MCQ
Let $\vec{b}=\hat{i}+\hat{j}+\lambda \hat{k}, \lambda \in R$. If $\vec{a}$ is a vector such that $\overrightarrow{ a } \times \overrightarrow{ b }=13 \hat{ i }-\hat{ j }-4 \hat{ k } \quad$ and $\quad \overrightarrow{ a } \cdot \overrightarrow{ b }+21=0$, then $(\vec{b}-\vec{a}) \cdot(\hat{k}-\hat{j})+(\vec{b}+\vec{a}) \cdot(\hat{i}-\hat{k})$ is equal to
  • A
    $36$
  • B
    $22$
  • $14$
  • D
    $19$

Answer

Correct option: C.
$14$
c
$(\overrightarrow{ a } \times \overrightarrow{ b }) \cdot \overrightarrow{ b }=0$

$\Rightarrow 13-1-4 \lambda=0 \Rightarrow \lambda=3$

$\Rightarrow \overrightarrow{ b }=\hat{ i }+\hat{ j }+3 \hat{ k } \Rightarrow \overrightarrow{ a } \times \overrightarrow{ b }=13 \hat{ i }-\hat{ j }-4 \hat{ k }$

$\Rightarrow(\overrightarrow{ a } \times \overrightarrow{ b }) \times \overrightarrow{ b }=(13 \hat{ i }-\hat{ j }-4 \hat{ k }) \times(\hat{ i }+\hat{ j }+3 \hat{ k })$

$\Rightarrow-21 \overrightarrow{ b }-11 \overrightarrow{ a }=\hat{ i }-43 \hat{ j }+14 \hat{ k }$

$\Rightarrow \overrightarrow{ a }=-2 \hat{ i }+2 \hat{ j }-7 \hat{ k }$

Now $(\overrightarrow{ b }-\overrightarrow{ a }) \cdot(\hat{ k }-\hat{ j })+(\overrightarrow{ b }+\overrightarrow{ a }) \cdot(\hat{ i }-\hat{ k })=14$

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

Similar questions

The graphs of sine and cosine functions, intersect each other at a number of points and between two consecutive points of intersection, the two graphs enclose the same area $A$. Then $A ^{4}$ is equal to ............
The function $\text{f}(\text{x})=\cot^{-1}\text{x}+\text{x}$ increases in the interval:
$\int\frac{\sin^2\text{x}-\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{dx}$ is equal to:
If $a, b, c $ are three non-coplanar vectors such that $a + b + c = \alpha \,d$ and $b + c + d = \beta \,a,$ then $a + b + c + d$ is equal to
Area enclosed by the curve $y = f(x)$ that is being defined parametrically as $x = \frac{{1 - {t^2}}}{{1 + {t^2}}},\,y = \frac{{2t}}{{1 + {t^2}}}$ (where $t \in R$ ) is equal to
If matrix $\text{A}=\big[\text{a}_{\text{ij}}\big]_{2\times2'}$ where $\text{a}_\text{ij}=\begin{cases}1,&\text{if }\text{i }\neq\text{j}\\0,&\text{if }\text{i }=\text{j}\end{cases},$ then $A^2$ is equal to:
Two adjacent sides of a parallelogram $\mathrm{ABCD}$ are given by $\overrightarrow{A B}=2 \hat{i}+10 \hat{j}+11 \hat{k}$ and $\overrightarrow{A D}=-\hat{i}+2 \hat{j}+2 \hat{k}$. The side $\mathrm{AD}$ is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that $\mathrm{AD}$ becomes $\mathrm{AD}^{\prime}$. If $\mathrm{AD}^{\prime}$ makes a right angle with the side $\mathrm{AB}$, then the cosine of the angle $\alpha$ is given by
The point on the curve $y ^2= x$ where tangent makes $45^{\circ}$ angle with $x -$ axis is :
What is integrating factor of $\frac{\text{dy}}{\text{dx}}+\text{y}\sec\text{x}=\tan\text{x}?$
The curve $y-x:$