Question

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ , $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$ be three vectors. $A$ vector $v$ in the plane of $a$ and $b$ , whose projection on $c$ is $\frac{1}{\sqrt{3}}$ is given by

• A. $3\hat{i}+\hat{j}-3\hat{k}$
• B. $3\hat{i}-\hat{j}+3\hat{k}$
• C. $3\hat{i}+\hat{j}+3\hat{k}$
• D. $-3\hat{i}+\hat{j}+3\hat{k}$

Answer 1

Answer: (B)

Since, $v$ is the coplanar to $a$ and $b$
$\therefore v=a+t{b}^{-1}$
$=(i+j+k)+t(i-j+k)$
$\Rightarrow r=(1+t)i+(1-t)j+(1+t)k.....$ (i)
$\Rightarrow r=(1+t)l+(1-t)j+(1+t)k$ (given)
$\Rightarrow \frac{v\cdot c}{|c|}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{|(1+t)1-1(1-t)-1(1+t)|}{\sqrt{3}}=\frac{1}{\sqrt{3}}$
$\Rightarrow 1+t-1+t-1-t=1$
$\Rightarrow t=2$
On putting the value of $t$ in Eq. (i), we get
$r=3i+(-1)j+(3)k$
$\Rightarrow v=3i-j+3k$