Question

If $\overrightarrow{a}=2\hat{i}+2\hat{j}+3\hat{k},\overrightarrow{b}=-\hat{i}+2\hat{j}+\hat{k}\mathrm{and}\overrightarrow{c}=3\hat{i}+\hat{j}$ are such that $\overrightarrow{a}+\lambda \overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$, then find the value of λ.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have} \\ \overrightarrow{a}=2\hat{i}+2\hat{j}+3\hat{k} \\ \overrightarrow{b}=-\hat{i}+2\hat{j}+\hat{k} \\ \mathrm{and} \\ \overrightarrow{c}=3\hat{i}+\hat{j} \\ \therefore \overrightarrow{a}+\lambda \overrightarrow{b}=2\hat{i}+2\hat{j}+3\hat{k}+\lambda \left(-\hat{i}+2\hat{j}+\hat{k}\right)=\left(2-\lambda \right)\hat{i}+\left(2+2\lambda \right)\hat{j}+\left(3+\lambda \right)\hat{k} \\ \text{Given that}\overrightarrow{a}+\lambda \overrightarrow{b}\text{is perpendicular to}\overrightarrow{c}\text{.} \\ \Rightarrow \left(\overrightarrow{a}+\lambda \overrightarrow{b}\right).\overrightarrow{c}=0 \\ \Rightarrow \left[\left(2-\lambda \right)\hat{i}+\left(2+2\lambda \right)\hat{j}+\left(3+\lambda \right)\hat{k}\right].\left(3\hat{i}+\hat{j}+0\hat{k}\right)=0 \\ \Rightarrow 3\left(2-\lambda \right)+1\left(2+2\lambda \right)+0=0 \\ \Rightarrow 6-3\lambda +2+2\lambda =0 \\ \therefore \lambda =8 \end{gathered}$$