></mrow><mo>×</mo><mrow><mo>(</mo><mover><mi>b</mi><mo>→</mo></mover><mo>−</mo><mover><mi>a</mi><mo>→</mo></mover><mo>)</mo></mrow><mo>]</mo></mrow><mo>.</mo><mover><mi>c</mi><mo>→</mo></mover></math><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>=</mo><mrow><mo>[</mo><mover><mi>a</mi><mo>→</mo></mover><mo>×</mo><mover><mi>b</mi><mo>→</mo></mover><mo>−</mo><mn>0</mn><mo>+</mo><mn>0</mn><mo>−</mo><mover><mi>b</mi><mo>→</mo></mover><mo>×</mo><mover><mi>a</mi><mo>→</mo></mover><mo>+</mo><mover><mi>c</mi><mo>→</mo></mover><mo>×</mo><mover><mi>b</mi><mo>→</mo></mover><mo>-</mo><mover><mi>c</mi><mo>→</mo></mover><mo>×</mo><mover><mi>a</mi><mrow></mrow><mo>→</mo></mover></mrow><mo>]</mo></mrow><mo>.</mo><mover><mi>c</mi><mo>→</mo></mover></math><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mrow><mo>[</mo><mn>2</mn><mrow><mo>(</mo><mover><mi>a</mi><mo>→</mo></mover><mo>×</mo><mover><mi>b</mi><mo>→</mo></mover><mo>)</mo></mrow><mo>.</mo><mover><mi>c</mi><mo>→</mo></mover><mo>+</mo><mrow><mo>(</mo><mover><mi>c</mi><mo>→</mo></mover><mo>×</mo><mover><mi>b</mi><mo>→</mo></mover><mo>)</mo></mrow><mo>.</mo><mover><mi>c</mi><mo>→</mo></mover><mo>−</mo><mrow><mo>(</mo><mover><mi>c</mi><mo>→</mo></mover><mo>×</mo><mover><mi>a</mi><mo>→</mo></mover><mo>)</mo></mrow><mo>.</mo><mover><mi>c</mi><mo>→</mo></mover><mo>]</mo></mrow></math><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>2</mn><mrow><mo>(</mo><mover><mi>a</mi><mo>→</mo></mover><mo>×</mo><mover><mi>b</mi><mo>→</mo></mover><mo>)</mo></mrow><mo>.</mo><mover><mi>c</mi><mo>→</mo></mover><mo>+</mo><mn>0</mn><mo>−</mo><mn>0</mn></math><br /><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>2</mn><mrow><mo>[</mo><mrow><mover><mi>a</mi><mo>→</mo></mover><mo> </mo><mover><mi>b</mi><mo>→</mo></mover><mo> </mo></mrow><mover><mi>c</mi><mo>→</mo></mover><mo>]</mo></mrow></math><br /><math> <mrow> <mo>=</mo> <mn>2</mn> <mo>·</mo> <mn>1</mn> <mo>·</mo> <mn>2</mn> <mo>·</mo> <mn>3</mn> </mrow> </math><br /><math> <mrow> <mo>=</mo> <mn>1</mn> <mn>2</mn> </mrow> </math></p>
|
["<math><mn>0</mn></math>", "<math><mn>6</mn></math>", "<math><mn>12</mn></math>", "<math><mn>18</mn></math>"]
|
[2]
| null |
Practise Problem
|
66d9afbffe823fe0780e2f7eec544c69
|
BITSAT_MATH
|
Vector Algebra
|
If <math><mover accent="true"><mrow><mi>a</mi></mrow><mo>-</mo></mover><mo>+</mo><mover accent="true"><mrow><mi>b</mi></mrow><mo>-</mo></mover><mo>,</mo><mi></mi><mover accent="true"><mrow><mi></mi><mi>b</mi></mrow><mo>-</mo></mover><mo>+</mo><mover accent="true"><mrow><mi>c</mi></mrow><mo>-</mo></mover></math> and <math><mover accent="true"><mrow><mi>c</mi></mrow><mo>-</mo></mover><mo>+</mo><mover accent="true"><mrow><mi>a</mi></mrow><mo>-</mo></mover></math> are coterminous edges of a parallelepiped, then its volume is ________
|
singleCorrect
| 1 |
<math><mo>∵</mo></math> Volume of a parallelelopiped whose, Coterminous edges are <math><mover accent="true"><mrow><mi>a</mi></mrow><mo>-</mo></mover><mo>,</mo><mover accent="true"><mrow><mi>b</mi></mrow><mo>-</mo></mover><mo>,</mo><mover accent="true"><mrow><mi>c</mi></mrow><mo>-</mo></mover><mo> </mo><mo> </mo><mi>i</mi><mi>s</mi></math> <br/><math><mi>v</mi><mo>=</mo><mfenced close="]" open="[" separators="|"><mrow><mover accent="true"><mrow><mi>a</mi></mrow><mo>-</mo></mover><mi></mi><mover accent="true"><mrow><mi>b</mi></mrow><mo>-</mo></mover><mi></mi><mover accent="true"><mrow><mi>c</mi></mrow><mo>-</mo></mover></mrow></mfenced></math> … (1) <br/>Then, required volume = <math><mfenced close="]" open="[" separators="|"><mrow><mover accent="true"><mrow><mi>a</mi></mrow><mo>-</mo></mover><mo>+</mo><mover accent="true"><mrow><mi>b</mi></mrow><mo>-</mo></mover><mo>,</mo><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mover accent="true"><mrow><mi>b</mi></mrow><mo>-</mo></mover><mo>+</mo><mover accent="true"><mrow><mi>c</mi></mrow><mo>-</mo></mover><mo>,</mo><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mover accent="true"><mrow><mi>c</mi></mrow><mo>-</mo></mover><mo>+</mo><mover accent="true"><mrow><mi>a</mi></mrow><mo>-</mo></mover></mrow></mfenced><mo>=</mo><mn>2</mn><mfenced close="]" open="[" separators="|"><mrow><mover accent="true"><mrow><mi>a</mi></mrow><mo>-</mo></mover><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mover accent="true"><mrow><mi>b</mi></mrow><mo>-</mo></mover><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mover accent="true"><mrow><mi>c</mi></mrow><mo>-</mo></mover></mrow></mfenced></math>
|
["<math><mn>3</mn><mfenced close=\"]\" open=\"[\" separators=\"|\"><mrow><mtable><mtr><mtd><mover accent=\"true\"><mrow><mi>a</mi></mrow><mo>-</mo></mover></mtd><mtd><mover accent=\"true\"><mrow><mi>c</mi></mrow><mo>-</mo></mover></mtd><mtd><mover accent=\"true\"><mrow><mi>b</mi></mrow><mo>-</mo></mover></mtd></mtr></mtable></mrow></mfenced></math>", "<math><mn>0</mn></math>", "<math><mn>2</mn><mfenced close=\"]\" open=\"[\" separators=\"|\"><mrow><mtable><mtr><mtd><mover accent=\"true\"><mrow><mi>a</mi></mrow><mo>-</mo></mover></mtd><mtd><mover accent=\"true\"><mrow><mi>b</mi></mrow><mo>-</mo></mover></mtd><mtd><mover accent=\"true\"><mrow><mi>c</mi></mrow><mo>-</mo></mover></mtd></mtr></mtable></mrow></mfenced></math>", "<math><mn>4</mn><mfenced close=\"]\" open=\"[\" separators=\"|\"><mrow><mtable><mtr><mtd><mover accent=\"true\"><mrow><mi>b</mi></mrow><mo>-</mo></mover></mtd><mtd><mover accent=\"true\"><mrow><mi>a</mi></mrow><mo>-</mo></mover></mtd><mtd><mover accent=\"true\"><mrow><mi>c</mi></mrow><mo>-</mo></mover></mtd></mtr></mtable></mrow></mfenced></math>"]
|
[2]
| null |
Practise Problem
|
301cdbdc73a69ddb1c3df61bfd56e881
|
BITSAT_MATH
|
Vector Algebra
|
If <math><mover accent="true"><mrow><mi>p</mi></mrow><mo>-</mo></mover><mo>,</mo><mi></mi><mi></mi><mover accent="true"><mrow><mi>q</mi></mrow><mo>-</mo></mover></math> and <math><mover accent="true"><mrow><mi>r</mi></mrow><mo>-</mo></mover></math> are non-zero, non-coplanar vectors, then <math><mfenced close="]" open="[" separators="|"><mrow><mtable><mtr><mtd><mover accent="true"><mrow><mi>p</mi></mrow><mo>-</mo></mover><mo>+</mo><mover accent="true"><mrow><mi>q</mi></mrow><mo>-</mo></mover><mo>-</mo><mover accent="true"><mrow><mi>r</mi></mrow><mo>-</mo></mover></mtd><mtd><mover accent="true"><mrow><mi>p</mi></mrow><mo>-</mo></mover><mo>-</mo><mover accent="true"><mrow><mi>q</mi></mrow><mo>-</mo></mover></mtd><mtd><mover accent="true"><mrow><mi>q</mi></mrow><mo>-</mo></mover><mo>-</mo><mover accent="true"><mrow><mi>r</mi></mrow><mo>-</mo></mover></mtd></mtr></mtable></mrow></mfenced><mo>=</mo></math> ________
|
singleCorrect
| 2 |
Since, <math><mi>p</mi><mo>,</mo><mi></mi><mi>q</mi><mo>,</mo><mi></mi><mi>r</mi></math> non-zero, non-coplanar vectors then, <math><mfenced close="]" open="[" separators="|"><mrow><mover accent="false"><mrow><mi>p</mi></mrow><mo>¯</mo></mover><mo>+</mo><mover accent="false"><mrow><mi>q</mi></mrow><mo>¯</mo></mover><mo>-</mo><mover accent="false"><mrow><mi>r</mi><mo>,</mo></mrow><mo>¯</mo></mover><mover accent="false"><mrow><mi mathvariant="normal"></mi><mi>p</mi></mrow><mo>¯</mo></mover><mo>-</mo><mover accent="false"><mrow><mi>q</mi><mo>,</mo></mrow><mo>¯</mo></mover><mover accent="false"><mrow><mi mathvariant="normal"></mi><mi>q</mi></mrow><mo>¯</mo></mover><mo>-</mo><mover accent="false"><mrow><mi>r</mi></mrow><mo>¯</mo></mover></mrow></mfenced><mo>=</mo><mfenced close="|" open="|" separators="|"><mrow><mtable columnalign="left"><mtr><mtd><mn>1</mn><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mn>1</mn><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mo>-</mo><mn>1</mn></mtd></mtr><mtr><mtd><mn>1</mn><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mo>-</mo><mn>1</mn><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mn>0</mn></mtd></mtr><mtr><mtd><mn>0</mn><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mn>1</mn><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mi mathvariant="normal"></mi><mo>-</mo><mn>1</mn></mtd></mtr></mtable></mrow></mfenced><mfenced close="]" open="[" separators="|"><mrow><mtable><mtr><mtd><mi>p</mi></mtd><mtd><mi>q</mi></mtd><mtd><mi>r</mi></mtd></mtr></mtable></mrow></mfenced></math> <br/><math><mo>=</mo><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mn>1</mn><mo>-</mo><mn>1</mn></mrow></mfenced><mfenced close="]" open="[" separators="|"><mrow><mtable><mtr><mtd><mi>p</mi></mtd><mtd><mi>q</mi></mtd><mtd><mi>r</mi></mtd></mtr></mtable></mrow></mfenced></math> <br/><math><mo>=</mo><mfenced close="]" open="[" separators="|"><mrow><mtable><mtr><mtd><mi>p</mi></mtd><mtd><mi>q</mi></mtd><mtd><mi>r</mi></mtd></mtr></mtable></mrow></mfenced></math>
|
["<math><mn>3</mn><mfenced close=\"]\" open=\"[\" separators=\"|\"><mrow><mtable><mtr><mtd><mover accent=\"true\"><mrow><mi>p</mi></mrow><mo>-</mo></mover></mtd><mtd><mover accent=\"true\"><mrow><mi>q</mi></mrow><mo>-</mo></mover></mtd><mtd><mover accent=\"true\"><mrow><mi>r</mi></mrow><mo>-</mo></mover></mtd></mtr></mtable></mrow></mfenced></math>", "<math><mn>0</mn></math>", "<math><mfenced close=\"]\" open=\"[\" separators=\"|\"><mrow><mtable><mtr><mtd><mover accent=\"true\"><mrow><mi>p</mi></mrow><mo>-</mo></mover></mtd><mtd><mover accent=\"true\"><mrow><mi>q</mi></mrow><mo>-</mo></mover></mtd><mtd><mover accent=\"true\"><mrow><mi>r</mi></mrow><mo>-</mo></mover></mtd></mtr></mtable></mrow></mfenced></math>", "<math><mn>2</mn><mfenced close=\"]\" open=\"[\" separators=\"|\"><mrow><mtable><mtr><mtd><mover accent=\"true\"><mrow><mi>p</mi></mrow><mo>-</mo></mover></mtd><mtd><mover accent=\"true\"><mrow><mi>q</mi></mrow><mo>-</mo></mover></mtd><mtd><mover accent=\"true\"><mrow><mi>r</mi></mrow><mo>-</mo></mover></mtd></mtr></mtable></mrow></mfenced></math>"]
|
[2]
| null |
Practise Problem
|
12e1c7d18d1a882792c226173e2667cd
|
BITSAT_MATH
|
Vector Algebra
|
If the vectors $\vec{\alpha}=\hat{\mathbf{i}}+a \hat{\mathbf{j}}+\mathbf{a}^{2} \hat{\mathbf{k}}, \vec{\beta}=\hat{\mathbf{i}}+b \hat{\mathbf{j}}+\mathbf{b}^{2} \hat{\mathbf{k}}$, and $\vec{\gamma}=\hat{\mathbf{i}}+c \hat{\mathbf{j}}+c^{2} \hat{k}$ are three non-coplanar vectors and $\left|\begin{array}{lll}\mathbf{a} & \mathbf{a}^{2} & 1+\mathrm{a}^{3} \\ \mathbf{b} & \mathbf{b}^{2} & 1+\mathrm{b}^{3} \\ \mathrm{c} & \mathrm{c}^{2} & 1+\mathrm{c}^{3}\end{array}\right|=0$,
then the value of abc is
|
singleCorrect
| 1 |
Hint:
$\left|\begin{array}{lll}\mathrm{a} & \mathrm{a}^{2} & 1 \\ \mathrm{~b} & \mathrm{~b}^{2} & 1 \\ \mathrm{c} & \mathrm{c}^{2} & 1\end{array}\right|(1+\mathrm{abc})=0$
$a b c=-1[\because \vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are non-coplanar vector $]$
|
["1", "0", "$-1$", "2"]
|
[2]
| null |
Practise Problem
|
4d74a75abb75b8dbc051b7fb1ef966ce
|
BITSAT_MATH
|
Vector Algebra
|
$\overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{i}}=\overrightarrow{\mathbf{a}} \cdot(2 \hat{\mathbf{i}}+\hat{\mathbf{j}})=\overrightarrow{\mathbf{a}}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})=1$, then $\overrightarrow{\mathbf{a}}$ is equal to :
|
singleCorrect
| 1 |
Let $\overrightarrow{\mathbf{a}}=a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}$
$\because \quad \overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{i}}=1$
$\therefore \quad\left(a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}\right) \cdot \hat{\mathbf{i}}=1$
$\Rightarrow \quad a_1=1$
Now, $\quad \overrightarrow{\mathbf{a}} \cdot(2 \hat{\mathbf{i}}+\hat{\mathbf{j}})=1$
$\Rightarrow \quad\left(a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}\right) \cdot(2 \hat{\mathbf{i}}+\hat{\mathbf{j}})=1$
$\Rightarrow \quad 2 a_1+a_2=1$
$\Rightarrow \quad a_2=1-2 \quad\left(\because a_1=1\right)$
$\Rightarrow \quad a_2=-1$
and $\quad \overrightarrow{\mathbf{a}} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})=1$
$\Rightarrow \quad\left(a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}\right) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})=1$
$\Rightarrow \quad a_1+a_2+3 a_3=1$
$\Rightarrow \quad 1-1+3 a_3=1$
$\Rightarrow \quad a_3=\frac{1}{3}$
$\therefore \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}=\frac{1}{3}(3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}})$
|
["$\\hat{\\mathbf{i}}-\\hat{\\mathbf{k}}$", "$1 / 3(3 \\hat{\\mathbf{i}}+3 \\hat{\\mathbf{j}}+\\hat{\\mathbf{k}})$", "$1 / 3(\\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}+\\hat{\\mathbf{k}})$", "$1 / 3(3 \\hat{\\mathbf{i}}-3 \\hat{\\mathbf{j}}+\\hat{\\mathbf{k}})$"]
|
[3]
| null |
Practise Problem
|
8369ce29f1e9983079732577e36fe3a0
|
BITSAT_MATH
|
Vector Algebra
|
If the points whose position vectors are $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, 6 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $14 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+p \hat{\mathbf{k}}$ are collinear, then the value of $p$ is
are collinear, then the value of $p$ is
|
singleCorrect
| 1 |
Given vectors $\mathbf{a}=2 \mathbf{i}+\mathbf{j}+\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=6 \mathbf{i}-\mathbf{j}+2 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{c}}=14 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+p \hat{\mathbf{k}}$ are collinear, therefore
$[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]=0$
$\begin{array}{cccc} & & \left|\begin{array}{ccc}2 & 1 & 1 \\ 6 & -1 & 2 \\ 14 & -5 & p\end{array}\right|=0 \\ \Rightarrow & 2[-p+10]-1[6 p-28]+1[-30+14]=0 \\ \Rightarrow & -2 p+20-6 p+28-16=0 \\ \Rightarrow & & -8 p+32=0 \\ \Rightarrow & & p=4\end{array}$
|
["$2$", "$4$", "$6$", "$8$"]
|
[1]
| null |
Practise Problem
|
42cdf380940ba00ea7ea180ac818a19c
|
BITSAT_MATH
|
Vector Algebra
|
The volume (in cubic units) of the tetrahedron with edges $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ is
|
singleCorrect
| 1 |
We know that, volume of tetrahedron
$\begin{aligned} & =\frac{1}{6}[\overrightarrow{\mathbf{A B}} \overrightarrow{\mathbf{A C}} \overrightarrow{\mathbf{A D}}] \\ & =\frac{1}{6}\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & -1\end{array}\right|\end{aligned}$
$\begin{aligned} & =\frac{1}{6}[1(1-2)-1(-1-1)+1(2+1)] \\ & =\frac{1}{6}[-1+2+3] \\ & =\frac{4}{6}=\frac{2}{3}\end{aligned}$
|
["$4$", "$2 / 3$", "$1 / 6$", "$1 / 3$"]
|
[1]
| null |
Practise Problem
|
034ab5a4a7cd0f385a8974e9a32dcefb
|
BITSAT_MATH
|
Vector Algebra
|
If $\bar{a}=\hat{\imath}+5 \hat{k}, \bar{b}=2 \hat{\imath}+3 \hat{k}, \bar{c}=4 \hat{\imath}-\hat{\jmath}+2 \hat{k}$ and $\bar{d}=\hat{\imath}-\hat{\jmath}$,
then $(\bar{c}-\bar{a}) \cdot(\bar{b} \times \bar{d})=$
|
singleCorrect
| 2 |
$\bar{b} \times \bar{d}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 3 \\ 1 & -1 & 0\end{array}\right|=\hat{i}(3)-\hat{j}(0-3)+\hat{k}(-2)$
$\bar{b} \times \bar{d}=3 \hat{i}+3 \hat{j}-2 \hat{k}$
and $\bar{c}-\bar{a}=3 \hat{i}-\hat{j}-3 \hat{k}$
then $(\bar{c}-\bar{a}) \cdot(\bar{b} \times \bar{d})=3(3)+(-1)(3)+(-3)(-2)$
$=9-3+6=12$
|
["$12$", "$20$", "$30$", "$10$"]
|
[0]
| null |
Practise Problem
|
31690ac1fc0c77cc77934f5126aa125c
|
BITSAT_MATH
|
Vector Algebra
|
The two vector $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ represent the two sides $\overline{A B}$ and $\overline{A C}$ respectively of a $\triangle A B C$. The length of the median through $A$ is
|
singleCorrect
| 1 |
We know that, the sum of three vectors of triangle is zero.
<br><img src="https://cdn-question-pool.getmarks.app/pyq/kcet/tU1onFK1ltK56Ba07KbZT2iGxYbTeoSRo4PvQcOvmdk.original.fullsize.png"><br>
$\begin{array}{ll}\therefore & \mathbf{A B}+\mathbf{B C}+\mathbf{C A}=0 \\ \Rightarrow \quad & \mathbf{B C}=\mathbf{A C}-\mathbf{A B} \\ \Rightarrow \mathbf{B} \mathbf{M}= & \frac{\mathbf{A C}-\mathbf{A B}}{2}(\text { since, } M \text { is a mid-point of } B C)\end{array}$
And also, $\mathbf{A B}+\mathbf{B} \mathbf{M}+\mathbf{M A}=0$
$\Rightarrow \quad \mathbf{A B}+\frac{\mathbf{A C}-\mathbf{A B}}{2}=\mathbf{A M}$
$\Rightarrow \quad \mathbf{A M}=\frac{\mathbf{A B}+\mathbf{A C}}{2}$
$\Rightarrow \quad \mathbf{A M}=\frac{(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})+(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})}{2}$
$\Rightarrow \quad \mathbf{A M}=\frac{2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}}{2}$
$\Rightarrow \quad \mathbf{A M}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
$\therefore \quad|\mathbf{A M}|=\sqrt{(1)^{2}+(2)^{2}+(3)^{2}}$
$=\sqrt{1+4+9}=\sqrt{14}$
|
["$\\frac{\\sqrt{14}}{2}$", "14", "7", "$\\sqrt{14}$"]
|
[3]
| null |
Practise Problem
|
ddf77622818298becf905ef6458786cc
|
BITSAT_MATH
|
Vector Algebra
|
The curve passing through the point $(1,2)$ given that the slope of the tangent at any point $(x, y)$ is $\frac{3 x}{y}$ represents
|
singleCorrect
| 1 |
We have,
$\begin{aligned}
\frac{d y}{d x} &=\frac{3 x}{y} \\
\Rightarrow & \int y d y &=\int 3 x d x \\
\Rightarrow & & \frac{y^{2}}{2} &=3 \cdot \frac{x^{2}}{2}+C...(i)
\end{aligned}$
Since, Eq (i) passing through point $(1,2)$
$\begin{aligned}
&\therefore \quad \quad \frac{4}{2}=\frac{3}{2}(1)^{2}+c \\
&\Rightarrow \quad c=2-\frac{3}{2}=\frac{1}{2} \\
&\therefore \text { Equation of curve, } \frac{y^{2}}{2}=\frac{3 x^{2}}{2}+\frac{1}{2} \\
&\Rightarrow \quad y^{2}=3 x^{2}+1 \Rightarrow y^{2}-3 x^{2}=1 \\
&\Rightarrow \quad \frac{y^{2}}{(1)^{2}}-\frac{x^{2}}{(1 / \sqrt{3})^{2}}=1
\end{aligned}$
Which represents a Hyperbola.
|
["Circle", "Parabola", "Ellipse", "Hyperbola"]
|
[3]
| null |
Practise Problem
|
b76cebedc02f82631effb7026b50a0b2
|
BITSAT_MATH
|
Vector Algebra
|
The projection of $=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ on $\hat{v}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{R}}$ is
|
singleCorrect
| 1 |
Projection of $a$ on $b=\frac{a \cdot b}{|b|}$
$$
\begin{aligned}
&=\frac{(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}})}{|2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}|} \\
&=\frac{6-3+5}{\sqrt{4+9+1}}=\frac{8}{\sqrt{14}}
\end{aligned}
$$
|
["$\\frac{8}{\\sqrt{35}}$", "$\\frac{8}{\\sqrt{39}}$", "$\\frac{8}{\\sqrt{14}}$", "$\\sqrt{14}$"]
|
[2]
| null |
Practise Problem
|
ccd97fabc9cb1de735895978ab1a7f00
|
BITSAT_MATH
|
Vector Algebra
|
A unit vector perpendicular to both the vectors $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\hat{\mathbf{j}}+\hat{\mathbf{k}}$ is
|
singleCorrect
| 2 |
Let $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\mathbf{b}=\hat{\mathbf{j}}+\hat{\mathbf{k}}$
Then, unit vector perpendicular to both $\mathbf{a}$ and $b$ is $\frac{a \times b}{|a \times b|}$
Now, $\mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right|=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$
and $|\mathbf{a} \times \mathbf{b}|=\sqrt{(1)^{2}+(-1)^{2}+(1)^{2}}=\sqrt{3}$
Thus, required vector $=\frac{\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{3}}$
|
["$\\frac{-\\hat{\\mathbf{i}}-\\hat{\\mathbf{j}}+\\hat{\\mathbf{k}}}{\\sqrt{3}}$", "$\\frac{\\hat{\\mathrm{i}}+\\hat{\\mathrm{j}}-\\hat{\\mathrm{k}}}{3}$", "$\\frac{\\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}+\\hat{\\mathbf{k}}}{\\sqrt{3}}$", "$\\frac{\\hat{\\mathrm{i}}-\\hat{\\mathrm{j}}+\\hat{\\mathrm{k}}}{\\sqrt{3}}$"]
|
[3]
| null |
Practise Problem
|
bfb389d69e8d6e416023d0c4a5ad3ee4
|
BITSAT_MATH
|
Vector Algebra
|
If $a$ and $b$ are vectors such that $|a \times b|=|a-b|$, then the angle between a and b is
|
singleCorrect
| 1 |
We have,
$$
\begin{aligned}
& &|a+b| &=|a-b| \\
& \Rightarrow &|a+b|^{2} &=|a-b|^{2} \\
\Rightarrow & &(a+b) \cdot(a+b) &=\langle a-b) \cdot(a-b) \\
& \Rightarrow &|a|^{2}+|b|^{2}+2 a \cdot b &=|a|^{2}+|b|^{2}-2 a \cdot b \\
\Rightarrow & & 4 a \cdot b &=0 \\
& \Rightarrow & a \cdot b &=0 \\
& \Rightarrow & a & \perp b
\end{aligned}
$$
So, angle between a and $\mathrm{b}$ is $90^{\circ}$.
|
["$60^{\\circ}$", "$120^{\\circ}$", "$30^{\\circ}$", "$90^{\\circ}$"]
|
[3]
| null |
Practise Problem
|
72f6d2ab1faffa4a3f0b53178d6e6af3
|
BITSAT_MATH
|
Vector Algebra
|
$\mathbf{O A}$ and $\mathbf{B O}$ are two vectors of magnitudes 5 and 6 respectively. If $\angle B O A=60^{\circ}$, then OA $\cdot \mathbf{O B}$ is equal to
|
singleCorrect
| 1 |
We have,
$\begin{aligned}|\mathrm{OA}|=5,|\mathrm{OB}| &=6, \angle \mathrm{BOA}=60^{\circ} \\ \text { Now, } \mathrm{OA} \cdot \mathrm{OB} &=|\mathrm{OA}||\mathrm{OB}| \cos (\angle B O A) \\ &=5 \times 6 \cos 60^{\circ}=5 \times 6 \times \frac{1}{2}=15 \end{aligned}$
|
["15", "0", "$15 \\sqrt{3}$", "$-15$"]
|
[0]
| null |
Practise Problem
|
22ec29cb4498ea01f1a947ebe6ac65d1
|
BITSAT_MATH
|
Vector Algebra
|
If $|\mathbf{a}|=8,|\mathbf{b}|=3$ and $|\mathbf{a} \times \mathbf{b}|=12$, then find the angle between $\mathbf{a}$ and $\mathbf{b}$.
|
singleCorrect
| 1 |
Given, $|\mathfrak{a}|=8,|\vec{b}|=3$ and $|\mathbf{a} \times \mathbf{b}|=12$
We know that, $\sin \theta=\frac{|a \times b|}{|a| b \mid}$
$\Rightarrow \quad \sin \theta=\frac{12}{8 \times 3}=\frac{1}{2} \Rightarrow \sin \theta=\sin \frac{\pi}{6}$
$\Rightarrow \quad \theta=\frac{\pi}{6} \quad\left[\sin \frac{\pi}{6}=\frac{1}{2}\right]$
Hence, angle between and b is $\frac{\pi}{6}$.
|
["$\\frac{\\pi}{3}$", "$\\frac{\\pi}{6}$", "$\\frac{\\pi}{4}$", "None of these"]
|
[1]
| null |
Practise Problem
|
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