Question

The area of the triangle formed by the points whose position vectors are $3i+j,5i+2j+k,i-2j+3k$ is

• A. $\sqrt{23}$ sq units
• B. $\sqrt{21}$ sq units
• C. $\sqrt{29}$ sq units
• D. $\sqrt{33}$ sq units

Answer 1

The correct option is C $\sqrt{29}$ sq units

Let,

$A=3i+j$
$B=5i+2j+k$
$C=i-2j+3k$
So, $\overrightarrow{AB}=(5i+2j+k)-(3i+j)=2i+j+k$
$\overrightarrow{AC}=(i-2j+3k)-(3i+j)=-2i-3j+3k$
Now, area of $D=\frac{1}{2}\left|\begin{array}{ccc}i& j& k\\ 2& 1& 1\\ -2& -3& 3\end{array}\right|$
$=\frac{1}{2}|(3-(-3))i-(6-(-2))j+(-6-(-2)\left)k\right|$
$=\frac{1}{2}|6i-8j-4k|$
$=\frac{1}{2}\sqrt{36+64+16}$
$=\sqrt{29}$