MHT CET · Maths · Vector Algebra
If vectors \(\bar{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \bar{b}=-\hat{i}+2 \hat{j}+\hat{k}\) and \(\bar{c}=-3 \hat{i}+\hat{j}+2 \hat{k}\) are such that, \(\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}\) is perpendicular to \(\overline{\mathrm{c}}\), then \(\lambda=\)
- A -14
- B 14
- C 2
- D -2
Answer & Solution
Correct Answer
(A) -14
Step-by-step Solution
Detailed explanation
From given data, we write
\(
\bar{a}+\lambda \bar{b}=(2-\lambda)
\hat{i}+(2+2 \lambda)
\hat{j}+(3+\lambda)
\hat{k}
\)
Since (1) is \(\perp\) er to \(\bar{c}\), we write
\(
\begin{aligned}
& (2-\lambda)(3)+(2+2 \lambda)(1)+(3+\lambda)(2)=0 \\
& \therefore 6-3 \lambda+2+2 \lambda+6+2 \lambda=0 \Rightarrow \lambda=-14
\end{aligned}
\)
\(
\bar{a}+\lambda \bar{b}=(2-\lambda)
\hat{i}+(2+2 \lambda)
\hat{j}+(3+\lambda)
\hat{k}
\)
Since (1) is \(\perp\) er to \(\bar{c}\), we write
\(
\begin{aligned}
& (2-\lambda)(3)+(2+2 \lambda)(1)+(3+\lambda)(2)=0 \\
& \therefore 6-3 \lambda+2+2 \lambda+6+2 \lambda=0 \Rightarrow \lambda=-14
\end{aligned}
\)
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