MHT CET · Maths · Vector Algebra
If \(\bar{a}=2 \hat{i}+3 \hat{j}+2 \hat{k}, \bar{b}=2 \hat{i}+\hat{j}-\hat{k}\) and \(\bar{c}=3 \hat{i}-\hat{j} \quad\) are such that \(\bar{a}+\lambda \bar{b}\) is perpendicular to \(\bar{c}\), then the value of \(\lambda\) is
- A \(\frac{-1}{5}\)
- B \(3\)
- C \(\frac{3}{5}\)
- D \(\frac{-3}{5}\)
Answer & Solution
Correct Answer
(D) \(\frac{-3}{5}\)
Step-by-step Solution
Detailed explanation
According to the given condition, we get
\((\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}) \cdot \overline{\mathrm{c}}=0 \)
\( \therefore {[(2+2 \lambda) \hat{\mathrm{i}}+(3+\lambda) \hat{\mathrm{j}}+(2-\lambda) \hat{\mathrm{k}}] \cdot(3 \hat{\mathrm{i}}-\hat{\mathrm{j}})=0} \)
\( \therefore 3(2+2 \lambda)-(3+\lambda)=0 \)
\( \therefore 6+6 \lambda-3-\lambda=0 \)
\( \therefore 3+5 \lambda=0 \)
\( \therefore \lambda=\frac{-3}{5}\)
\((\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}) \cdot \overline{\mathrm{c}}=0 \)
\( \therefore {[(2+2 \lambda) \hat{\mathrm{i}}+(3+\lambda) \hat{\mathrm{j}}+(2-\lambda) \hat{\mathrm{k}}] \cdot(3 \hat{\mathrm{i}}-\hat{\mathrm{j}})=0} \)
\( \therefore 3(2+2 \lambda)-(3+\lambda)=0 \)
\( \therefore 6+6 \lambda-3-\lambda=0 \)
\( \therefore 3+5 \lambda=0 \)
\( \therefore \lambda=\frac{-3}{5}\)
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