MHT CET · Maths · Vector Algebra
If the vector \(\bar{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \bar{b}=-\hat{i}+2 \hat{j}+\hat{k}\) and \(\bar{c}=3 \hat{i}+\hat{j}\) are such that \((\bar{a}+\lambda \bar{b})\) is perpendicular to \(\bar{c}\), then the value of \(\lambda\) is
- A -8
- B 10
- C 8
- D \(\frac{8}{3}\)
Answer & Solution
Correct Answer
(C) 8
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \because \vec{a}+\lambda \vec{b} \perp \vec{c} \\ & \Rightarrow(\vec{a}+\lambda \vec{b}) \cdot \vec{c}=0 \\ & \Rightarrow\{(2 \widehat{i}+2 \widehat{j}+3 \widehat{k})+\lambda(-\hat{i}+2 \hat{j}+\hat{k})\} \cdot(3 \hat{i}+\hat{j})=0 \\ & \Rightarrow\{(2-\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(3+\lambda) \widehat{k}\} \cdot(3 \hat{i}+\hat{j})=0 \\ & \Rightarrow(2-\lambda) \times 3+(2+2 \lambda) \times 1+(3+\lambda) \times 0=0 \\ & \Rightarrow \lambda=8\end{aligned}\)
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