Question

The area of the triangle whose sides are given by $2i-7j-k$ and $4j-3k$ is

• A. $\frac{\sqrt{725}}{2}sq.units$
• B. $172$ $sq$ units
• C. $174$ $sq$ units
• D. $12\sqrt{389}$ $sq$ units

Answer 1

The correct option is A $\frac{\sqrt{725}}{2}sq.units$

We have

$\overrightarrow{a}=2\hat{i}-7\hat{j}-\hat{k}$

$\overrightarrow{b}=4\hat{j}-3\hat{k}$

Area of triangle $=\frac{1}{2}\left|\overrightarrow{a}\times \overrightarrow{b}\right|$

Now,

$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -7& -1\\ 0& 4& -3\end{array}\right|$

$=\hat{i}\left(21+4\right)-\hat{j}\left(-6\right)+\hat{k}\left(8\right)$

$=25\hat{i}+6\hat{j}+8\hat{k}$

Now,

Area of triangle,

=$\frac{1}{2}$$\sqrt{625+36+64}$

=$\frac{\sqrt{725}}{2}sq.units$

Hence, the option $A$ is the correct answer.