Question

If $\overrightarrow{a}=2\hat{i}-\hat{j}+\hat{k}$, $\overrightarrow{b}=\hat{i}+\hat{j}-2\hat{k}$ and $\overrightarrow{c}=\hat{i}+3\hat{j}-\hat{k}$, find λ such that $\overrightarrow{a}$ is perpendicular to $\lambda \overrightarrow{b}+\overrightarrow{c}$. [NCERT EXEMPLAR]

Answer 1

The given vectors are $\overrightarrow{a}=2\hat{i}-\hat{j}+\hat{k}$, $\overrightarrow{b}=\hat{i}+\hat{j}-2\hat{k}$ and $\overrightarrow{c}=\hat{i}+3\hat{j}-\hat{k}$.

Now,

$\lambda \overrightarrow{b}+\overrightarrow{c}=\lambda \left(\hat{i}+\hat{j}-2\hat{k}\right)+\left(\hat{i}+3\hat{j}-\hat{k}\right)=\left(\lambda +1\right)\hat{i}+\left(\lambda +3\right)\hat{j}-\left(2\lambda +1\right)\hat{k}$

It is given that

$$\begin{gathered} \overrightarrow{a}\perp \left(\lambda \overrightarrow{b}+\overrightarrow{c}\right) \\ \Rightarrow \overrightarrow{a}.\left(\lambda \overrightarrow{b}+\overrightarrow{c}\right)=0 \\ \Rightarrow \left(2\hat{i}-\hat{j}+\hat{k}\right).\left[\left(\lambda +1\right)\hat{i}+\left(\lambda +3\right)\hat{j}-\left(2\lambda +1\right)\hat{k}\right]=0 \\ \Rightarrow 2\left(\lambda +1\right)-\left(\lambda +3\right)-\left(2\lambda +1\right)=0 \\ \Rightarrow 2\lambda +2-\lambda -3-2\lambda -1=0 \\ \Rightarrow \lambda =-2 \end{gathered}$$

Thus, the value of λ is −2.