Question:
If \(\vec{a}=6 \hat{i}+2 \hat{j}-\lambda_{1} \hat{k}\), \(\vec{b}=3 \hat{i}+\left(3-\lambda_{2}\right) \hat{j}+2 \hat{k}\) and \(\vec{c}=2 \hat{i}+\left(1+\lambda_{3}\right) \hat{j}-3 \hat{k}\) are three vectors such that \(\vec{a}=3 \vec{c}\) and \(\vec{b}\) is perpendicular to \(\vec{c}\) then \(\left(\lambda_{1}, \lambda_{2}, \lambda_{3}\right)\) is
If \(\vec{a}=6 \hat{i}+2 \hat{j}-\lambda_{1} \hat{k}\), \(\vec{b}=3 \hat{i}+\left(3-\lambda_{2}\right) \hat{j}+2 \hat{k}\) and \(\vec{c}=2 \hat{i}+\left(1+\lambda_{3}\right) \hat{j}-3 \hat{k}\) are three vectors such that \(\vec{a}=3 \vec{c}\) and \(\vec{b}\) is perpendicular to \(\vec{c}\) then \(\left(\lambda_{1}, \lambda_{2}, \lambda_{3}\right)\) is
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Compare components of \(\vec{a}=3\vec{c}\) to get \(\lambda_1,\lambda_3\), then use \(\vec{b}\cdot\vec{c}=0\) for \(\lambda_2\).
Updated On: Nov 12, 2026
- \(\left(9, 21, -\frac{1}{3}\right)\)
- \(\left(9, 3, -\frac{1}{3}\right)\)
- \(\left(\frac{1}{3}, -\frac{1}{3}, \frac{1}{3}\right)\)
- \((3, 3, 1)\)
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The Correct Option is B
Solution and Explanation
Since \(\vec{a}=3\vec{c}\): \(6=3\cdot 2\) (true), \(2=3(1+\lambda_3)\Rightarrow \lambda_3=-\frac{1}{3}\), and \(-\lambda_1=-9\Rightarrow \lambda_1=9\).
\(\vec{b}\perp\vec{c}\Rightarrow \vec{b}\cdot\vec{c}=0\): \(3\cdot 2+(3-\lambda_2)(1+\lambda_3)+2(-3)=0\).
\(6+(3-\lambda_2)\cdot\frac{2}{3}-6=0\Rightarrow 3-\lambda_2=0\Rightarrow \lambda_2=3\).
So \((\lambda_1,\lambda_2,\lambda_3)=\left(9,3,-\frac{1}{3}\right)\).
Answer: (2) \(\left(9, 3, -\frac{1}{3}\right)\)
\(\vec{b}\perp\vec{c}\Rightarrow \vec{b}\cdot\vec{c}=0\): \(3\cdot 2+(3-\lambda_2)(1+\lambda_3)+2(-3)=0\).
\(6+(3-\lambda_2)\cdot\frac{2}{3}-6=0\Rightarrow 3-\lambda_2=0\Rightarrow \lambda_2=3\).
So \((\lambda_1,\lambda_2,\lambda_3)=\left(9,3,-\frac{1}{3}\right)\).
Answer: (2) \(\left(9, 3, -\frac{1}{3}\right)\)
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