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