Question

Let $\overrightarrow{a}=\hat{i}+2\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{c}=\hat{i}+\hat{j}-\hat{k}$.The vector in the plane of $\overrightarrow{a}$ and $\overrightarrow{b}$ , whose projection on C is

• A. $-2\hat{i}+5\hat{j}-2\hat{k}$
• B. $3\hat{i}+\hat{j}-3\hat{k}$
• C. $2\hat{i}+\hat{j}+2\hat{k}$
• D. $4\hat{i}+\hat{j}-4\hat{k}$

Answer 1

Answer: (C)

$2\hat{i}+\hat{j}+2\hat{k}$

$\overline{a}=\overline{i}+2\overline{j}+\overline{k}$

$\overline{b}=\overline{i}-\overline{j}+\overline{k}$

$\overline{c}=\overline{i}+\overline{j}-\overline{k}$

The value in plane of $\overline{a}$ and $\overline{b}$ where projection on C

$\frac{\left[\right(\overline{a}+\lambda \overline{b}\left)\overline{c}\right]}{\left|\overline{c}\right|}=\frac{1}{\sqrt{5}}$

$\frac{[\hat{i}2\hat{j}+\hat{k}+\lambda (\hat{i}-\hat{j}+\hat{k}]\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}=\frac{1}{\sqrt{3}}$

$\Rightarrow \lambda +1+2-\lambda -\lambda +1=1$

$\Rightarrow \lambda =3$

Required equation is

$2\hat{i}+\hat{j}+2\hat{k}$

OPTION : C