MHT CET · Maths · Vector Algebra
If \(\bar{a}=(2 \hat{i}+2 \hat{j}+3 \hat{k}), \bar{b}=(-\hat{i}+2 \hat{j}+\hat{k}) \quad\) and \(\overline{\mathrm{c}}=(3 \hat{\mathrm{i}}+\hat{\mathrm{j}})\) such that \((\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}})\) is perpendicular to \(\bar{c}\), then the value of \(\lambda\) is
- A -8
- B 8
- C 10
- D \(\frac{8}{3}\)
Answer & Solution
Correct Answer
(B) 8
Step-by-step Solution
Detailed explanation
\((\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}})=(2-\lambda) \hat{\mathrm{i}}+(2+2 \lambda) \hat{\mathrm{j}}+(3+\lambda) \hat{\mathrm{k}}\)
Since \(\bar{a}+\lambda \bar{b}\) is perpendicular to \(\bar{c}\)
\(\begin{aligned}
\therefore \quad & (\bar{a}+\lambda \bar{b}) \cdot \bar{c}=0 \\
& \Rightarrow(2-\lambda)(3)+(2+2 \lambda)(1)+(3+\lambda)(0)=0 \\
& \Rightarrow 6-3 \lambda+2+2 \lambda=0 \\
& \Rightarrow \lambda=8
\end{aligned}\)
Since \(\bar{a}+\lambda \bar{b}\) is perpendicular to \(\bar{c}\)
\(\begin{aligned}
\therefore \quad & (\bar{a}+\lambda \bar{b}) \cdot \bar{c}=0 \\
& \Rightarrow(2-\lambda)(3)+(2+2 \lambda)(1)+(3+\lambda)(0)=0 \\
& \Rightarrow 6-3 \lambda+2+2 \lambda=0 \\
& \Rightarrow \lambda=8
\end{aligned}\)
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