Question:

Let \(\vec{a}=2\hat{i}+\lambda_1\hat{j}+3\hat{k}\), \(\vec{b}=4\hat{i}+(3-\lambda_2)\hat{j}+6\hat{k}\) and \(\vec{c}=3\hat{i}+6\hat{j}+(\lambda_3-1)\hat{k}\) be three vectors such that \(\vec{b}=2\vec{a}\) and \(\vec{a}\) is perpendicular to \(\vec{c}\). Then a possible value of \((\lambda_1,\lambda_2,\lambda_3)\) is :

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\(\vec b=2\vec a\) gives \(3-\lambda_2=2\lambda_1\); \(\vec a\cdot\vec c=0\) gives a second relation. Test the options.
Updated On: Nov 5, 2026
  • (1, 5, 1)
  • (1, 3, 1)
  • \(\left(-\frac{1}{2},4,0\right)\)
  • \(\left(\frac{1}{2},4,-2\right)\)
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The Correct Option is C

Solution and Explanation

\(\vec b=2\vec a\): \(3-\lambda_2=2\lambda_1\).
\(\vec a\cdot\vec c=0\): \(2(3)+\lambda_1(6)+3(\lambda_3-1)=0\Rightarrow 2\lambda_1+\lambda_3+1=0\).
Option (3): \(\lambda_1=-\tfrac12,\lambda_2=4,\lambda_3=0\): \(3-4=-1=2(-\tfrac12)\) true, and \(-1+0+1=0\) true.
Option (1): \(3-5=-2\ne2\). Option (2): \(0\ne2\). Option (4): \(3-4=-1\ne1\).
Answer: (3) \(\left(-\frac{1}{2},4,0\right)\)
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