Let a = 2 i + _1 j + 3 k, b = 4 i + ( 3 - _2 ) j + 6 k c = 3 i + … - Vidyadip

MCQ

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
$(1, 3, 1)$
✓
$\left( {-\frac{1}{2},4, 0} \right)$
C
$\left( {\frac{1}{2},4, - 2} \right)$
D
$(1, 5, 1)$

✓

Answer

Correct option: B.

$\left( {-\frac{1}{2},4, 0} \right)$

b
Because ${\rm{b}} = 2\vec a$ so $3-\lambda_{2}=2 \lambda_{1}$ ...$(i)$

Because a is perpendicular to $\mathrm{c}$ so $6+6 \lambda_{1}+3\left(\lambda_{3}-1\right)=0$ .........$(ii)$

$\Rightarrow\left(\lambda_{1}, \lambda_{2}, \lambda_{3}\right)=\left(\lambda_{1}, 3-2 \lambda_{1},-1-2 \lambda_{1}\right)$ where $\lambda_{1} \in R$

$\Rightarrow\left(-\frac{1}{2}, 4,0\right)$ satisfied above triplet.

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