ISRO-DEC2017-10

Question

If vectors $\vec{a}=2\hat{i}+\lambda \hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-2\hat{j}+3\hat{k}$ are perpendicular to each other, then value of $\lambda$ is

• A. $\dfrac{2}{5}$
   
• B. $2$
   
• C. $3$
   
• D. $\dfrac{5}{2}$

Answer 1

Dot product between two vectors $\vec{u}$ and $\vec{v}$ is given by

$\vec{u}.\vec{v}=|\vec{u}||\vec{v}|cos\theta $ where $\theta$ is the angle between both the vectors

If vectors are perpendicular then $\theta=90°$ and $cos90°=0$, i.e $\vec{u}.\vec{v}=0$

Given $\vec{u}=2i+\lambda j+k$ and $\vec{v}=i-2j+3k$

$\vec{u}.\vec{v}$= $2-2\lambda+3=0$

$5-2\lambda=0$

$\lambda= \frac{5}{2}$

Hence option D) is correct

Answer 2

The dot product v.w = 0 when v is perpendicular to w

$\vec{a} = (2, \lambda, 1)$
$\vec{b} = (1, -2, 3)$

$\vec{a}.\vec{b}$ = $2 - 2$$\lambda$$+3$ $  = 0$

$2 - 2$$\lambda$$+3$ $  = 0$

$ \lambda = \large \frac{5}{2} $