MCQ Questions for Class 12 Commerce Maths Vector Algebra Quiz 4

Let

$\overrightarrow { b } =4\hat { i } +3\hat { j }$ and

$\overrightarrow{c}$ be two vector perpendicular to each other in the

$xy$ -plane. Then a vector in the same plane having projections

$1$ and

$2$ along

$\overrightarrow{b}$ and

$\overrightarrow{c}$ , respectively, is

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$\hat { i } +2\hat { j }$
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$2\hat { i } -\hat { j }$
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$2\hat { i } +\hat { j }$
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None of these

Explanation

$\overrightarrow { c } =x\hat { i } +y\hat { j }$

Now,

$\overrightarrow { b } \bot \overrightarrow { c } \Rightarrow \overrightarrow { b } .\overrightarrow { c } =0$

$\displaystyle \Rightarrow 4x+3y=0\Rightarrow \dfrac { x }{ 3 } =\frac { y }{ -4 } =\lambda \\ \Rightarrow x=3\lambda ,y=-4\lambda \\ \therefore \overrightarrow { c } =\lambda \left( 3\hat { i } -4\hat { j } \right)$

Let the required vector be

$\alpha =p\hat { i } +q\hat { j }$ .

Then the projections of

$\overrightarrow{\alpha}$ on

$\overrightarrow{b}$ and

$\overrightarrow{c}$ are

$\displaystyle \frac { \overrightarrow { \alpha } .\overrightarrow { b } }{ \left| b \right| }$ and

$\displaystyle \dfrac { \overrightarrow { \alpha } .\overrightarrow { c } }{ \left| c \right| }$ respectively.

$\displaystyle \therefore \dfrac { \overrightarrow { \alpha } .\overrightarrow { b } }{ \left| \overrightarrow { b } \right| } =1$ and

$\displaystyle \dfrac { \overrightarrow { \alpha } .\overrightarrow { c } }{ \left| \overrightarrow { c } \right| } =2$

$\Rightarrow 4p+3q=5$ and

$3p-4q=10$

$\Rightarrow p=2,q=-1$

Hence, the required vector is

$2\hat{i}-\hat{j}$ .

The angle between the Vectors

$\vec{a}\times\vec{b}$ and

$\vec{b}\times\vec{a}$ is

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$0^{0}$
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$45^{0}$
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$90^{0}$
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$180^{0}$

Explanation

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

$(\vec{a}\times \vec{b})(\vec{b}\times \vec{a})=|\vec{a}\times \vec{b}||\vec{b}\times \vec{a}|cos \theta$

$\Rightarrow \dfrac{-|\vec{a}\times \vec{b}|}{|\vec{a}\times\vec{b}|}=cos \theta$

$\Rightarrow \theta =\pi$

The vectors

$\vec{a},\ \vec{b},\ \vec{a}\times \vec{b}$ form

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A right handed system
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A left handed system
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A set of coplanar vectors
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A set of mutually perpendicular vectors

Explanation

$\vec{a}.(\vec{a}\times \vec{b})=0$

$\vec{b}.(\vec{a}\times \vec{b})=0$

$\vec{a}$ &

$\vec{b}$ are

$\perp$ to

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

By Right hand rule, thumb point in direction of

$\vec{a}\times \vec{b}$ when we curl our finger from

$\vec{a}$ to

$\vec{b}$

In a triangle

$ABC, D$ and

$E$ are points on

$BC$ and

$AC$ respectively, such that

$BD=2DC, AE=3EC,$ Let

$P$ be the point of intersection of

$AD$ and

$BE$ . Then

$\dfrac{BE}{PE}=$

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$2:3$
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$3:8$
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$8:3$
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$1:2$

Explanation

Let p.v of

$\vec{a}=\vec{a}$
p.v of

$\vec{B}=\vec{b}$
p.v of

$\vec{C}=\vec{c}$
p.v of

$E=\dfrac{3\vec{c}}{4}$
p.v of

$D=\dfrac{3\vec{c}}{4}$

$=\dfrac{2\vec{c}+\vec{b}}{3}$

$p$ lie on line

$AD$ so p.v of

$p=b\left ( \dfrac{2\hat{c}+\hat{b}}{3} \right )$ ..............(1)
Let

$\dfrac{BP}{PA}=K$
So p.v of

$p=\dfrac{K\dfrac{3\hat{c}}{4}+\hat{b}}{K+1}$ .............(2)
equate (1) and (2)

$K=\dfrac{8}{3}$

$\overline{a}+\overline{b}+\overline{c}=\alpha\overline{d},\overline{b}+\overline{c}+\overline{d}=\beta\overline{a}$ , then

$\overline{a}+\overline{b}+\overline{c}+\overline{d}$ is

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$(\beta+1)\overline{a}$
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$\alpha\overline{a}$
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$(\alpha+1)\overline{b}$
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$\alpha\overline{a}+\beta\overline{b}$

$OABC$ is a parallelogram with

$\vec{OB}=\vec{a},\vec{AB}=\vec{b}$ then

$\vec{OA}=$

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$\vec{a}+\vec{b}$
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$\vec{a}-\vec{b}$
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$\displaystyle \dfrac{1}{2}(\vec{a}+\vec{b})$
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$\displaystyle \dfrac{1}{2}(\vec{a}-\vec{b})$

Explanation

From figure we get

$\overrightarrow{OA}+\overrightarrow{AB}=\overrightarrow{OB}$

$\overrightarrow{OA}=\vec{a}-\vec{b}$

Let us define, the length of a vector

$a\overline{i}+b\overline{j}+c\overline{k}$ as

$|{a}|+|{b}|+|{c}|$ . This definition coincides with the usual definition of the length of a vector

$a\overline{i}+b\overline{j}+c\overline{k}$ if

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$a=b=c=0$
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Any one of $a, b, c$ , is zero
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Any two of $a, b, c$ are zero
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$a=b=c\neq 0$

Explanation

The definitions will coincides if

$|a|+|b|+|c|=\sqrt{a^2+b^2+c^2}$

$\Rightarrow a^2+b^2+c^2+2(|a||b|+|b||c|+|c||a|)=a^2+b^2+c^2$ , square both sides

$\Rightarrow |a||b|+|b||c|+|c||a|=0$

Clearly above equation will satisfy if any two of

$a,b,c$ are zero.

$\Delta ABC,\ D,\ E,\ F$ are midpoints of the sides

$BC, CA$ and

$AB$ respectively.

$O$ ' is the circumcentre,

$G$ ' is the centroid,

$H$ ' is the orthocentre and

$P$ is any point.
Match the following

List I	List II
$1) \vec{PA} +\vec{PB}+\vec{PC}$	$a)$ $0$
$2)\vec {GA}+\vec{GB}+\vec{GC}$	$b) \vec{OH}$
$3)\displaystyle \vec{AD}+\dfrac{2}{3}\vec{BE}+\dfrac{1}{3}\vec{CF}$	$c)\vec{ PD}+\vec{PE}+\vec{PF}$
$4)\vec{OA}+\vec{OB}+\vec{OC}$	$d){\displaystyle\dfrac{1}{2}}\vec{AC}$

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$1-a,2- b,3- c,4- d$
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$1-c,2- a,3- b,4- c$
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$1-c,2- a,3- d,4- b$
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$1-a,2- b,3- d,4- c$

Explanation

1. Since there is an arbitrary point

$P$ in the question, the answer needs to be dependent on

$P$ and so, there's only one option which has that, implying the answer to be

$c$ .
2.

$G = \dfrac{a + b + c}{3}$ where

$a,b,c$ are the position vectors of

$A,B,C$ respectively.
Thus,

$\vec {GA} + \vec{ GB} + \vec {GC} = G - a + G - b + G - c = 3G - a - b - c = 0$
3. We have

$D = \dfrac{b + c}{2}, E = \dfrac{c + a}{2}, F = \dfrac{a + b}{2}$
Thus,

$\vec{AD} + \dfrac{2}{3}\vec{BE} + \dfrac{1}{3}\vec{CF} = D - a + \dfrac{2}{3}E - \dfrac{2}{3}b + \dfrac{1}{3}F - \dfrac{1}{3}c$

$= \dfrac{1}{6}\left[3b + 3c - 6a + 2c + 2a - 4b + a + b - 2c \right] = \dfrac{1}{2}\left[c - a\right]$ , hence option

$d$ .
4. Option

$b$ is the remaining option!

Vector area is a vector quantity associated with each plane figure whose magnitude is

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Any quantity and direction parallel to the plane
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Any quantity and direction perpendicular to the plane
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Equal to the area and direction parallel to the plane
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Equal to the area and direction perpendicular to the plane

Let

$OABC$ be a parallelogram and

$D$ the midpoint of

$OA$ . The ratio in which

$OB$ divides

$CD$ in the ratio

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$1:2$
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$1:3$
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$1:4$
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$2:1$

Explanation

P.v. of

$D =\dfrac{\vec{a}}{2}$

P.v. of

$E=\dfrac{\lambda (\vec{b}+\vec{a})}{}$
Let

$E$ divide

$CD$ in ratio

$K:1$
P.v. of

$E=\dfrac{K\dfrac{\vec{a}}{2}+\vec{b}}{K+1}$

$\lambda (\vec{b}+\vec{a})=\dfrac{\dfrac{k\vec{a}}{2}+\vec{b}}{k+1}$

$\lambda =\dfrac{1}{k+1} \lambda =\dfrac{k}{2k+1}$
So

$K=2$

$\Delta OAB$ , if

$\vec{OA}=\vec{a},\ \vec{OB}=\vec{b}. L$ is mid point of

$\vec{OA}$ and

$M$ is point on

$\vec{OB}$ such that

$\vec{OM}:\vec{MB}=2:1$ . If

$P$ is mid point of

$LM$ then

$\vec{AP}=$

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${\dfrac{1}{3}\vec{b}-\dfrac{3}{4}\vec{a}}$
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$\displaystyle \dfrac{1}{3}\vec{b}+\dfrac{3}{4}\vec{a}$
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${\dfrac{1}{3}\vec{a}-\dfrac{3}{4}\vec{b}}$
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$\displaystyle \dfrac{1}{3}\vec{a}+\dfrac{3}{4}\vec{b}$

Explanation

P.v. of

$L=\dfrac{\vec{a}}{2}$ (mid point of

$\vec{OA}$ )
P.v. of

$M=\dfrac{2\vec{b}}{3}$ (use section formula)
p.v. of

$P=\dfrac{\dfrac{a}{2}+\dfrac{2\vec{b}}{3}}{2}$ (mid point of LM)

$\vec{AP}=\dfrac{\vec{a}}{4}+\dfrac{\vec{b}}{3}-\vec{a}$

$=\dfrac{-3\vec{a}}{4}+\dfrac{\vec{b}}{3}$

If the vectors

$\vec{c},\vec{a}=x\hat{i}+y\hat{j}+z\hat{k}$ and

$\vec{b}=\hat{j}$ are such that

$\vec{a},\vec{c}$ and

$\vec{b}$ form a right handed system, then

$\vec{c}=$

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$z\hat{i}$
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$y\hat{i}$
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$z\hat{i}-x\hat{k}$
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${-z\hat{i}+x\hat{k}}$

Explanation

$\vec{b}\times \vec{a}=\vec{c}$ (as it follow RHS)

$\vec{b}\times \vec{a}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 0 & 1 & 0\\ x & y & z \end{vmatrix}=z\hat{i}-x\hat{k}=\vec{c}$

The vector

$\vec{AB}=3\hat{i}+4\hat{k}$ and

$\vec{AC}=5\hat{i}-2\hat{j}+4\hat{k}$ are the sides of a

$\Delta ABC$ where

$A$ is the origin. The length of median through

${A}$ is

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$\sqrt{72}$
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$\sqrt{33}$
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$\sqrt{288}$
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$\sqrt{18}$

Explanation

Let

$A$ be the origin.

$D$ is mid of

$BC=(4,-1,4)$
So Length

$AD=\sqrt{16+16+1}$

$=\sqrt{33}$

$C$ is the mid point of

$AB$ and

$P$ is any point out side

$AB$ , then

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$\vec{PA}+\vec{PB}+2\vec{PC}=\vec{0}$
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$\vec{PA}+\vec{PB}+\vec{PC}=\vec{0}$
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$\vec{PA}+\vec{PB}=2\vec{PC}$
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$\vec{PA}+\vec{PB}=\vec{PC}$

Explanation

From above figure

$\vec{PA}=\vec{CA}+\vec{BC}+\vec{PB}$

$=2\vec{BC}+\vec{PB}$

$=2(\vec{PC}-\vec{PB})+\vec{PB}$

$\vec{PA}+\vec{PB}=2\vec{PC}$

$ABC$ is a triangle and

$P$ is any point on

$BC$ . If

$PQ$ is the resultant of the vectors

$\vec {AP},\ \vec {PB}$ and

$\vec{PC}$ then

$ACQB$ is

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Rectangle
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Square
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Rhombus
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Parallelogram

Explanation

$\vec{PQ}=\vec{AP}+\vec{PB}+\vec{PC}$

$\vec{PQ}-\vec{PB}=\vec{AP}+\vec{PC}$

$\vec{BQ}=\vec{AC}$ --------(1)

$\vec{PQ}-\vec{PC}=\vec{AP}+\vec{PB}$

$\vec{CQ}=\vec{AB}$ ---------(2)
So

$ACQB$ is Parallelogram.

$\vec{b}$ is the vector whose initial point divides the joining

$5\hat{i}$ and

$5\hat{j}$ in the ratio

$\lambda :$

$1$ and terminal point is at origin. lf

$|\vec{b}|\leq\sqrt{37}$ , then

$\lambda\in$ .

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$(-\displaystyle \infty, -6]\cup \left [-\dfrac{1}{6}, \infty \right)$
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$(-\displaystyle \infty, -3)\cup \left [-\dfrac{1}{4}, \infty \right)$
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$(-\infty, 0)\cup \left(\dfrac{1}{2} , \infty\right)$
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$\left [-6, -\displaystyle \dfrac{1}{6}\right]$

Explanation

Apply section formula and we get,

$\vec b = \dfrac{5 \hat i + 5 \lambda \hat j}{\lambda + 1}$

$|\vec b| \leq \sqrt{37}$

$\therefore$

$25 \times {\lambda}^2 + 25 \leq ({\lambda + 1}^2) \times 37$

$\therefore$

$25 \times {\lambda}^2 + 25 \leq 37 \times ({\lambda}^2 + 2\lambda + 1)$

$\therefore$

$25 \times {\lambda}^2 + 25 \leq 37\times {\lambda}^2 + 74 \lambda + 37$

$\therefore 12 \times {\lambda}^2 + 74 \lambda + 12 \geq 0.$

$\therefore 6 \times {\lambda}^2 + 37 \lambda + 6 \geq 0$

$\therefore \lambda \in ( - \infty , -6] \cup \left [ \dfrac{-1}{6}, \infty \right)$

lf the Vector

$\overline{\mathrm{c}},\ \vec{\mathrm{a}}=\mathrm{x}\hat{\mathrm{i}}+\mathrm{y}\hat{\mathrm{j}}+\mathrm{z}\hat{\mathrm{k}},\ \vec{\mathrm{b}}=\hat{\mathrm{j}}$ are such that

$\vec{\mathrm{a}},\ \vec{\mathrm{c}},\ \vec{\mathrm{b}}$ form

$\mathrm{R}.\mathrm{H}.\ \mathrm{S}$ then

$\vec{\mathrm{c}}=$

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$\mathrm{z}\hat{\mathrm{i}}-\mathrm{x}\hat{\mathrm{k}}$
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$\mathrm{x}\hat{\mathrm{i}}-\mathrm{z}\hat{\mathrm{k}}$
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$\mathrm{x}\hat{\mathrm{j}}-\mathrm{y}\hat{\mathrm{k}}$
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$\mathrm{y}\hat{\mathrm{j}}-\mathrm{x}\hat{\mathrm{k}}$

Explanation

$\vec{b}\times \vec{a}=\vec{c} (As \vec{a}, \vec{c}, \vec{b} are in R.H.S)$

$\vec{b} \times \vec{a}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 0 & 1 & 0\\ x & y & z \end{vmatrix} =z\hat{i}-x\hat{k}$

$\vec{c}=z\vec{i}-x\vec{k}$

$\vec{r}=3\hat{i}+2\hat{j}-5\hat{k},\vec{a}=2\hat{i}-\hat{j}+\hat{k}$ ,

$\vec{b}=\hat{i}+3\hat{j}-2\hat{k},\ \vec{c}=-2\hat{i}+\hat{j}-3\hat{k}$ such that

$\vec{r}=\lambda\vec{a}+\mu\vec{b}+v\vec{c}$ , then

$\mu,\ \displaystyle \frac{\lambda}{2}$ ,

$v$ are in

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$H.P$
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$G.P$
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$A.G.P$
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$A.P$

Explanation

$(3,2,-5)=\lambda (2,-1,1)+\mu (1,3,-2)+\gamma (-2,1,-3)$

$3=2\lambda +\mu -2\gamma$

$2=-\lambda +3\mu +\gamma$

$-5=\lambda -2\mu -3\gamma$

$\begin{bmatrix} 2 & 1 & -2\\ -1 & 3 & 1\\ 1 & 2 & -3 \end{bmatrix} \begin{bmatrix} \lambda\\ \mu\\ \nu \end{bmatrix}=\begin{bmatrix} 3\\ 2\\ -5 \end{bmatrix}$

$R_{2}\rightarrow R_{2}+R_{3}$

$\begin{bmatrix} 2 & 1 & -2\\ 0 & 1 & -2\\ 1 & -2 & -3 \end{bmatrix} \begin{bmatrix} \lambda\\ \mu\\ \nu \end{bmatrix}=\begin{bmatrix} 3\\ -3\\ -5 \end{bmatrix}$

$R_{1}\rightarrow R_{1}-R_{2}$

$\begin{bmatrix} 2 & 0 & 0\\ 0 & 1 & -2\\ 1 & -2 & -3 \end{bmatrix} \begin{bmatrix} \lambda\\ \mu\\ \nu \end{bmatrix}=\begin{bmatrix} 6\\ -3\\ -5 \end{bmatrix}$

$\lambda =3 \gamma =2 \mu =1$

$\overline{a}=x\hat {i}+y\hat {j}+z\hat {k},\ \overline{b}=\hat {j}$ , then the vector

$\overline{c}$ for which

$\overline{a},\overline{b},\ \overline{c}$ form a right hand triangle

0%
$x(\hat {i}-\hat {k})$
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$\hat {0}$
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$-z\hat {i}+x\hat {k}$
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$y\hat {j}$

Explanation

$\overline{a},\overline{b},\overline{c}$ form a right handed triangle.
When this holds true,

$\overline{c} = \overline{a} \times \overline{b}$
Thus, taking cross product , we get

$\overline{c} = -z \overline{i} + x \overline{k}$

$I$ is the center of a circle inscribed in a triangle

$ABC$ , then

$|BC|IA+|CA|IB+|AB|IC$

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$0$
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$IA+IB+IC$
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$\displaystyle \dfrac{IA+IB+IC}{3}$
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$\displaystyle \dfrac{IA+IB+IC}{2}$

Explanation

We know that the position vaector of the incenter of a

$\triangle ABC$ is given by

$\displaystyle \dfrac { BC.\overrightarrow { a } +CA.\overrightarrow { b } +AB.\overrightarrow { c } }{ BC+CA+AB }$

where

$\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c }$ are the affixes of the vertices of the triangle

$\therefore$ , in the given triangle

$ABC$ , the affix of

$I$ is given by

$\displaystyle \dfrac { \left| \overrightarrow { BC } \right| \overrightarrow { IA } +\left| \overrightarrow { CA } \right| \overrightarrow { IB } +\left| \overrightarrow { AB } \right| \overrightarrow { IC } }{ \left| \overrightarrow { BC } \right| +\left| \overrightarrow { CA } \right| +\left| \overrightarrow { AB } \right| }$

But is the position vector of

$I$ with respect to

$I$ .

$\displaystyle \therefore \overrightarrow { 0 } =\dfrac { \left| \overrightarrow { BC } \right| \overrightarrow { IA } +\left| \overrightarrow { CA } \right| \overrightarrow { IB } +\left| \overrightarrow { AB } \right| \overrightarrow { IC } }{ \left| \overrightarrow { BC } \right| +\left| \overrightarrow { CA } \right| +\left| \overrightarrow { AB } \right| }$

$\displaystyle \Rightarrow \left| \overrightarrow { BC } \right| \overrightarrow { IA } +\left| \overrightarrow { CA } \right| \overrightarrow { IB } +\left| \overrightarrow { AB } \right| \overrightarrow { IC } =\overrightarrow { 0 }$

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

$\vec {b}=2\hat {i}-\hat {j}-\hat {k}$ then the ratio between the projection of

$\vec b$ on

$\vec {a}$ and the projection of

$\vec {a}$ on

$\vec {b}$ is

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$\sqrt{5}:\sqrt{7}$
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$\sqrt{3}:\sqrt{7}$
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$\sqrt{7}:\sqrt{3}$
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$\sqrt{7}:\sqrt{5}$

Explanation

Projection of

$\overline{b}$ on

$\overline{a}=\dfrac{\overline{b}\cdot\overline{a}}{|\overline{a}|}$
Projection of

$\overline{a}$ on

$\overline{b}=\dfrac{\overline{b}\cdot\overline{a}}{|\overline{b}|}$
Ratios

$=|\overline{b}|:|\overline{a}|$

$=\sqrt{6} :\sqrt{14} = \sqrt{3} : \sqrt{7}$

Given,

$|\vec {a}|=|\vec {b}|=1$ and

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

$\vec {c}$ be a vector such that

$\vec {c}-\vec {a}-2\vec {b}=3(\vec {a}\times\vec {b})$ , then

$\vec {c}.\vec {b}$ is equal to

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$-\displaystyle \dfrac{1}{2}$
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$\displaystyle \dfrac{1}{2}$
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$\displaystyle \dfrac{3}{2}$
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$\displaystyle \dfrac{5}{2}$

Explanation

We have

$\overline{c}-\overline{b}=\overline{a}-\overline{b}+2$

$\left | \overline{a}+\overline{b} \right |=\sqrt{\bar a^{2}+\bar b^{2}+2\bar \cdot \bar b}$

$3=1+1+2\left ( \overline{a\cdot} \overline{b}\right )$

$\dfrac {1}{2}=\overline{a}-\overline{b}$

$\overline{c}-\overline b=\dfrac {5}{2}$

Which of the following is a true statement.

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$\vec {a}\times\vec {b})\times\vec {c}$ is coplanar with $\vec {c}$
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$(\vec {a}\times\vec {b})\times\vec {c}$ is perpendicular to $\vec {a}$
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$(\vec {a}\times\vec {b})\times\vec {c}$ is perpendicular to $\vec {b}$
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$(\vec {a}\times\vec {b})\times\vec {c}$ is perpendicular to $\vec {c}$

Explanation

We know that

$\vec {a}\times\vec {b}$ is perpendicular to both

$\vec {a}$ and

$\vec {b}$

Hence

$(\vec {a}\times \vec {b})\times \vec {c}$ is perpendicular to both

$\vec {a}\times \vec {b}$ and

$\vec {c}$

If the position vector of a point

$A$ is

$\overrightarrow { a } +\overrightarrow { 2b }$ and

$\overrightarrow { a }$ divides

$\overrightarrow{AB}$ in the ratio

$2:3$ , then the position vector of

$B$ is

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$\overrightarrow { 2a } -\overrightarrow { b }$
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$\overrightarrow { b } -\overrightarrow { 2a }$
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$\overrightarrow { a } -\overrightarrow { 3b }$
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$\overrightarrow { b }$

Explanation

$\overrightarrow{x}$ is the position vector of

$B$ , then

$\overrightarrow{a}$ divides

$\overrightarrow{AB}$ in the ration

$42:3$

$\displaystyle \therefore \overrightarrow { a } =\dfrac { 2\overrightarrow { x } +3\left( \overrightarrow { a } +2\overrightarrow { b } \right) }{ 2+3 }$

$\Rightarrow 5\overrightarrow { a } -3\overrightarrow { a } -6\overrightarrow { b } =2\overrightarrow { x } \Rightarrow \overrightarrow { x } =\overrightarrow { a } -3\overrightarrow { b }$

In a tetrahedron if two pairs of opposite edges are at a right angles then the third pair is inclined at an angle of

0%
$30^{0}$
0%
$60^{0}$
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$90^{0}$
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$120^{0}$

Explanation

Let the position vectors of the vertices be

$\overline o, \overline a, \overline b\ and\ \overline c.$
One opposite pair=

$\overline a.(\overline b- \overline c) = 0$
hence,

$a.b = a.c$
Similarly from other pair

$b.a = b.c$
Hence , from these 2 equations,

$a.b = b.c = a.c$

Hence

$c.(a-b) = a.c - b.c = 0$
Hence proved

$\overrightarrow { a } .\overrightarrow { b } =0$ and also

$\overrightarrow { a } \times \overrightarrow { b } =0,$ then

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$\overrightarrow{a}$ is parallel to $\overrightarrow{b}$
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$\overrightarrow{a}$ is perpendicular to $\overrightarrow{b}$
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Either $\overrightarrow{a}$ or $\overrightarrow{b}$ is a non-zero vector
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None of these

Explanation

$\overrightarrow{a}$ and

$\overrightarrow{b}$ cannot be perpendicular both at the same time.

So in order to hold

$\overrightarrow { a } .\overrightarrow { b } =0$ and

$\overrightarrow { a }\times\overrightarrow { b } =0$ , simultaneously, either

$\overrightarrow { a }$ or

$\overrightarrow{b}$ must be zero.

Let

$G$ be the centroid of

$\triangle ABC$ . If

$\overrightarrow{AB}=\overrightarrow{a}$ and

$\overrightarrow{AC}=\overrightarrow{b},$ then

$\overrightarrow{AG}$ , in terms of

$\overrightarrow{a}$ and

$\overrightarrow{b}$ is

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$\displaystyle \dfrac { 2 }{ 3 } \left( \overrightarrow { a } +\overrightarrow { b } \right)$
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$\displaystyle \dfrac { 1 }{ 6 } \left( \overrightarrow { a } +\overrightarrow { b } \right)$
0%
$\displaystyle \dfrac { 1 }{ 3 } \left( \overrightarrow { a } +\overrightarrow { b } \right)$
0%
$\displaystyle \dfrac { 1 }{ 2 } \left( \overrightarrow { a } +\overrightarrow { b } \right)$

Explanation

Let

$A$ be the origin.

then

$\overrightarrow { AB } =\overrightarrow { a } ,\overrightarrow { AC } =\overrightarrow { b }$ implies that the position vectors of

$B$ and

$C$ are

$\overrightarrow{b}$ and

$\overrightarrow{c}$ respectively.

Let

$AD$ be the median and

$G$ be the centroid. Then,

$P.V.$ of

$\displaystyle D=\dfrac { \overrightarrow { a } +\overrightarrow { b } }{ 2 }$ ,

$P.V.$ of

$\displaystyle G=\dfrac { \overrightarrow { a } +\overrightarrow { b } }{ 3 }$

$\displaystyle \therefore \overrightarrow { AG } =\dfrac { \overrightarrow { a } +\overrightarrow { b } }{ 3 }$

$\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c }$ are the position vectors of the vertices of an equilateral triangle whose orthocentre is at the origin, then the centroid of the triangle satisfies which of the following relation?

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$\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c }=0$
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$\overrightarrow { { a }^{ 2 } } =\overrightarrow { { b }^{ 2 } } +\overrightarrow { { c }^{ 2 } }$
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$\overrightarrow { a } +\overrightarrow { b } -\overrightarrow { c } =0$
0%
None of these

Explanation

The position vector of the centroid of the triangle is

$\displaystyle \dfrac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } }{ 3 }$

Since the triangle is equilateral, therefore the orthocenter coincides with the centroid and hence

$\displaystyle \dfrac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } }{ 3 } =0$

$\therefore \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } =0$

$\alpha(\vec a \times \vec b)+\beta(\vec b \times \vec c)+\gamma(\vec c \times \vec {a})=\vec{0}$ and at least one of the scalars

$\alpha,\ \beta,\gamma$ is non-zero, then the vectors

$\vec{a},\vec{b},\vec{c}$ are

0%
Collinear
0%
Coplanar
0%
Non-coplanar
0%
Cannot be determined.

Explanation

As per the question, and the definition of linearly dependent vectors, we get that

$(\overline a \times \overline b), (\overline a \times \overline c)$ and

$(\overline b \times \overline c)$ are linearly dependent.
Now, since they are linearly dependent, any one of them can be written in terms of other two and thus we get that they are coplanar.

lf the four points

$\overline{a},\overline{b},\overline{c},\overline{d}$ are coplanar then

$[\overline{b}\overline{c}\overline{d}]+[\overline{c}\overline{a}\overline{d}]+[\overline{a}\overline{b}\overline{d}]=$

0%
0
0%
$[\overline{a},\overline{b},\overline{c}]$
0%
2 $[\overline{a},\overline{b},\overline{c}]$
0%
3 $[\overline{a},\overline{b},\overline{c}]$

A point

$O$ is the centre of a circle circumscribed about a triangle

$ABC$ , then

$\vec{OA}\sin 2A + \vec{OB}\sin 2B + \vec{OC} \sin 2C$ is equal to

0%
$(\vec{OA} + \vec{OB} + \vec{OC})\sin 2A$
0%
$3\vec{OG}$ , where $G$ is the centroid of triangle $ABC$
0%
$\vec 0$
0%
None of these

Explanation

The position vector of the point

$O$ with respect to itself is

$\displaystyle \dfrac{\vec{OA} \sin 2A + \vec{OB} \sin 2B + \vec{OC} \sin 2C}{\sin 2A + \sin 2B + \sin 2C}$

$\displaystyle \Rightarrow \dfrac{\vec{OA} \sin 2A + \vec{OB} \sin 2B + \vec{OC} \sin 2C}{\sin 2A + \sin 2B + \sin 2C} = \vec{0}$

or

$\vec{OA} \sin 2A + \vec{OB}\sin 2B + \vec{OC} \sin 2C = \vec{0}$

$\vec {a}=x\hat {i}+12\hat {j}-\hat {k},\vec {b}=2\hat {i}+2x\hat {j}+\hat {k}$ and

$\vec {c}=\hat {i}+\hat {k}$ and given that the vectors

$\vec {a},\vec {b},\vec {c}$ form a right handed system, then the range of

$x$ is

0%
$R-[-3,2]$
0%
$(-4,3)$
0%
$R-(-3,2)$
0%
$(-2,3)$

Explanation

For Right handed system

$[abc]\neq 0$

$[abc]=\begin{vmatrix}x &12 &-1 \\2 &2x &1 \\1 &0 &1 \end{vmatrix}$

$0\not\equiv 2x^{2}-12+2x$

$\neq x^{2}-6+x$

$0\neq (x+3)(x-2)$

$x=R-[-3,2]$

$G$ is the centroid of a

$\Delta ABC$ , then

$\vec{GA} + \vec{GB} + \vec{GC}$ is equal to

0%
$\vec{0}$
0%
$3\vec{GA}$
0%
$3\vec{GB}$
0%
$3\vec{GC}$

Explanation

We have

$\vec{GB}+\vec{GC} = (1+1) \vec{GD} = 2 \vec{GD}$ , where

$D$ is the midpoint of

$BC$ . Therefore,

$\vec{GA}+ \vec{GB} + \vec{GC} = \vec{GA} + 2\vec{GD}$

$= \vec{GA} - \vec{GA} =0$

$(\because G$ divides

$AC$ in the ration

$2 : 1, \therefore 2 \vec{GD} = - \vec{GA})$

In triangle ABC, which of the following is not true.

0%
$\overrightarrow{AB} + \overrightarrow{BC}+\overrightarrow{CA} = \overrightarrow{0}$
0%
$\overrightarrow{AB} + \overrightarrow{BC}- \overrightarrow{AC} = \overrightarrow{0}$
0%
$\overrightarrow{AB} + \overrightarrow{BC}- \overrightarrow{CA} = \overrightarrow{0}$
0%
$\overrightarrow{AB} - \overrightarrow{CB} + \overrightarrow{CA} = \overrightarrow{0}$

Explanation

${\textbf{Hint: Use Triangle Law of vector addition}}$

${\textbf{Step 1: Find the resultant of two vetors}}$

$\Rightarrow \overrightarrow {AB} + \overrightarrow {BC} = \overrightarrow {AC}$

${\textbf{Step 2: Solve the vector sum and Compare with the options}}$

$\overrightarrow {AB} + \overrightarrow {BC} - \overrightarrow {AC} = \overrightarrow 0$

${\Rightarrow}\overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CA} = \overrightarrow 0$

${\Rightarrow}\overrightarrow {AB} - \overrightarrow {CB} + \overrightarrow {CA} = \overrightarrow 0$

${\text{But }}$ $\overrightarrow {AB} + \overrightarrow {BC} - \overrightarrow {CA} \ne \overrightarrow 0$

${\textbf{Final Answer: Option (C) is incorrect}}$

Six vectors,

$\vec a$ to

$\vec f$ , all of magnitude 1 and directions indicated in the figure ( Consider all of them to be originating at origin ). Which of the following statement is true?

0%
$\vec b+\vec e=\vec f$
0%
$\vec b+\vec c=\vec f$
0%
$\vec d+\vec c=\vec f$
0%
$\vec d+\vec e=\vec f$

Explanation

We have, assuming unit magnitude for all vectors:

$\overrightarrow{a}= \overrightarrow{i}$

$\overrightarrow{b}= \overrightarrow{-j}$

$\overrightarrow{c}= \overrightarrow{i}$

$\overrightarrow{d}= \overrightarrow{j}$

$\overrightarrow{e}= \overrightarrow{-i}$

$\overrightarrow{f}= \overrightarrow{-i} + \overrightarrow{j}$

$\therefore \overrightarrow{d} + \overrightarrow{e} = \overrightarrow{j} + \overrightarrow{-i}$

If C is the mid point of AB and P is any point outside AB, then

0%
$\vec {\mathrm{P}\mathrm{A}}+\vec {\mathrm{P}\mathrm{B}}=2\vec {\mathrm{P}\mathrm{C}}$
0%
$\vec {\mathrm{P}\mathrm{A}}+\vec {\mathrm{P}\mathrm{B}}=\vec {\mathrm{P}\mathrm{C}}$
0%
$\vec {\mathrm{P}\mathrm{A}}+\vec {\mathrm{P}\mathrm{B}}+2\vec {\mathrm{P}\mathrm{C}}=0$
0%
$\vec {\mathrm{P}\mathrm{A}}+\vec {\mathrm{P}\mathrm{B}}=\vec {\mathrm{P}\mathrm{C}}=0$

Explanation

$\overrightarrow { PA } +\overrightarrow { AC } +\overrightarrow { CP } =0$

$\overrightarrow { PB } +\overrightarrow { BC } +\overrightarrow { CP } =0$
Adding we get

$\overrightarrow { PA } +\overrightarrow { PB } +\overrightarrow { AC } +\overrightarrow { BC } +2\overrightarrow { CP } =0$
Since

$\overrightarrow { AC } =-\overrightarrow { BC }$ and

$\overrightarrow { CP } =-\overrightarrow { PC }$

$\Rightarrow \overrightarrow { PA } +\overrightarrow { PB } -2\overrightarrow { PC } =0$

$\vec a$ is a non-zero vector of modulus

$a$ and

$m$ is a non-zero scalar, then

$m \vec a$ is a unit vector if

0%
$m = \pm 1$
0%
$a = |m|$
0%
$a = \dfrac{1}{|m|}$
0%
$a = \dfrac{1}{m}$

Explanation

$m\vec a$ is a unit vector if and only if

$|m \vec a| = 1$

$\Rightarrow |m||\vec a| = 1$

$\Rightarrow |m| a = 1$

$\Rightarrow \displaystyle \dfrac{1}{|m|}$

If vector

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

$\vec{b} = - 2\hat i + 2\hat j - \hat k$ , then ratio of Projection of

$\vec a$ on vector

$\vec b$ to Projection of

$\vec b$ on

$\vec a$ is equal to

0%
$\displaystyle \dfrac{3}{7}$
0%
$\displaystyle \dfrac{7}{3}$
0%
$3$
0%
$7$

Explanation

From the condition given we get

$=\dfrac{\dfrac{a.b}{|b|}}{\dfrac{b.a}{|a|}}$

Now we know that

$a.b=b.a$ ...(by property of scalar/dot product of vectors).

Hence the above expression simplifies as

$\dfrac{|a|}{|b|}$

$=\dfrac{7}{3}$

$'I'$ is the incentre of triangle of

$ABC$ whose corresponding sides are

$a, b, c,$ respectively,

$a\vec{IB} + b\vec{IB} + c\vec{IC}$ is always equal to

0%
$\vec{0}$
0%
$(a+b+c) \vec{BC}$
0%
$(\vec{a} + \vec{b} + \vec{c}) \vec{AC}$
0%
$(a + b + c)\vec{AB}$

Explanation

Let the incentre be at the origin and be

$A(\vec{p}), B(\vec{q})$ and

$C(\vec{r})$ ., then

$\vec{IA} = \vec{p}, \vec{IB} = \vec{q}$ and

$\vec{IC} = \vec{r}$ .

Incentre

$I$ is

$\displaystyle \dfrac{a\vec{p} + b\vec{q}+c\vec{r}}{a + b + c}$ , where

$p = BC, q = AC$ and

$r = AB$

Incentre is at the origin. Therefore,

$\displaystyle \dfrac{a \vec{p} + b\vec{q} + c\vec{r}}{a + b + c} = \vec{0}$ , or

$a\vec{p} + b\vec{q} + c\vec{r} = 0$

$\Rightarrow a\vec{IA} + b\vec{IB} + c\vec{IC} = \vec{0}$

Let

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

$\vec{c}$ be unit vectors such that

$\vec{a} + \vec{b} - \vec{c} = 0$ . If the area of triangle formed by vectors

$\vec{a}$ and

$\vec{b}$ is

$A$ , then what is the value of

$4A^2$ ?

0%
$3$
0%
$9$
0%
$\dfrac { 3 }{ 4 }$
0%
$\dfrac { 9 }{ 4 }$

Explanation

Given,

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

Now vector

$\vec{c}$ is along the diagonal of the parallelogram which has adjacent side vectors

$\vec{a}$ and

$\vec{b}$ .

Since

$\vec{c}$ is also a unit vector, triangle formed by vectors

$\vec{a}$ and

$\vec{b}$ is an equilateral triangle, then
Area of triangle

$A= \displaystyle \dfrac{\sqrt{3}}{4}$

$4A^2=4 \left(\dfrac{\sqrt{3}}{4} \right)^2=\dfrac{3}{4}$

In a trapezium, vector

$\vec{BC} = \alpha \vec{AD}$ . Also,

$\vec p = \vec{AC} + \vec{BD}$ is collinear with

$\vec{AD}$ and

$\vec p = \mu \vec{AD}$ , then which of the following is true?

0%
$\mu = \alpha + 2$
0%
$\mu + \alpha =1$
0%
$\alpha = \mu + 1$
0%
$\mu = \alpha +1$

Explanation

Let

$A$ be the origin.

$\vec BC = \vec c - \vec b = \alpha \vec d$ and

$\vec p = \vec{AC} + \vec{BD} = \mu AD$
Hence,

$\vec p = \vec c + \vec d - \vec b = \mu \vec d$
Using

$\vec c - \vec b = \alpha \vec d$

$\Rightarrow \alpha + 1 = \mu$

In triangle

$ABC$ ,

$\angle A = 30^o$ ,

$H$ is the orthocentre and

$D$ is the midpoint of

$BC$ . Segment

$HD$ is produced to

$T$ such that

$HD = DT$ . The length

$AT$ is equal to

0%
$2BC$
0%
$3BC$
0%
$\displaystyle \dfrac{4}{3}BC$
0%
None of these

Explanation

Let us assume that the circumcenter of

$\triangle ABC$ is

$O$ and is situated at the origin of the coordinate plane. Hence, the vector notation for all the vertices is

$\overrightarrow{OA}=\overrightarrow{a}$

$\overrightarrow{OB}=\overrightarrow{b}$

$\overrightarrow{OC}=\overrightarrow{c}$

Also, as

$O$ is the circumcenter of

$\triangle ABC$ ,

$\overrightarrow{a}=\overrightarrow{b}=\overrightarrow{c}= R$ (circum-radius)

Since

$D$ is mid-point of

$BC$ ,

$\overrightarrow{d}=\dfrac{\overrightarrow{b}+\overrightarrow{c}}{2}$

$\therefore \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{a}+2\overrightarrow{d}$

Also,

$2\overrightarrow{OD}=\overrightarrow{AH}$ as

$H$ is the orthocentre and it divides height in 2:1 ratio.

$\therefore \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{a}+2\overrightarrow{d}=\overrightarrow{OA}+\overrightarrow{AH} = \overrightarrow{OH}=\overrightarrow{h}\ \dots\ (1)$

According to given information,

$HD=DT$

$\implies \dfrac{\overrightarrow{b}+\overrightarrow{c}}{2}=\dfrac{\overrightarrow{h}+\overrightarrow{t}}{2}$

$\implies \dfrac{\overrightarrow{b}+\overrightarrow{c}}{2}=\dfrac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{t}}{2}$

$\implies \overrightarrow{a}=-\overrightarrow{t}$

$\implies \overrightarrow{AT}=2\overrightarrow{a}$

$\implies |\overrightarrow{AT}|=2|\overrightarrow{a}|=2R$

Also, from property of side-length of triangle,

$BC=2RsinA=R$ (

$\because\angle A = 30^{\circ}$ )

$\therefore AT=2BC$

$P(\vec{p})$ and

$Q(\vec{q})$ are the position vectors of two fixed points and

$R(\vec{r})$ is the position vector of a variable point. If

$R$ moves such that

$(\vec{r} - \vec{p}) \times (\vec{r} - \vec{q}) = \vec{0}$ , then the locus of

$R$ is

0%
A plane containing the origin $O$ and parallel to two non-collinear vectors $\vec{OP}$ and $\vec{OQ}$
0%
The surface of a sphere described on $PQ$ as its diameter
0%
A line passing through points $P$ and $Q$
0%
A set of lines parallel to line $PQ$

Explanation

$(\vec{r}-\vec{p})\times(\vec{r}-\vec{q})=0$

$\vec{r}\times(\vec{r}-\vec{q})-\vec{p}\times(\vec{r}-\vec{q})=0$

$\vec{r}\times\vec{r}-\vec{r}\times\vec{q}-\vec{p}\times\vec{r}+\vec{p}\times\vec{q}=0$

$\vec{p}\times\vec{q}-\vec{r}\times\vec{q}-\vec{p}\times\vec{r}=0$

$\vec{p}\times\vec{q}-\vec{r}\times\vec{q}+\vec{r}\times\vec{p}=0$

$\vec{p}\times\vec{q}+\vec{r}\times(\vec{p}-\vec{q})=0$

$\vec{p}\times\vec{q}=\vec{r}\times(\vec{q}-\vec{p})$

Therefore

$\vec{r}=\vec{p}+t(\vec{q}-\vec{p})$

This represents a line passing through P and Q.

$A, B, C$ and

$D$ have position vectors

$\vec{a}, \vec{b}, \vec{c}$ and

$\vec{d}$ , respectively, such that

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

0%
$AB$ and $CD$ bisect each other
0%
$AB$ and $CD$ trisect each other
0%
$BD$ and $AC$ bisect each other
0%
$BD$ and $AC$ trisect each other

Explanation

Given that

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

$\therefore \displaystyle \frac{\vec a + 2 \vec c}{2 + 1} = \frac{\vec b + 2 \vec d}{2 + 1}$

Hence,

$AC$ and

$BD$ trisect each other as

$L.H.S$ is the position vector of a point trisecting

$A$ and

$C$ , and

$R.H.S$ . that of

$B$ and

$D$ .

Let

$ABC$ be a triangle whose centroid is

$G$ , orthocentre is

$H$ and circumcentre is the origin '

$O$ '. If

$D$ is any point in the plane of the triangle such that no three of

$O, A, C$ and

$D$ are collinear satisfying the relation

$\vec{AD} + \vec{BD} + \vec{CH} + 3 \vec{HG} = \lambda \vec{HD}$ , then what is the value of the scalar

$'\lambda'$ ?

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

Explanation

We know that in vector

$\vec{AB}=\vec{OB}-\vec{OA}$ where O is the origin

SO from ques

$\vec{AD}+\vec{BD}+\vec{CH}+3\vec{HG}=\lambda\vec{HD}$

$\vec{OD}-\vec{OA}+\vec{OD}-\vec{OB}+\vec{OH}-\vec{OC}+3\vec{OG}-3\vec{OH}=\lambda\vec{HD}$

$2\vec{OD}-2\vec{OH}-\vec{OA}-\vec{OB}-\vec{OC}+3\vec{OG}=\lambda\vec{HD}$

$2(\vec{OD}-\vec{OH})-\vec{OA}-\vec{OB}-\vec{OC}+3\vec{OG}=\lambda\vec{HD}$

$2\vec{HD}-\vec{OA}-\vec{OB}-\vec{OC}+3\vec{OG}=\lambda\vec{HD}$

Comparing both sides

$\lambda=2$

The projection of the line segment joining the points A(-1, 0, 3) and B(2, 5, 1) on the line whose direction ratios are proportional to 6, 2, 3, is

0%
$\dfrac{10}{7}$
0%
$\dfrac{22}{7}$
0%
$\dfrac{18}{7}$
0%
none of these

Explanation

Given points of line segment

$A(-1,0,3)\ and\ B(2,5,1)$

direction ratio of line be

$R(2+1,5-0,1-3)=R(3,5,-2)$

$a_{1}=3,b_{1}=5,c_{1}=-2$

$a_{2}=6,b_{2}=2,c_{2}=3$

projection of lines

$l_{1}\ on\ l_{2} =\dfrac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}}$

$\Rightarrow projection=\dfrac{3\times6+5\times2+3\times(-2)}{\sqrt{6^2+2^2+3^2}}$

$\Rightarrow projection=\dfrac{18+10-6}{\sqrt{36+4+9}}$

$\Rightarrow projection=\dfrac{22}{\sqrt{49}}$

$\Rightarrow projection=\dfrac{22}{7}$

Let

$\vec {p}$ is the position vector of the orthocentre &

$\vec {g}$ is the position vector of the centroid of the triangle

$ABC$ where circumcentre is the origin. If

$\vec {p}= K\vec{g},$ then

$K=$

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

Explanation

$\vec p = \vec a + \vec b + \vec c$ as per the orthocentre definition.
Also, centroid

$\vec g = \dfrac{\vec a + \vec b + \vec c}{3}$ where,

$\vec a, \vec b, \vec c$ are the position vectors of points

$A, B, C$ of the triangle respectively.

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

$\vec {c}$ are three non- coplanar vectors, then the length of projection of vector

$\vec{a}$ in the plane of the vectors

$\vec{b}$ and

$\vec{c}$ may be given by

0%
$\dfrac{| \vec{a}.(\vec{b}\times \vec{c})|}{|\vec{b} \times \vec{c}|}$
0%
$\dfrac{| \vec{a}\times(\vec{b}\times \vec{c})|}{|\vec{b} \times \vec{c}|}$
0%
$\dfrac{[\vec{a} \vec{b} \vec{c}]}{(\vec{b}. \vec{c})}$
0%
none of these

Explanation

if the planes contains

$\vec{b}\ and\ \vec{c}$ then normal vector of plane

$\vec{b}\times\vec{c}$

projection formula

projection of

$\vec{a}$ on plane

$=\vec{a}\times\hat{(\vec{b}\times\vec{c})}$

$length=\dfrac{\left |\vec{a}\times(\vec{b}\times\vec{c})\right |}{\left |\vec{b}\times\vec{c}\right |}$

Five coplanar forces (each of magnitude

$20N$ ) are acting on a body. The angle between two neighboring forces have the same value. The resultant of these forces is necessarily equal to

0%
$20N$
0%
$20\sqrt{2}N$
0%
Zero
0%
none of these

Explanation

As the angle between any two neighbouring forces have the same value which is

$\dfrac{360^{\circ}}{5}=72^{\circ}$
and all the forces are of equal magnitude.
Hence, their resultant is zero.

If the scalar projection of the vector

$x\hat i - \hat j + \hat k$ on the vector

$2 \hat i - \hat j + 5\hat k$ is

$\dfrac{1}{\sqrt{30}}$ , then value of

$x$ is equal to

0%
$\dfrac{-5}{2}$ units
0%
$6$ units
0%
$- 6$ units
0%
$3$ units

Explanation

Projection of

$x\hat i -\hat j + \hat k$ on

$2\hat i - \hat j + 5\hat k$

$= \displaystyle \dfrac{(x\hat i - \hat j + \hat k) (2\hat i - \hat j + 5\hat k)}{\sqrt{4 + 1 + 25}} = \dfrac{2x + 1 + 5}{\sqrt{30}}$

But given,

$\displaystyle \dfrac{2x + 6}{\sqrt{30}} = \dfrac{1}{\sqrt{30}}$

$\Rightarrow 2x + 6 = 1$

$\displaystyle \Rightarrow x = \dfrac{-5}{2}$

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 4 - 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 4 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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