Question

The area of the triangle whose vertices are $A(1,2,3),B(2,-1,1)$ and $C(1,2,-4)$ is

• A. $7\sqrt{10}$sq units
• B. $\frac{1}{2}\sqrt{10}$sq units
• C. $\frac{7}{2}\sqrt{10}$sq units
• D. None of these

Answer 1

The correct option is C $\frac{7}{2}\sqrt{10}$sq units

Let $A_x{A}_y$ and $A_z$ be the area of projection of $\Delta ABC$ on yz, zx and xy-planes. Then,
$A_x=\frac{1}{2}\times$ Absolute values of $\left|\begin{array}{ccc}2& 3& 1\\ -1& 1& 1\\ 2& -4& 1\end{array}\right|=\frac{1}{2}\left(2\right(5)-3(-3)+1(2\left)\right)=\frac{21}{2}$
$A_y=\frac{1}{2}\times$ Absolute values of $\left|\begin{array}{ccc}3& 1& 1\\ 1& 2& 1\\ -4& 1& 1\end{array}\right|=\frac{1}{2}(3-1(5)+9)=\frac{7}{2}$
$A_z=\frac{1}{2}\times$ Absolute values of $\left|\begin{array}{ccc}1& 2& 1\\ 2& -1& 1\\ 1& 2& 1\end{array}\right|=\frac{1}{2}(-3-2+5)=0$
$\therefore$ Area of $\Delta ABC=\sqrt{{A_x}^2+{A_y}^2+{A_z}^2}$

$=\sqrt{\frac{441}{4}+\frac{49}{4}+0}=\frac{7}{2}\sqrt{10}$ sq units