Question

Let $m$ be a vector of magnitude $\sqrt{3}$ and perpendicular to the vectors $\hat{ i }+\hat{ j }$ and $\hat{ j }-\hat{ k }$ Let $n$ be another vector of magnitude $2 \sqrt{6}$ and perpendicular to the vectors $2 \hat{ i }-\hat{ j }$ and $\hat{ j }+2 \hat{ k }$. The area (in sq. units) of the triangle formed with $m$ and $n$ as sides is

• A. $\sqrt{2}$
• B. $\sqrt{6}$
• C. $2 \sqrt{3}$
• D. $3 \sqrt{2}$

Answer 1

Answer: (D)

Given,
$ m =\sqrt{3} \times \text { unit vectors of }[(\hat{ i }+\hat{ j }) \times(\hat{ j }-\hat{ k })] $
$\because(\hat{ i }+\hat{ j }) \times(\hat{ j }-\hat{ k })=\begin{bmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & 1 & 0 \\ 0 & 1 & -1\end{bmatrix}$
$ =\hat{ i }(-1-0)-\hat{ j }(-1-0)+\hat{ k }(1) $
$=-\hat{ i }+\hat{ j }+\hat{ k }$
$\therefore m =\sqrt{3} \times \frac{-\hat{ i }+\hat{ j }+\hat{ k }}{\sqrt{1+1+1}} $
$=\sqrt{3} \times \frac{-\hat{ i }+\hat{ j }+\hat{ k }}{\sqrt{3}}=-\hat{ i }+\hat{ j }+\hat{ k } $
and $n =2 \sqrt{6} \times$ unit vectors of $[(2 \hat{ i }-\hat{ j }) \times(\hat{ j }+2 \hat{ k })]$
$\therefore (\hat{ i }-\hat{ j }) \times(\hat{ j }+2 \hat{ k })$
$=\begin{bmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & -1 & 0 \\ 0 & 1 & 2\end{bmatrix}$
$=\hat{ i }(-2-0)-\hat{ j }(4-0)+\hat{ k }(2+0)=-2 \hat{ i }-4 \hat{ j }+2 \hat{ k }$
$n =2 \sqrt{6} \times \frac{-2 \hat{ i }-4 \hat{ j }+2 \hat{ k }}{\sqrt{4+16+4}}=2 \sqrt{6} \times \frac{-2 \hat{ i }-4 \hat{ j }+2 \hat{ k }}{2 \sqrt{6}}$
$=-2 \hat{ i }-4 \hat{ j }+2 \hat{ k }$
Now, required area
$=\frac{1}{2}( m \times n )=\frac{1}{2}|(-\hat{ i }+\hat{ j }+\hat{ k }) \times(-2 \hat{ i }-4 \hat{ j }+2 \hat{ k })|$
$=\frac{1}{2}\begin{bmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ -1 & 1 & 1 \\ -2 & -4 & 2\end{bmatrix}$
$=\frac{1}{2} \times \mid \hat{ i }(2+4)-\hat{ j }(-2+2)+\hat{ k }(4+2)|$
$=\frac{1}{2} \times|6 \hat{ i }+6 \hat{ k }|=\frac{1}{2} \sqrt{6^{2}+6^{2}}=\frac{1}{2} \times 6 \times \sqrt{2}=3 \sqrt{2}$