MCQ Questions for Class 12 Commerce Maths Vector Algebra Quiz 9

If vectors

$\vec {A} = 2\hat {i} + 3\hat {j} + 4\hat {k}, \vec {B} = \hat {i} + \hat {j} + 5\hat {k}$ and

$\vec {C}$ form a left-handed system, then

$\vec {C}$ is

0%
$11\hat {i} - 6\hat {j} - \hat {k}$
0%
$-11\hat {i}+ 6\hat {j} + \hat {k}$
0%
$11\hat {i} - 6\hat {j} + \hat {k}$
0%
$-11\hat {i} + 6\hat {j} - \hat {k}$

$2\overrightarrow{a}+3\overrightarrow{b}+4\overrightarrow{c}=0 \Rightarrow \overrightarrow{a}\times \overrightarrow{b} + \overrightarrow{b}\times \overrightarrow{c} +\overrightarrow{c}\times \overrightarrow{a}=$

0%
$0$
0%
$3\overrightarrow{a}\times \overrightarrow{b}$
0%
$3\overrightarrow{b}\times \overrightarrow{c}$
0%
$3\overrightarrow{c}\times \overrightarrow{a}$

Explanation

$2\overrightarrow{a}+3\overrightarrow{b}+4\overrightarrow{c}=0$

$\Rightarrow \left(2\overrightarrow{a}+3\overrightarrow{b}+4\overrightarrow{c}\right)\times \overrightarrow{a}=0$

$\Rightarrow 4\overrightarrow{c}\times \overrightarrow{a}=3\overrightarrow{a}\times \overrightarrow{b}$ ..................

$\left(1\right)$

$\left(2\overrightarrow{a}+3\overrightarrow{b}+4\overrightarrow{c}\right)\times\overrightarrow{b}=0$

$\overrightarrow{a}\times \overrightarrow{b}=2\overrightarrow{b}\times \overrightarrow{c}$ ............

$\left(2\right)$

$\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}$

$= \overrightarrow{a}\times \overrightarrow{b}+\dfrac{1}{2}\overrightarrow{a}\times \overrightarrow{b}+\dfrac{3}{4}\overrightarrow{a}\times \overrightarrow{b}$

$= \overrightarrow{a}\times \overrightarrow{b}\left(1+\dfrac{1}{2}+\dfrac{3}{4}\right)$

$= \dfrac{9}{4}\overrightarrow{a}\times \overrightarrow{b}$

$=3\times \dfrac{3}{4}\overrightarrow{a}\times \overrightarrow{b}$

$=3\overrightarrow{c}\times\overrightarrow{a}$ using

$\left(1\right)$ and

$\left(2\right)$

$|\vec{a}|=5, |\vec{a}-\vec{b}|=8$ and

$|\vec{a}+\vec{b}|=10$ , then

$|\vec{b}|$ is equal to :

0%
$1$
0%
$\sqrt{57}$
0%
$13$
0%
$14$

Explanation

$\begin{array}{l} \left| { \vec { a } } \right| =5,\left| { \vec { a } -\vec { b } } \right| =8,\left| { \vec { a } +\vec { b } } \right| =10 \\ { \left| { \vec { a } -\vec { b } } \right| ^{ 2 } }=64 \\ { \left| { \vec { a } } \right| ^{ 2 } }+{ \left| { \vec { b } } \right| ^{ 2 } }-2\vec { a } \cdot \vec { b } =64 \\ { \left| { \vec { a } +\vec { b } } \right| ^{ 2 } }=100 \\ { \left| { \vec { a } } \right| ^{ 2 } }+{ \left| { \vec { b } } \right| ^{ 2 } }-2\vec { a } \cdot \vec { b } =100 \\ { \left| { \vec { b } } \right| ^{ 2 } }-2\vec { a } \cdot \vec { b } =39 \\ { \left| { \vec { b } } \right| ^{ 2 } }+2\vec { a } \cdot \vec { b } =75 \\ 2{ \left| { \vec { b } } \right| ^{ 2 } }=114 \\ { \left| { \vec { b } } \right| ^{ 2 } }=57 \\ \left| { \vec { b } } \right| =\sqrt { 57 } \\ Hence,\, correct\, option\, is\, \left( B \right) \end{array}$

Let

$\overrightarrow{a}=\hat{i}+2\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$ and

$\overrightarrow{c}=\hat{i}+\hat{j}-\hat{k}$ . A vector in the plane of

$\overrightarrow{a}$ and

$\overrightarrow{b}$ , where projection on

$\overrightarrow{c}$ is

$\dfrac{1}{\sqrt{3}}$ , is

0%
$4\hat{i}-\hat{j}+4\hat{k}$
0%
$3\hat{i}+\hat{j}-3\hat{k}$
0%
$2\hat{i}+\hat{j}$
0%
$4\hat{i}+\hat{j}-4\hat{k}$

Explanation

A vector in the plane of

$\overrightarrow{a}$ and

$\overrightarrow{b}$ is

$\overrightarrow{a}+\lambda \overrightarrow{b} =\left(\hat{i}+2\hat{j}+\hat{k}\right)+\lambda \left(\hat{i}-\hat{j}+\hat{k}\right)$

$=\left(1+\lambda\right)\hat{i}+\left(2-\lambda\right)\hat{j}+\left(1+\lambda\right)\hat{k}$
If projection on

$\overrightarrow{c}$ is

$\dfrac{1}{\sqrt{3}}$

$\Rightarrow \left(\overrightarrow{a}+\lambda \overrightarrow{b}\right).\dfrac{\overrightarrow{c}}{\left|\overrightarrow{c}\right|}=\pm \dfrac{1}{\sqrt{3}}$

$\Rightarrow \left(\left(1+\lambda\right)\hat{i}+\left(2-\lambda\right)\hat{j}+\left(1+\lambda\right)\hat{k}\right)\dfrac {\left(\hat{i}+\hat{j}-\hat{k}\right)} {\sqrt{1+1+1}} =\pm \dfrac{1}{\sqrt{3}}$

$\Rightarrow \dfrac{1+\lambda+2-\lambda-1-\lambda} {\sqrt{3}} =\pm \dfrac{1}{\sqrt{3}}$

$\Rightarrow 2-\lambda=\pm 1$

$\Rightarrow \lambda=2\pm 1$

$\therefore \lambda=1$ or

$3$
If

$\lambda =1, \overrightarrow{a}+\lambda \overrightarrow{b}=\left(1+1\right)\hat{i}+\left(2-1\right)\hat{j}+\left(1+1\right)\hat{k}$

$=2\hat{i}+\hat{j}+2\hat{k}$
If

$\lambda=3, \overrightarrow{a}+\lambda \overrightarrow{b} = \left(1+3\right)\hat{i}+\left(2-3\right)\hat{j}+\left(1+3\right)\hat{k}$

$=4\hat{i}-\hat{j}+4\hat{k}$

$\bar{a}, \bar{b}, \bar{c}$ are position vectors of the non-collinear points A, B, C respectively, the shortest distance of A and BC is?

0%
$\bar{a}\cdot(\bar{b}-\bar{c})$
0%
$\bar{b}\cdot(\bar{c}-\bar{a})$
0%
$|\bar{b}-\bar{a}|$
0%
$\sqrt{|\vec{b}-\vec{a}|^2-[\cfrac{(\vec{a}-\vec{b}\cdot (\vec{b}-\vec{c})}{|\vec{b}-\vec{c}|}}]^2$

Explanation

vector equation

$\vec{r}=\vec{b}+\lambda(\vec{c}-\vec{b})$

Let

$P$ be the foot of perpendicular of the point

$A$ on

$BC$

Thus, the required distance is

$AP$ .

$AP=\sqrt{AB^2-BP^2}$

$AP=|\vec{b}-\vec{a}|^2$

$BP=\cfrac{(\vec{a}-\vec{b}\cdot (\vec{b}-\vec{c})}{|\vec{b}-\vec{c}|}$

$\therefore$ The requiried distance

$AP$

$AP=\sqrt{|\vec{b}-\vec{a}|^2-[\cfrac{(\vec{a}-\vec{b}\cdot (\vec{b}-\vec{c})}{|\vec{b}-\vec{c}|}}]^2$

1152360_1136960_ans_ca617c56882e4854a3aef11d4896a8ce.PNG

$A,B,C,D$ be any four points and

$E$ and

$F$ be the mid-points of

$AC$ and

$BD$ , respectively, then

$\vec{AB}+\vec{CB}+\vec{CD}+\vec{AD}$ is equal to

0%
$3\vec{EF}$
0%
$4\vec{EF}$
0%
$4\vec{FE}$
0%
$3\vec{FE}$

Explanation

Solution:- (C)

$4 \vec{EF}$

$ABCD$ is a quadrilateral.

Given that

$F$ is the mid-point of

$BD$ , we get

$\vec{AB} + \vec{AD} = 2 \vec{AF}..... \left( 1 \right)$

Similarly

$\vec{CB} + \vec{CD} = 2 \vec{CF} ..... \left( 2 \right) \; \left[ \text{E is the mid point of AC} \right]$

Adding

${eq}^{n} \left( 1 \right) \& \left( 2 \right)$ , we have

$\vec{AB} + \vec{AD} + \vec{CB} + \vec{CD} = 2 \left( \vec{AF} + \vec{CF} \right)$

$\vec{AB} + \vec{AD} + \vec{CB} + \vec{CD} = 2 \left( \left( -\vec{FA} \right) + \left( - \vec{FC} \right) \right)$

$\vec{AB} + \vec{AD} + \vec{CB} + \vec{CD} = -2 \left( \vec{FA} + \vec{FC} \right)$

$\vec{AB} + \vec{AD} + \vec{CB} + \vec{CD} = -2 \left( 2 \vec{FE} \right)$

$\vec{AB} + \vec{AD} + \vec{CB} + \vec{CD} = -2 \left( -2 \vec{EF} \right)$

$\vec{AB} + \vec{AD} + \vec{CB} + \vec{CD} = 4 \vec{EF}$

Let

$\overrightarrow{a}=2\hat{i}-3\hat{j}+6\hat{k}$ and

$\overrightarrow{b}=-2\hat{i}+3\hat{j}-\hat{k}$ then projection of

$\overrightarrow{a}$ on

$\overrightarrow{b} :$ projection of

$\overrightarrow{b}$ on

$\overrightarrow{a}=$

0%
$3:7$
0%
$7:3$
0%
$-4$
0%
$3$

Explanation

Projection of

$\overrightarrow{a}$ on

$\overrightarrow{b}$

$=$ scalar components of

$\overrightarrow{a}$ along

$\overrightarrow{b}$

$=\dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{b}\right|}$
Projection of

$\overrightarrow{b}$ on

$\overrightarrow{a}$

$=$ scalar components of

$\overrightarrow{b}$ along

$\overrightarrow{a}$

$= \dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|}$
Ratio

$= \dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{b}\right|}:\dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|}$

$=\left|\overrightarrow{a}\right|:\left|\overrightarrow{b}\right|=7:3$
where

$\left|\overrightarrow{a}\right|=\sqrt{4+9+36}=\sqrt{49}=7$

$\left|\overrightarrow{b}\right|=\sqrt{4+4+1}=\sqrt{9}=3$
Thus, the ratio is

$7:3$

Let

$\overrightarrow{u}=\hat{i}+\hat{j},\overrightarrow{v}=\hat{i}-\hat{j},\overrightarrow{w}=\hat{i}+2\hat{j}+3\hat{k}$ If

$\hat{n}$ is a unit vector such that

$\overrightarrow{u}.\hat{n}=0$ and

$\overrightarrow {v}.\hat{n}=0$ , then

$\left|\overrightarrow {w}.\hat{n}\right|=$

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Explanation

$\overrightarrow{u}.\hat{n}=0$

$\overrightarrow{v}.\hat{n}=0$

$\Rightarrow\hat{n}$ is along

$\overrightarrow{u}\times\overrightarrow{v}$

$=\begin{bmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&0\\1&-1&0\end{bmatrix}=-2k$

$\therefore\hat{n}=\pm k$ (unit vector)

$\overrightarrow{w}.\hat{n}=\left(\hat{i}+2\hat{j}+3\hat{k}\right).\left(\pm\hat{k}\right)$

$=\pm 3$

$\Rightarrow\left|\overrightarrow{w}.\hat{n}\right|=3$

Let

$\overrightarrow{a}=\hat{i}+2\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$ and

$\overrightarrow{c}=\hat{i}+\hat{j}-\hat{k}$ . A vector in the plane of

$\overrightarrow{a}$ and

$\overrightarrow{b}$ , where projection on

$\overrightarrow{c}$ is

$\dfrac{1}{\sqrt{3}}$ , is

0%
$4\hat{i}-\hat{j}+4\hat{k}$
0%
$3\hat{i}+\hat{j}-3\hat{k}$
0%
$2\hat{i}+\hat{j}$
0%
$4\hat{i}+\hat{j}-4\hat{k}$

Explanation

A vector in the plane of

$\overrightarrow{a}$ and

$\overrightarrow{b}$ is

$\overrightarrow{a}+\lambda \overrightarrow{b} =\left(\hat{i}+2\hat{j}+\hat{k}\right)+\lambda \left(\hat{i}-\hat{j}+\hat{k}\right)$

$=\left(1+\lambda\right)\hat{i}+\left(2-\lambda\right)\hat{j}+\left(1+\lambda\right)\hat{k}$
If projection on

$\overrightarrow{c}$ is

$\dfrac{1}{\sqrt{3}}$

$\Rightarrow \left(\overrightarrow{a}+\lambda \overrightarrow{b}\right).\dfrac{\overrightarrow{c}}{\left|\overrightarrow{c}\right|}=\pm \dfrac{1}{\sqrt{3}}$

$\Rightarrow \dfrac{1+\lambda+2-\lambda-1-\lambda} {\sqrt{3}} =\pm \dfrac{1}{\sqrt{3}}$

$\Rightarrow 2-\lambda=\pm 1$

$\Rightarrow \lambda=2\pm 1$

$\therefore \lambda=1$ ,or

$3$
If

$\lambda =1, \overrightarrow{a}+\lambda \overrightarrow{b}=\left(1+1\right)\hat{i}+\left(2-1\right)\hat{j}+\left(1+1\right)\hat{k}$

$=2\hat{i}+\hat{j}+2\hat{k}$
If

$\lambda=3, \overrightarrow{a}+\lambda \overrightarrow{b} = \left(1+3\right)\hat{i}+\left(2-3\right)\hat{j}+\left(1+3\right)\hat{k}$

$=4\hat{i}-\hat{j}+4\hat{k}$

let

$\overrightarrow{u},\overrightarrow{v},\overrightarrow{w}$ be such that

$\left|\overrightarrow{u}\right|=1,\left|\overrightarrow{v}\right|=2,\left|\overrightarrow{w}\right|=3$ . If the projection of

$\overrightarrow{v}$ along

$\overrightarrow{u}$ is equal to the projection of

$\overrightarrow{w}$ along

$\overrightarrow{u}$ and

$\overrightarrow{v},\overrightarrow{w}$ are perpendicular to each other, then

$\left|\overrightarrow{u}-\overrightarrow{v}+\overrightarrow{w}\right|=$

0%
$2$
0%
$\sqrt{17}$
0%
$\sqrt{14}$
0%
$\sqrt{15}$

Explanation

Projection of

$\overrightarrow{v}$ along

$\overrightarrow{u} =\dfrac{\overrightarrow{u}.\overrightarrow{v}}{\left|\overrightarrow{u}\right|}=\overrightarrow{u}.\overrightarrow{v}$
Projection of

$\overrightarrow{w}$ along

$\overrightarrow{u} =\overrightarrow{w}.\overrightarrow{u}$

$\overrightarrow{v}\perp \overrightarrow{w} \Rightarrow \overrightarrow{v}.\overrightarrow{w}=0$ and

$\overrightarrow{u}.\overrightarrow{v}=\overrightarrow{u}.\overrightarrow{w} \left(given\right)$
Now,

${\left|\overrightarrow{u}-\overrightarrow{v}+\overrightarrow{w}\right|}^{2} ={\left|\overrightarrow{u}\right|}^{2}+{\left|\overrightarrow{v}\right|}^{2}+{\left|\overrightarrow{w}\right|}^{2}-2 \overrightarrow{u}.\overrightarrow{v}-2 \overrightarrow{v}.\overrightarrow{w}+2 \overrightarrow{u}.\overrightarrow{w}$

$= {1}^{2}+{2}^{2}+{3}^{2} = 14$ on simplification

$\therefore \left|\overrightarrow{u}-\overrightarrow{v}+\overrightarrow{w}\right|=\sqrt{14}$

$\vec{a}+\vec{b}+\vec{c}=vec{0}$ then

$\vec{a}\times \vec{b}=?$

0%
$\vec{c}\times \vec{b}$
0%
$\vec{b}\times \vec{c}$
0%
$\vec{a}\times \vec{c}$
0%
$2\vec{b}\times \vec{c}$

Explanation

$\vec { a } +\vec { b } +\vec { c } =\vec { 0 }\ \dots(1) \, \, \, \, then,\, \, \\ \vec { a } \times \vec { b } =? \\ take\, \, (1)\, \, cross\, \, product\, \, with\, \, \vec { b } \, \,. \\ \Rightarrow \left( { \vec { a } +\vec { b } +\vec { c } } \right) \times \vec { b } =\vec { 0 } \times \vec { b } \\ \Rightarrow \vec { a } \times \vec { b } +\vec { b } \times \vec { b } +\vec { c } \times \vec { b } =0 \\ \to \vec { a } \times \vec { b } +0-\vec { b } \times \vec { c } =0 \\ \therefore \vec { a } \times \vec { b } =\vec { b } \times \vec { c } \, \, \, \,$

The ratio in which

$i+2j+3k$ divides the join of

$-2i+3j+5k$ and

$7i-k$ is?

0%
$-3:2$
0%
$1:2$
0%
$2:3$
0%
$-4:3$

Explanation

$\begin{array}{l} i+2j+3k=\left( { 1,2,3 } \right) \\ -2i+3j+5k=\left( { -2,3,5 } \right) \\ 7i-k=\left( { 7,0,-1 } \right) \\ 1=\frac { { 7m-2 } }{ { m+1 } } \\ m+1=7m-2 \\ 3=6m \\ m=\frac { 1 }{ 2 } \\ So,ratio\, \, is\, \, 1:2 \\ Option\, \, B\, \, \, is\, \, correct\, \, \, answer. \end{array}$
1239754_1148802_ans_56a34a3e0efc4c81bb3f827e5450e061.JPG

1239754_1148802_ans_56a34a3e0efc4c81bb3f827e5450e061.JPG

$\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}$ and

$\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}$ , then

0%
$\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|$
0%
$\left|\overrightarrow{a}\right|=\left|\overrightarrow{c}\right|,\left|\overrightarrow{b}\right|=1$
0%
$\overrightarrow{c}\times \overrightarrow{a}=\overrightarrow{b}$
0%
$\overrightarrow{c}\times \overrightarrow{a}=\left[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\right]\overrightarrow{b}$

Explanation

$\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}$

$\Rightarrow \overrightarrow{a}$ is perpendicular to

$\overrightarrow{c}$ and

$\overrightarrow{b}$ is perpendicular to

$\overrightarrow{c}$ and

$\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}$

$\Rightarrow \overrightarrow{c}\bot\overrightarrow{a}$

$\overrightarrow{b}\bot \overrightarrow{a}$

$\Rightarrow \left|\overrightarrow{a}\times \overrightarrow{b}\right|=\left|\overrightarrow{c}\right| , \left|\overrightarrow{b}\times \overrightarrow{c}\right|=\left|\overrightarrow{a}\right|$

$\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right| , \left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|=\left|\overrightarrow{a}\right|$

$\therefore \left|\overrightarrow{a}\right|=\left|\overrightarrow{c}\right|,\left|\overrightarrow{b}\right|=1$
Further

$\overrightarrow{c}\times \overrightarrow{a}=\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \left(\overrightarrow{b}\times \overrightarrow{c}\right)$

$\left(\overrightarrow{a}\times \overrightarrow{b}.\overrightarrow{c}\right)\overrightarrow{b}-\left(\overrightarrow{a}\times \overrightarrow{b}\right).\overrightarrow{b}\overrightarrow{c}$

$=\left[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}\right]\overrightarrow{b}$

Let

$\left| \overline { a } +\overline { b } \right| =\left| \overline { a } -\overline { b } \right|$ . If

$\left| \overline { a } \times \overline { b } \right| =\lambda \left| \overline { a } \right|$ , then

$\lambda=$

0%
$\left| \overline { a } \right|$
0%
$\left| \overline { b } \right|$
0%
$1$
0%
$2$

Explanation

$|\bar{a}+\bar{b}|=|\bar{a}-\bar{b}|$

$|a|^2+|b|^2+2\bar{a}\cdot \bar{b}=|a|^2+|b|^2-2\bar{a}\cdot\bar{b}$

$4\bar{a}\cdot \bar{b}=0$

$\bar{a}\cdot \bar{b}=0$

$\Rightarrow \bar{a}$ &

$\bar{b}$ are perpendicular to each other.

$|\bar{a}\times \bar{b}|=|\bar{a}||\bar{b}|\sin\theta$

Since

$\bar{a}$ &

$\bar{b}$ are perpendicular,

$\theta =90^o$

$=|\bar{a}|\bar{b}|\sin 90^o$

$=|a||b|$

$\therefore \lambda =|b|$ .

1049914_1156556_ans_5373cbf8284a4f568105b0780e2484e7.jpeg

The projection of the vector

$\hat{i}-2\hat{j}+\hat{k}$ on he vector

$4\hat{i}-4\hat{j}+7\hat{k}$ is equal to:

0%
$\dfrac{19}{9}$
0%
$\dfrac{9}{19}$
0%
$\dfrac{\sqrt{3}}{19}$
0%
$\dfrac{19}{\sqrt{3}}$

Explanation

The projection of the vector

$\overrightarrow A$ on the vector

$\overrightarrow B$ is

$\dfrac{{\overrightarrow A \cdot \overrightarrow B }}{{\left| {\overrightarrow B } \right|}} = \dfrac{{1 \times 4 + \left( { - 2} \right) \times \left( { - 4} \right) + 1 \times 7}}{{\sqrt {{4^2} + {{\left( { - 4} \right)}^2} + {7^2}} }} = \dfrac{{19}}{{\sqrt {81} }} = \dfrac{{19}}{9}$

$\left|\overrightarrow{a} \right|=1$ , the projection of

$\overrightarrow{r}$ along

$\overrightarrow{a}$ is

$2$ and

$\overrightarrow{a}\times \overrightarrow{r}+\overrightarrow{b}=\overrightarrow{r}$ , then

$\overrightarrow{r}=$

0%
$\dfrac{1}{2}\left[\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{b}\right]$
0%
$\dfrac{1}{2}\left[2\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{b}\right]$
0%
$\overrightarrow{a}+\overrightarrow{a}\times \overrightarrow{b}$
0%
$\overrightarrow{a}-\overrightarrow{a}\times \overrightarrow{b}$

Explanation

Projection of

$\overrightarrow{r}$ on

$\overrightarrow{a}$ is

$2$

$\Rightarrow \overrightarrow{r}.\overrightarrow{a}=2$ -----

$\left(1\right)$

$\overrightarrow{a}\times \overrightarrow{r}+\overrightarrow{b}=\overrightarrow{r}$ ------

$\left(2\right)$
Dot

$\left(1\right)$ with

$\overrightarrow{a}$

$\Rightarrow\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{r}.\overrightarrow{a}=2$ by

$\left(1\right)$
cross

$\left(1\right)$ with

$\overrightarrow{a}$

$\Rightarrow \overrightarrow{a}\times \left(\overrightarrow{a}\times \overrightarrow{r}\right)+\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{a}\times \overrightarrow{r}$

$\Rightarrow 0+\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{a}\times \overrightarrow{r}$

$\therefore 2\overrightarrow{r}-\overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{b}=2\overrightarrow{r}$

$\Rightarrow \overrightarrow{r}=\dfrac{1}{2}\left[2\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{b}\right]$

Let

$A,B,C$ be distinct point with position vectors

$\hat{i}+\hat{j}$ ,

$\hat{i}-\hat{j}$ ,

$p\hat{i}-q\hat{j}+r\hat{k}$ respectively. Points

$A,B,C$ are collinear, then which of the following can be correct:

0%
$p=q=r=1$
0%
$p=q=r=0$
0%
$p=q=2,r=0$
0%
$p=1,q=2,r=0$

Explanation

The three points A, B and C are collinear if they satisfy one of the conditions

$1) AB+BC=AC, 2). AC+BC=AB$ and

$3). AB+AC=BC$

considering the first condition, we have

$AB=2$ ,

$AC=\sqrt{(1-p)^2+(1+q)^2+r^2}$ and

$BC=\sqrt{(1-p)^2+(-1+q)^2+r^2}$

To satisfy the condition

$AB+BC=AC$ ,

$p, q$ and

$r$ take the values

$1, 2$ and

$0$ respectively.

Option D gives the correct answer.

$D,E\ and \ F$ are the mid-point of the sides

$BC,CA \ and \ AB$ respectively of the triangle

$ABC$ . Which of the following is true?

0%
${ \xrightarrow { AB= } \xrightarrow { 2ED } }$
0%
${ \xrightarrow { AB= } \xrightarrow { 2DE } }$
0%
${ \xrightarrow { AB= } \xrightarrow { ED } }$
0%
${ \xrightarrow { AB= } \xrightarrow { 2DF } }$

Explanation

$From\,the\,figure$

$ED=DG\,\,\,............\left( i \right)$

$EG=AB\,\,\,\,...........\left( ii \right)$

$ED+DG=AB$

$\Rightarrow ED+ED=AB\,\,\,\,\,\,\,\,\,\,\left( from\left( i \right) \right)$

$2ED=AB$

1913491_1150771_ans_5364526531e34402b9b67275f4502fb0.JPG

$\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{c}\times \overrightarrow{a}\neq 0$ then

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=$

0%
$-\overrightarrow{b}$
0%
$2\overrightarrow{a}$
0%
$0$
0%
none of these

A unit vector perpendicular to the plane of the triangle ABC with the position vectors $\vec a\,\,\,\,\vec b\,\,\vec c$ of the vectors A,B,C, is

0%
$\frac{{(\vec a\,\, \times \,\,\vec b + \,\vec b\,\, \times \,\,\vec c\,\, + \vec c \times \vec a)\,\,\,}}{\Delta }$
0%
$\frac{{(\vec a\,\, \times \,\,\vec b + \,\vec b\,\, \times \,\,\vec c\,\, + \vec c \times \vec a)\,\,\,}}{{2\Delta }}$
0%
$\frac{{(\vec a\,\, \times \,\,\vec b + \,\vec b\,\, \times \,\,\vec c\,\, + \vec c \times \vec a)\,\,\,}}{{4\Delta }}$
0%
none of these

$\overrightarrow { a } =2\hat { i } +\hat { j } +\hat { k } ,\overrightarrow { b } =3\hat { i } -4\hat { j } +2\hat { k } ,\overrightarrow { c } =\hat { i } -2\hat { j } +2\hat { k }$ then the projection of

$\overrightarrow { a } +\overrightarrow { b }$ on

$\overrightarrow { c }$ is

0%
$\cfrac{17}{3}$
0%
$\cfrac{5}{3}$
0%
$\cfrac{4}{3}$
0%
$\cfrac{17}{\sqrt{43}}$

Explanation

$\vec{a}+\vec{b}=5\hat{i}-3\hat{j}+3\hat{k}$

Projection of

$\vec{a}+\vec{b}$ on

$\vec{c}$ is

$\dfrac{(\vec{a}+\vec{b}).\vec{c}}{|\vec{c}|}$

$\dfrac{5+6+6}{\sqrt{1+4+4}}=\dfrac{17}{3}$

$|a|=5.|\vec{b}|=4$ , and

$|c|=3$ . then what will be the value of

$\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}$ given that

$\vec{a}+\vec{b}+\vec{c}=0$

0%
$25$
0%
$50$
0%
$-25$
0%
$-50$

Explanation

Given

$\left | \vec{a} \right |=5\left | \vec{b} \right |=4$

$\left | \vec{c} \right |=3$

$\vec{a}+\vec{b}+\vec{c} =0$

$\left | \vec{a}+\vec{b}+\vec{c} \right |=0$

$\Rightarrow \left | \vec{a}+\vec{b}+\vec{c} \right |^{2}=0=(\vec{a}+\vec{b}+\vec{c}).(\vec{a}+\vec{b}+\vec{c})$

$\Rightarrow \vec{a}.\vec{a}+\vec{a}.\vec{b}+\vec{a}.\vec{c}+\vec{b}.\vec{a}+\vec{b}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}+\vec{c}.\vec{b}+\vec{c}.\vec{c}=0$

$\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}+\left | \vec{c} \right |^{2}+ 2(\vec{a}\vec{b}+\vec{b}\vec{c}+\vec{c}\vec{a})=0$

$25+16+9+2(\vec{a}\vec{b}+\vec{b}\vec{c}+\vec{c}\vec{a})=0$

$\Rightarrow \vec{a}\vec{b}+\vec{b}\vec{c}+\vec{c}\vec{a}=-25$

$\therefore$ Option C is correct

$A(6,3,2),B(5,1,4),C(3,-4,7), D(0,2,5)$ are four points, then projection of

$CD$ on

$AB$ is

0%
$-\cfrac{13}{3}$
0%
$-\cfrac{13}{7}$
0%
$-\cfrac{3}{13}$
0%
$-\cfrac{7}{13}$

$\vec a,\vec b$ and

${\vec c}$ are unit vectors, then

${\left| {\vec a + \vec b} \right|^2} + {\left| {\vec b - \vec c} \right|^2} + {\left| {\vec c - \vec a} \right|^2}$ does NOT exceed

0%
4
0%
9
0%
8
0%
6

Explanation

Given

$\bar {a}, \bar {b}$ and

$\bar {c}$ are unit vector

we need maximum value, so :-

Let

$\bar {a}= \hat {i} \bar {b}= \hat {j}$ and

$\bar {c} = \hat {k}$

$|\bar {a}+ \bar {b}|^{2}+ |\bar {b}-\bar {c}|^{2}+ |\bar {c}-\bar {a}|^{2}$

$|\hat{i}+ \hat{j}|^{2}+ |\hat{j}- \hat{k}|^{2}+ |\hat{k}- \hat{i}|^{2}$

$(\sqrt{1^{2}+1^{2}})^{2}+ (\sqrt{1^{2}+1^{2}})^{2}+ (\sqrt{1^{2}+1^{2}})^{2}$

$\Rightarrow 2+2+2=6$

$\therefore 6$ is the maximum value.

$\therefore |\bar {a}+ \bar {b}|+ |\bar {b} - \bar {c}| + |\bar {c}- \bar {a}|^{2}$ does not exceed

$6$

$\overline { a }$ and

$\overline { b }$ include an angle of

${120}^{o}$ and their magnitudes are

$2$ and

$\sqrt{3}$ then

$\overline { a } .\overline { b }$ is

0%
$3$
0%
$-\sqrt{3}$
0%
$\sqrt{3}$
0%
$-3$

Explanation

Given,

$|\vec{a}| = 2, \, |\vec{b}| = \sqrt{3}$

$\theta = 120^o$

$\vec{a} . \vec{b} = |\vec{a}| |\vec{b}| cos \theta$

$= 2 \times \sqrt{3} \times cos \, 120^o$

$= 2 \sqrt{3} \times \left(\dfrac{-1}{2} \right)$

$\therefore \vec{a} . \vec{b} = - \sqrt{3}$

$\left| {\vec a} \right| = 2,\left| {\vec b} \right| = 3$ and

$\left| {2\vec a - \vec b} \right| = 5,$ then

$\left| {2\vec a + \vec b} \right|$ equals:

0%
$17$
0%
$7$
0%
$5$
0%
$1$

Explanation

We have

$\left| {\overrightarrow a } \right| = 2\,\,\,\,\,\left| {\overrightarrow b } \right| = 3\,\,\,$

And,

$\left| {2\overrightarrow a - \overrightarrow b } \right| = 5$

We have to find the value of

$\left| {2\overrightarrow a + \overrightarrow b } \right|$

Now

$\left| {2\overrightarrow a - \overrightarrow b } \right|.\left| {2\overrightarrow a - \overrightarrow b } \right| = 25$

$4{\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} - 4\overrightarrow a .\overrightarrow b = 25$

$25 - 4\overrightarrow a .\overrightarrow b = 25$

$\therefore \overrightarrow a .\overrightarrow b = 0$

Then for

$\overrightarrow a \bot \overrightarrow b$

${\left| {2\overrightarrow a + \overrightarrow b } \right|^2} = \left( {2\overrightarrow a + \overrightarrow b } \right).\left( {2\overrightarrow a + \overrightarrow b } \right)$

$= 4{\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + 4\overrightarrow a .\overrightarrow b$

$= 4 \times 4 + 9 = 25$

$\therefore \left| {2\overrightarrow a + \overrightarrow b } \right| = 5$

Hence, the option

$C$ is the correct answer.

If the vectors

$\overrightarrow { a } =\hat { i } -\hat { j } +2\hat { k } ;\overrightarrow { b } =2\hat { i } +4\hat { j } +\hat { k } ;\overrightarrow { c } =\lambda \hat { i } +\hat { j } +\mu \hat { k }$ are mutually orthogonal, then

$(\lambda,\mu)=$

0%
$(2,-3)$
0%
$(-2,3)$
0%
$(3,-2)$
0%
$(-3,2)$

Explanation

Given

$\vec{a} = \hat{i}-\hat{j}+2\hat{k}$

$\vec{b} = 2\hat{i}+4\hat{j}+\hat{k}$

$\vec{c} = \lambda \hat{i}+\hat{j}+\mu \hat{k}$

since, vector's are mutually orthogonal

$\therefore \vec{a}.\vec{b} = 0, \vec{b}.\vec{c} = 0, \vec{c}.\vec{a} = 0$

$\vec{b}.\vec{c} = 0$

$2\lambda +4+\mu = 0$ __ (1)

$\vec{c}.\vec{a} = 0$

$\lambda -1 +2\mu = 0$ __ (II)

$(1)\times 2$

$4\lambda +8+2\mu = 0$

$\lambda -1+2\mu = 0$

____________

$3\lambda +9 = 0$

$\Rightarrow \lambda = -3$

$-3-1+2\mu = 0$

$\Rightarrow \mu = 2$

Hence,

$(\lambda ,\mu ) =(-3,2)$

1072295_1183522_ans_44760a46546a44bb9736f4dea264460a.png

$\left| \overrightarrow { a } \right| =3,\left| \overrightarrow { b } \right| =4$ , if

$\left( \overrightarrow { a } +\lambda \overrightarrow { b } \right)$ is perpendicular to

$\left( \overrightarrow { a } -\lambda \overrightarrow { b } \right)$ then

$\lambda =$

0%
$\cfrac{9}{16}$
0%
$\cfrac{3}{5}$
0%
$\cfrac{3}{4}$
0%
$\cfrac{4}{3}$

Explanation

Given

$(\vec {a}+\lambda\vec {b})\bot (\vec {a}-\lambda {\vec{b}})$

$\therefore \quad (\vec {a}+\lambda \vec {b})(\vec {a}-\lambda \vec {b})=0$

$\Rightarrow \quad |\vec {a}|^{2}-\lambda^{2}|\vec {b}|^{2}=0$

$\Rightarrow \quad 9-\lambda^{2}\cdot 16^{2}=0$

$\Rightarrow \quad \dfrac {9}{16}=\lambda^{2}$

$\therefore \quad \lambda=\pm \dfrac {3}{4}$

$\therefore \quad \lambda =\dfrac {3}{4},\lambda=-\dfrac {3}{4}$

Let

$\bar{a},\bar{b}$ be two noncollinear vectors. If

$A=(x+4y)\bar{a}+(2x+y+1)\bar{b},$

$B=(y-2x+2)\bar{a}+(2x-3y-1)\bar{b} \quad and \quad 3A=2B$ then (x,y) =

0%
(1,2)
0%
(1,-2)
0%
(2,-1)
0%
(-2,-1)

Explanation

$3A=(3x+12y)\bar{a}+(6x+3y+3)\bar{b}$

$2B=(2y-4x+4)\bar{a}+(4x-6y-2)\bar{b}$

$3A=2B$

$\Rightarrow 3x+12y=2y-4x+4$ &

$6x+3y+3=4x-6y-2$

$7x+10y=4$ &

$2x+9y=-5$

Solving

$7x+10y=4$ &

$2x+9y=-5$

We get

$x=2$ ,

$y=-1$

$\therefore (x, y)=(2, -1)$ .

1191630_1195007_ans_bcb641551a3b497992e551ac9d29a4e7.jpg

The projection of the vector

$2\hat i + \hat j - 3\hat k$ on the vector

$\hat i - 2\hat j - \hat k$

0%
$- \dfrac{3}{{\sqrt {14} }}$
0%
$\dfrac{3}{{\sqrt {14} }}$
0%
$- \sqrt {\dfrac{3}{2}}$
0%
$\sqrt {\dfrac{3}{2}}$

Explanation

Given that:

$\vec{P}=2\hat i+\hat j -3\hat k$

$\vec Q=\hat i-2\hat j -\hat k$

Now

We know that projection of

$\vec P$ on

$\vec Q$ is :

$\frac{\vec P . \vec Q}{|\vec Q|}$

So,

$\vec P.\vec Q=(2\hat i+\hat j-3\hat k).(\hat i-2\hat j-\hat k)$

$\vec P.\vec Q=2\times1+(1\times(-2))+(-3\times-1)$

$\vec P.\vec Q=2-2+3$

$\vec P.\vec Q=3$

Now

$|\vec Q|=\sqrt{1+4+1}=\sqrt6$

So,

$proj{\vec Q}{\vec P}=\dfrac{3}{\sqrt6}$

$proj{\vec Q}{\vec P}=\dfrac{\sqrt3}{\sqrt2}$

The position vectors of the points A,B,C are

$\overline { i } + 2 \overline { j } - \overline { k } , \overline { i } + \overline { j } + \overline { k } , 2 \overline { i } + 3 \overline { j } + 2 \overline { k }$ respectively. If A is chosen as the origin then the position vectors of B and C are

0%
$\overline { i } + 2 \overline { k } , \overline { i } + \overline { j } + 3 \overline { k }$
0%
$\overline { j } + 2 \overline { k } , \vec { i } + \overline { j } + 3 \overline { k }$
0%
$- \overline { j } + 2 \overline { k } , \vec { i } - \overline { j } + 3 \overline { k }$
0%
$- \overline { j } + 2 \overline { k } , \overline { i } + \overline { j } + 3 \overline { k }$

Explanation

$\textbf{Step1-Find position vector from given point}$

$\text{The origin is shifted so the new position vectors of B and C would}$

$\text{ be the difference of the old position vectors and the new origin vector i.e.}$

$(i+j+k) - (i+2j-k) = -j+2k$ (

$\text{ New position vector of point B}$ )

$(2i+3j+2k) - (i+2j-k) = i+j+3k$ (

$\text{ New position vector of point A}$ )

$\textbf{Hence , option D is correct}$

Given

$\vec{\alpha} = 3\hat{i} + \hat{j} + 2\hat{k}\ ,\ \vec{\beta} = \hat{i} - 2\hat{j} - 4\hat{k}$ are the position vectors of the points

$A$ and

$B$ . Then the distance of the point

$-\hat{i} + \hat{j} + \hat{k}$ from the passing through

$B$ and perpendicular to

$AB$ is

0%
$-7$
0%
$10$
0%
$15$
0%
$20$

Explanation

We have,

$\overrightarrow{\alpha }=3\widehat{i}+\widehat{j}+2\widehat{k}$

$\overrightarrow{\beta }=\widehat{i}-2\widehat{j}-4\widehat{k}$

So,

$\overrightarrow{AB}=\overrightarrow{\beta }-\overrightarrow{\alpha }$

$\overrightarrow{AB}=\left( \widehat{i}-2\widehat{j}-4\widehat{k} \right)-\left( 3\widehat{i}+\widehat{j}+2\widehat{k} \right)$

$\overrightarrow{AB}=\widehat{i}-2\widehat{j}-4\widehat{k}-3\widehat{i}-\widehat{j}-2\widehat{k}$

$\overrightarrow{AB}=-2\widehat{i}-3\widehat{j}-6\widehat{k}$

Now, equation of plane passing through the point and perpendiular vector is

So,

$\left( -2\widehat{i}-3\widehat{j}-6\widehat{k} \right).\left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right)=\left( -2\widehat{i}-3\widehat{j}-6\widehat{k} \right).\left( -\widehat{i}+\widehat{j}+\widehat{k} \right)$

$\Rightarrow 2x+3y+6z=2-3-6$

$\Rightarrow 2x+3y+6z=-7$

Hence, this is the answer.

$M$ and

$N$ are the mid-points of the diagonals

$AC$ and

$BD$ respectively of a quadrilateral

$ABCD$ , then the value of

$\overline { AB } +\overline { AD } +\overline { CB } +\overline { CD }$

0%
$2\overline { MN }$
0%
$2\overline { NM }$
0%
$4\overline { NM }$
0%
$4\overline { MN }$

Explanation

$\vec{AB}+\vec{AD}+\vec{CB}+\vec{CD}$

$=\vec{b}-\vec{a}+\vec{d}-\vec{a}+\vec{b}+\vec{c}+\vec{d}-\vec{c}$

$=2\vec{b}+2\vec{d}-2\vec{a}-2\vec{c}$

$=2(\vec{b}+\vec{d}-\vec{a}-\vec{c})$

$=4\left(\dfrac{\vec{b}+\vec{d}}{2}-\dfrac{\vec{a}+\vec{c}}{2}\right)$

$=4(\vec{N}-\vec{M})$

$=4\vec{MN}$ .

1098361_1195900_ans_802f478fe248481d9db911486864f97e.jpg

$S$ is the circumcentre,

$O$ is the orthocentre of

$\triangle{ABC}$ , then

$\overline { SA } +\overline { SB } +\overline { SB }$ equals

0%
$\overline { SO }$
0%
$2\overline { SO }$
0%
$\overline { OS }$
0%
$2\overline { OS }$

Explanation

Given, 'S' is the circumcentre

'O' is the orthocentre

$\vec{SA}=\vec{a}$

$\vec{SB}=\vec{b}$

$\vec{SC}=\vec{c}$

$\vec{SO}=\vec{s}$

$\vec{SP}||\vec{AO}$

$\vec{SP}=\dfrac{\vec{SB}+\vec{SC}}{2}=\dfrac{\vec{b}+\vec{c}}{2}$

$\vec{SP}=\lambda \vec{AO}$

$\Rightarrow \dfrac{\vec{b}+\vec{c}}{2}=\lambda (\vec{s}-\vec{a})$

$\Rightarrow \vec{s}=\dfrac{\vec{b}+\vec{c}}{2\lambda}+\vec{a}$ …….

$(1)$

$\vec{SQ}=\mu \vec{BO}$

$\dfrac{\vec{a}+\vec{c}}{2}=\mu (\vec{s}-\vec{b})$

$\Rightarrow \vec{s}=\dfrac{\vec{a}+\vec{c}}{2\mu}+\vec{b}$ ……….

$(2)$

$\vec{a}+\dfrac{\vec{b}+\vec{c}}{2\lambda}=\vec{b}+\dfrac{\vec{a}+\vec{c}}{2\mu}$

$\Rightarrow \vec{a}+\dfrac{\vec{b^2}}{2\lambda}+\dfrac{\vec{c}}{2\lambda}=\vec{b}+\dfrac{\vec{a}}{2\mu}+\dfrac{\vec{c}}{2\mu}$

$\therefore 1=\dfrac{1}{2\mu}$

$\Rightarrow \mu =\dfrac{1}{2}$

$\therefore \vec{s}=\vec{a}+\vec{c}+\vec{b}$

$\therefore \vec{SO}=\vec{SA}+\vec{SB}+\vec{SC}$ .

1098450_1196903_ans_3064848e19474eb4a1e12083a28e079f.jpg

A (1,-1,-1) , B (2,1,-2) and C (-5,2,-6) are the position vectors of the vertices of triangle ABC The length of the bisector of its internal angle at A is:

0%
$\dfrac{\sqrt{10}}{4}$
0%
$\dfrac{3\sqrt{10}}{4}$
0%
$\sqrt{10}$
0%
none

Explanation

$\dfrac {AB}{AC}=\dfrac {BE}{EC}$

$\dfrac {\sqrt 6}{3\sqrt 6}=\dfrac {BE}{EC}=\dfrac {1}{3}$

So point

$E=\left (\dfrac {1}{4}+\dfrac {5}{4},-3\right)$

$AE=\dfrac {3\sqrt {10}}{4}$

1354087_1204209_ans_7fd18e4c883f4c0ab40324025026a4a5.png

$\overline { a } =\left( 2\overline { i } -10\overline { j } +6\overline { k } \right) ;\overline { b } =\left( 5\overline { i } -3\overline { j } +\overline { k } \right)$ . The ratio of projection of

$\overline { a }$ on

$\overline { b }$ to projection of

$\overline { b }$ on

$\overline { a }$ is

0%
$2:1$
0%
$1:2$
0%
$2:3$
0%
$3:2$

Explanation

$\cfrac { Projection\quad of\quad \vec { a } \quad on\quad \vec { b } \quad }{ Projection\quad of\quad \vec { b } \quad on\quad \vec { a } }$

$=\cfrac { \cfrac { \vec { a } .\vec { b } }{ \left| \vec { b } \right| } }{ \cfrac { \vec { b } .\vec { a } }{ \left| \vec { a } \right| } } =\cfrac { \left| \vec { a } \right| }{ \left| \vec { b } \right| } =\cfrac { \sqrt { 4+100+76 } }{ \sqrt { 25+9+1 } } =\cfrac { 2 }{ 1 }$

Let

$\vec a$ and

$\vec b$ be two unit vectors such that

$\left| {\vec a + \vec b} \right| = \sqrt 3$ . If

$\vec c = \vec a + 2\vec b + 3(\vec a \times \vec b)$ , then

$2\left| {\vec c} \right|$ is equal to:

0%
$\sqrt {55}$
0%
$\sqrt {51}$
0%
$\sqrt {43}$
0%
$\sqrt {37}$

Explanation

$\left | \vec{a}+\vec{b} \right | = \sqrt{3}$ [Given]

$\Rightarrow \left | \vec{a}+\vec{b} \right | = \left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}+2\vec{a}\vec{b} = 3$

$\Rightarrow 2+2cos\theta = 3$

$[ \therefore \vec{a}$

$\vec{b}$ are unit vector]

$cos\theta =1/2$

$\Rightarrow \theta = \pi /3$

$\Rightarrow$ Now,

$\vec{c} = \vec{a}+2\vec{b}+3(\vec{a}\times \vec{b})$

$= \vec{a}+2\vec{b}+3\left | \vec{a} \right |\left | \vec{b} \right |sin\theta \,\hat{n}$

$=\vec{a}+2\vec{b}+\frac{3\sqrt{3}}{2}\hat{n}$

$\Rightarrow \left | \vec{c} \right |^{2}=\left | \vec{a} \right |^{2}+4\left | \vec{b} \right |^{2}+\frac{27}{4}\left | \vec{n} \right |^{2}+$

$4\vec{a}.\vec{b}+2\times 3\sqrt{3}\hat{n}.\vec{b}+3\sqrt{3}\hat{n}.\vec{a}$

$[\because \vec{n}\perp \vec{b}$ and

$\vec{n}\perp \vec{a}]$

$= 5+\frac{27}{4}+2 = \frac{28+27}{4} = \frac{55}{4}$

$\Rightarrow \left | \vec{c} \right | = \frac{\sqrt{55}}{a}$

$\Rightarrow 2\left | \vec{c} \right | = \sqrt{55}$

1118858_1202397_ans_8d37321f69e246cd8deacc18a293aab5.jpg

The

$X$ &

$Y$ components of vector A have numerical values

$6$ each & that of

$(A+B)$ have numerical values

$10$ and

$9$ What is the numerical value of

$B$ ?

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

we know that, any vector

$\vec R$ is represented as

$\vec R=x\hat i+y\hat j+z\hat k$

we have

$\vec A=6\hat i+6\hat j$

and

$\vec A+\vec B=10\hat i+9\hat j$

$\Rightarrow \ (\vec A+\vec B)-\vec A=(10\hat i+9\hat j)-(6\hat i+6\hat j)$

$\Rightarrow \ \vec B=4\hat i+3\hat j$

$\Rightarrow \ |\vec B|=\sqrt {4^2+3^2}$

$\Rightarrow \ |\vec B|=5$

$ABC$ is an isosceles triangle right angled at

$A$ . Force of magnitude

$2\sqrt{2},5$ and

$6$ act along

$\overline { BC } ,\overline { CA } ,\overline { AB }$ respectively. The magnitude of their resultant force is

0%
$4$
0%
$5$
0%
$11+2\sqrt{2}$
0%
$30$

Explanation

REF.Image.

$AB = AC \Rightarrow \angle B = \angle C$

And

$\angle B+\angle C+90-180$

$\Rightarrow \angle B = \angle C = 45^{\circ}$

Taking components of

$2\sqrt{2}$

$2\sqrt{2}cos45 = 2\sqrt{2}\times \frac{1}{\sqrt{2}} = 2$

$2\sqrt{2}sin45 = 2\sqrt{2}\times \frac{1}{\sqrt{2}} = 2$

$\Rightarrow$ Resultant =

$\sqrt{16+9} = 5 N$

$\Rightarrow (B)$

1124024_1202826_ans_e332e2d57eef4477a9d49ead02c28b40.jpg

$\overline { a }$ is a vector of magnitude

$\sqrt{3}$ and

$\overline { b }$ is unit vector making an angle

$\tan ^{ -1 }{ \left( 1/\sqrt { 2 } \right) }$ with

$\overline { a }$ then projection of

$\overline { a }$ on

$\overline { b }$ is

0%
$\cfrac{\sqrt{3}}{2}$
0%
$\sqrt{2}$
0%
$\sqrt{3}$
0%
$\sqrt{6}$

Explanation

$\left| \vec { a } \right| =\sqrt 3$ ,

$\left| \vec { b } \right| =1$

$\tan \theta =\cfrac{1}{\sqrt 2}$

$\vec { a } .\vec { b } =\left| \vec { a } \right| \left| \vec { b } \right| \cos \theta$

$=\sqrt 3 \times 1\times \cfrac{\sqrt 2}{\sqrt 3}$

$=\sqrt 2$

Projection of

$\vec { a }$ on

$\vec { b }$

$=\cfrac { \vec { a } .\vec { b } }{ \left| \vec { b } \right| }$

$=\cfrac{\sqrt 2}{1}=\sqrt 2$

In the figure given below

$\bar { AE },$

$\bot\bar { DB },$

$\bar { CF }$

$\bot\bar { BD },$

$\bot\bar { DF }$

$=$

$\bar { BE },$

$\bar { AD },$

$=$

$\bar { BC }$
then

$\bar { DC }=\bar { AB }.$

0%
True
0%
False

Given a parallelogram

$ABCD$ . If

$|\vec{AB}=a, |\vec{AD}|=b$ and

$|vec{AC}|=c$ , then

$|\vec{DB}|.|\vec{AB}|$ has the

0%
$\dfrac{3a^{2}+b^{2}-c^{2}}{2}$
0%
$\dfrac{a^{2}+3b^{2}-c^{2}}{2}$
0%
$\dfrac{a^{2}+b^{2}-3c^{2}}{2}$
0%
$none$

$|\bar {a}-\bar {b}|=|\bar {a}|=|\bar {b}|$ , where

$\bar a$ and

$\bar b$ are non zero vecrors then the angle between

$\bar {a}-\bar {b}$ and

$\bar b$ is

0%
$120^{o}$
0%
$45^{o}$
0%
$60^{o}$
0%
$90^{o}$

The vector

$T + 2 \overline { y } + 2 k$ restated through an angle

$\theta$ and doubled in magnitude then it becomes

$2 T + ( 2 x + 2 ) \} + ( 6 x - 2 ) k$ values of

$x$ are

0%
$1,\frac { 1 } { 3 }$
0%
$- 1 , \frac { 1 } { 3 } =$
0%
$1,\frac { - 1 } { 3 }$
0%
$( 0,3 )$

Explanation

$\begin{array}{l} let,\, \, \, V=T+2\bar { y } +2k \\ \, \, \, \, \, \, \, \left| V \right| =\sqrt { 1+4{ x^{ 2 } }+4 } \\ And,\, \, \, U=\sqrt { 2T+(2x+2)y+(6x-2)k } \\ \, \, \, \, \, \, \, \, \left| U \right| =\sqrt { 4+4{ { (x+1) }^{ 2 } }+4{ { (3x-1) }^{ 2 } } } \\ According\, to\, the\, condition:2\left| V \right| =\left| U \right| \\ \Rightarrow 2\left( { \sqrt { 1+4{ x^{ 2 } }+4 } } \right) =\sqrt { 4+4{ { (x+1) }^{ 2 } }+4{ { (3x-1) }^{ 2 } } } \\ \Rightarrow { 2^{ 2 } }\left( { 1+4{ x^{ 2 } }+4 } \right) =4+4{ (x+1)^{ 2 } }+4{ (3x-1)^{ 2 } } \\ \Rightarrow 4\left( { 5+4{ x^{ 2 } } } \right) =4\left[ { 1+{ { (x+1) }^{ 2 } }+{ { (3x-1) }^{ 2 } } } \right] \\ \Rightarrow \left( { 5+4{ x^{ 2 } } } \right) =1+{ (x+1)^{ 2 } }+{ (3x-1)^{ 2 } } \\ \Rightarrow \left( { 5+4{ x^{ 2 } } } \right) =1+{ x^{ 2 } }+1+2x+9{ x^{ 2 } }+1-6x \\ \Rightarrow 5+4{ x^{ 2 } }=3+10{ x^{ 2 } }-4x \\ \Rightarrow 6{ x^{ 2 } }-4x-2=0 \\ \Rightarrow (x-1)\, (6x+2)=0 \\ \therefore \, \, x=1,\frac { { -2 } }{ 6 } =\frac { { -1 } }{ 3 } \\ \, \, \, \, \, \, \, x=1,\frac { { -1 } }{ 3 } \\ so\, that\, the\, correct\, option\, is\, C. \end{array}$

If AD, BE and CF are

$\Delta ABC$ , then

$\\ \vec { AD } +\vec { BE } \vec { +CF }$

0%
$\vec { 0 }$
0%
1
0%
0
0%
2

Let

$\overline a,\overline b, \overline c$ be three vectors such that

$\overline a \ne 0$ , and

$\overline a \times \overline b= 2\overline a \times \overline c, |\overline a|=|\overline c|=1, |\overline a|$

$|\overline b \times \overline c|= \sqrt 15$ . If

$\overline b - 2 \overline c= \lambda \overline a$ , then

$\lambda$ equals to

0%
$1$
0%
$± 4$
0%
$3$
0%
$-2$

Explanation

$\textbf{Step1-Find the value of}$

$\lambda$

$\text{Given}$

${|\overrightarrow a|=}$

${|\overrightarrow b|=}$

${|\overrightarrow c|}=4$

$|\ {\overrightarrow b \times \overrightarrow c|= }$

${|\overrightarrow b|}$

${|\overrightarrow c|}$

$sin{\theta=}$

$\sqrt15$

$\Rightarrow$

$Sin{\theta=}$

$\frac{\sqrt15}{4}$

$\Rightarrow$

$Cos{\theta=}$

$\frac{1}{4}$

$\text{Now}$

${\overrightarrow b}-$

${2\overrightarrow c=}$

$\lambda$

${\overrightarrow a}$

$\textbf{Step2-Squaring both side and simplify the above expression we get required result}$

$a^2-$

$b^2+$

$4c^2-$

$4bc$

$=$

$\lambda^2$

$a^2$

$\Rightarrow$

$16+4-4=$

$\lambda^2$

$\Rightarrow$

$\lambda$

$=$

$\pm4$

$\textbf{Hence , option B is correct}$

$\bar{a}$ ,

$\bar{b}$ are unit vectors such that

$\mid\bar{a}\times\bar{b}\mid = \bar{a}. \bar{b}$ , then

$\mid\bar{a}+\bar{b}\mid^2 =$

0%
$2$
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$2+\sqrt{2}$
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$2-\sqrt{2}$
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$\sqrt{2}$

The length of the projection of the line segment joining the points

$(5, -1, 4)$ and

$(4, -1, 3)$ on the plane ,

$x+ y+ z = 7$ is :

0%
$\dfrac{2}{3}$
0%
$\dfrac{1}{3}$
0%
$\sqrt{\dfrac{2}{3}}$
0%
$\dfrac{2}{\sqrt{3}}$

Explanation

Here

$PQ$ is the projection

Also

$PQ=BC$

$AC=\overline { AB } .\overline { BC } =\left( \hat { i } +\hat { k } \right) .\left( \dfrac { \hat { i } +\hat { j } +\hat { k } }{ \sqrt { 3 } } \right)$

$=2/\sqrt {3}$

Also,

$AB=\sqrt {2}$

$AB=AC^{2}+BC^{2}=2=4/3+BC^{2}$

$BC^{2}=2/3$

$BC=\sqrt {2/3}$

1347569_1232674_ans_4d2ec19d22c44ef4b88191063b8ed71b.png

$\vec{a},\vec{b},\vec{c},\vec{d}$ are the position vectors of four coplanar points A,B,C,D respectively. If no three of them are collinear and

$|\vec{a}-\vec{d}|=|\vec{b}-\vec{d}|=|\vec{c}-\vec{d}|$ then for triangle ABC, D is

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centroid
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orthocenter
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incenter
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circumcenter

If the position vectors of P, Q are respectively 5a + 4b and 3a - 2b then

$\vec {QP}$ =

0%
$2a + 6b$
0%
$2a - 6b$
0%
$2a + 5b$
0%
$2a - 5b$

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 9 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 9 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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