JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\vec a = 2\hat i + {\lambda _1}\hat j + 3\hat k\), \(\vec b = 4\hat i + \left( {3 - {\lambda _2}} \right)\hat j + 6\hat k\) \(\vec c = 3\hat i + 6\hat j + \left( {{\lambda _3} - 1} \right)\hat k\) be three vectors such that \(\vec b = 2\vec a\) and \(\vec a\) is perpendicular to \(\vec c\). Then a possible value of \(\left( {{\lambda _1},{\lambda _2},{\lambda _3}} \right)\) is
- A \((1, 3, 1)\)
- B \(\left( {-\frac{1}{2},4, 0} \right)\)
- C \(\left( {\frac{1}{2},4, - 2} \right)\)
- D \((1, 5, 1)\)
Answer & Solution
Correct Answer
(B) \(\left( {-\frac{1}{2},4, 0} \right)\)
Step-by-step Solution
Detailed explanation
Because \({\rm{b}} = 2\vec a\) so \(3-\lambda_{2}=2 \lambda_{1}\) ...\((i)\) Because a is perpendicular to \(\mathrm{c}\) so \(6+6 \lambda_{1}+3\left(\lambda_{3}-1\right)=0\) .........\((ii)\)…
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