Let a vector a be coplanar with vectors b=2 i+j+k and c=i-j+k . I… - Vidyadip

MCQ

Let a vector $\vec{a}$ be coplanar with vectors $\vec{b}=2 \hat{i}+\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}+\hat{k} .$ If $\vec{a}$ is perpendicular to $\vec{d}=3 \vec{i}+2 \hat{j}+6 \hat{k}$, and $|\vec{a}|=\sqrt{10} .$ Then a possible value of $[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]+[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \vec{d}]+[\overrightarrow{\mathrm{a}} \vec{c} \vec{d}]$ is equal to:

A
$-40$
✓
$-42$
C
$-29$
D
$-38$

✓

Answer

Correct option: B.

$-42$

b
$\vec{a}=\lambda \vec{b}+\mu \vec{c}=\hat{i}(2 \lambda+\mu)+\hat{j}(\lambda-\mu)+\hat{k}(\lambda+\mu)$

$\vec{a} \cdot \vec{d}=0=3(2 \lambda+\mu)+2(\lambda-\mu)+6(\lambda+\mu)$

$\Rightarrow 14 \lambda+7 \mu=0 \Rightarrow \mu=-2 \lambda$

$\Rightarrow \vec{a}=(0) \hat{i}-3 \lambda \hat{j}+(-\lambda) \hat{k}$

$\Rightarrow|\vec{a}|=\sqrt{10}|\lambda|=\sqrt{10} \Rightarrow|\lambda|=1$

$\Rightarrow \lambda=1 \text { or }-1$

${[\vec{a} \vec{b} \vec{c}]=0 \quad \text { (as vectors are coplanar) }}$

$[\vec{a} \vec{b} \vec{c}]+[\vec{a} \vec{b} \vec{d}]+[\vec{a} \vec{c} \vec{d}]=\left[\begin{array}{lll}\vec{a} & \vec{b}+\vec{c} & \vec{d}\end{array}\right]$

$=\left|\begin{array}{ccc}0 & -3 \lambda & \lambda \\ 3 & 0 & 2 \\ 3 & 2 & 6\end{array}\right|$

$=3 \lambda(12)+\lambda(6)=42 \lambda=-42$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 10. vector algebra→STD 12 - 10. vector algebra — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

Let $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}},\vec{\text{b}}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ and $\vec{\text{c}}=\text{c}_1\hat{\text{i}}+\text{c}_2\hat{\text{j}}+\text{c}_3\hat{\text{k}}$ be three zero vectors such that $\vec{\text{c}}$ is a unit vector perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}.$ If the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\frac{\pi}{6},$ then $\begin{vmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3 \end{vmatrix}^2$ is equal to:

Let $P Q R$ be a triangle with $R(-1,4,2)$. Suppose $M(2,1,2)$ is the mid point of $PQ$. The distance of the centroid of $\triangle \mathrm{PQR}$ from the point of intersection of the line $\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}$ and $\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}$ is

If $ a $ has magnitude $5$ and points north-east and vector $b $ has magnitude $ 5$ and points north-west, then $|\,\,a - b\,\,|\, = $

If $A = \left[ {\begin{array}{*{20}{c}}i&0\\0&i\end{array}} \right]$, then ${A^2} = $

If $A$ and $B$ are symmetric matrices of the same order, then:

$\int {\frac{{1 + {{\tan }^2}x}}{{1 - {{\tan }^2}x}}\,dx} $ equals to

Choose the correct answer in Exercise:
$\int^{\sqrt{3}}_{1}\frac{\text{dx}}{1+\text{x}^{2}}\text{equals}$

The distance of the point $(-3, 4, 5)$ from the origin:

He area bounded by $y = x^2, x = y^2$ is:

Let $f (x) = \frac{{2{{\sin }^2}x - 1}}{{\cos x}} + \frac{{\cos x(2\sin x + 1)}}{{1 + \sin x}}$ then $\int e^x (f(x)+f'(x))dx$ (where $c$ is the constant of integeration)