MCQ
Let $\vec a\, = \,\hat i\, + \,2\hat j\, + 4\hat k\,,\,\vec b\, = \,\hat i\, + \,\lambda \hat j\, + 4\hat k$ and $\vec c\, = \,2\hat i\, + \,4\hat j\, + ({\lambda ^2} - 1)\hat k$ be coplanar vectors. Then the non -zero vector $\vec a\times \vec c$ is
  • A
    $ - 10\,\,\hat i\, - 5\,j$
  • B
    $ - 14\,\,\hat i\, - 5\,j$
  • C
    $ - 14\,\,\hat i\, + 5\,j$
  • $ - 10\,\,\hat i\, + 5\,j$

Answer

Correct option: D.
$ - 10\,\,\hat i\, + 5\,j$
d
$\left[\begin{array}{lll}{\vec{a}} & {\vec{b}} & {\vec{c}}\end{array}\right]=0$

$\left|\begin{array}{lll}{1} & {2} & {4} \\ {1} & {\lambda} & {4} \\ {2} & {4} & {\lambda^{2}-1}\end{array}\right|_{R_{3} \rightarrow R_{3}-2 R_{1}}=0$

$\left|\begin{array}{ccc}{1} & {2} & {4} \\ {1} & {\lambda} & {4} \\ {0} & {0} & {\lambda^{2}-9}\end{array}\right|=0$

$\left(\lambda^{2}-9\right)(\lambda-2)=0$

$\Rightarrow \lambda=2 \quad$ OR $\quad \lambda^{2}=5$

$\vec{a} \times \vec{c}=\left|\begin{array}{lll}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {1} & {2} & {4} \\ {2} & {4} & {\lambda^{2}-1}\end{array}\right|$

$=\hat{i}\left(2 \lambda^{2}-2-16\right)-\hat{j}\left(\lambda^{2}-1-8\right)$

$\left.=\left(\lambda^{2}-9\right)(2 \hat{i}-\hat{j})=-5(2 \hat{i}-\hat{j}) \quad \text { (put } \lambda=2\right)$

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