Question

Let $\overrightarrow{a}=2\hat{i}-\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}+2\hat{j}-\hat{k}$ and $\overrightarrow{c}=\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k}$ be three vectors. A vector of the type $\overrightarrow{b}+\lambda \overrightarrow{c}$ for some scalar $\lambda$, whose projection on $\overrightarrow{a}$ is of magnitude $\sqrt{\frac{2}{3}}$. Thenthe value of $\lambda$ is

• A. $1$
  
• B. $0$
• C. $-1$
• D. $2$
  

Answer 1

The correct option is C $-1$

Consider the given vectors,

$\overrightarrow{a}=2\hat{i}-\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}+2\hat{j}-\hat{k},\overrightarrow{c}=\hat{i}+\hat{j}-2\overrightarrow{k}$

Given that projection of $\overrightarrow{b}+\lambda \overrightarrow{c}$ on $\overrightarrow{a}$ =$\sqrt{\frac{2}{3}}$

Hence,

$\frac{[\overrightarrow{b}+\lambda \overrightarrow{c}].\overrightarrow{a}}{\left|\overrightarrow{a}\right|}$ =$\sqrt{\frac{2}{3}}$

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

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

$\frac{(1+\lambda ).2+(2+\lambda )(-1)+(1-2\lambda )\left(1\right)}{\sqrt{6}}=\sqrt{\frac{2}{3}}$

$2+2\lambda -2-\lambda +1-2\lambda =\sqrt{\frac{2}{3}}.\sqrt{6}$

$-\lambda +1=2$

$-\lambda =1$

$\lambda =-1$

Hence, this is the answer.