Question

The area of the triangle formed by the point $A(2,3,4),B(3,4,2)$ and $C(4,2,3)$ is?

• A. $3\sqrt{3}$
• B. $\frac{3\sqrt{3}}{2}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{5\sqrt{3}}{2}$

Answer 1

The correct option is B $\frac{3\sqrt{3}}{2}$

Direction ratio of $AB$ = $(1,1,-2)$

$a_1,b_1,c_1$

Direction ratio of $ABC$ = $(1,1,-2)$

$a_2,b_2,c_2$

[$\because a_1,b_1,c_1$ are direction ratio]

Area of $\Delta ABC$ = $\frac{1}{2}\sqrt{\sum {\left(b_1{c}_1-b_2{c}_1\right)}^2}$

$\begin{array}{c}=\frac{1}{2}\sqrt{{\left(1-4\right)}^2+{\left(-2-1\right)}^2+{\left(-2-1\right)}^2}\\ =\frac{1}{2}\sqrt{9+9+9}\\ =\frac{3\sqrt{3}}{2}\end{array}$