Question

Let a vector $\vec{a}$ be coplanar with vectors $\vec{b}=2\hat{i}+\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}+\hat{k}$. If $\vec{a}$ is perpendicular to $\vec{d}=3\hat{i}+2\hat{j}+6\hat{k}$, and $|\vec{a}|=\sqrt{10}$. Then a possible equal to:

• A. -42
• B. -40
• C. -29
• D. -38

Answer 1

Answer: (A)

$\vec{a}=\lambda \vec{b}+\mu \vec{c}=\hat{i}(2\lambda +\mu )+\hat{j}(\lambda -\mu )+\hat{k}(\lambda +\mu )$
$\vec{a}\cdot \vec{d}=0=3(2\lambda +\mu )+2(\lambda -\mu )+6(\lambda +\mu )$
$\Rightarrow 14\lambda +7\mu =0$
$\Rightarrow \mu =-2\lambda$
$\Rightarrow \vec{a}=(0)\hat{i}-3\lambda \hat{j}+(-\lambda )\hat{k}$
$\Rightarrow |\vec{a}|=\sqrt{10}|\lambda |=\sqrt{10}\Rightarrow |\lambda |=1$
$\Rightarrow \lambda =1$ or $-1$
$[\vec{a}\vec{b}\vec{c}]=0$
$[\vec{a}\vec{b}\vec{c}]+\left[\begin{array}{ccc}\vec{a}& \vec{b}& \vec{d}\end{array}\right]+\left[\begin{array}{ccc}\vec{a}& \vec{c}& \vec{d}\end{array}\right]=[\vec{a}\vec{b}+\vec{c}\vec{d}]$
$=\left|\begin{array}{ccc}0& -3\lambda & \lambda \\ 3& 0& 2\\ 3& 2& 6\end{array}\right|$
$=3\lambda (12)+\lambda (6)=42\lambda =-42$