Question

The area of the ${\Delta }^{le}$ whose vertices are the point $(1,2,3),(-2,1,-4),(3,4,-2)$ is

• A. $\frac{\sqrt{609}}{2}$
• B. $\sqrt{1218}$
• C. $\sqrt{609}$
• D. $\frac{\sqrt{1214}}{2}$

Answer 1

The correct option is D $\frac{\sqrt{1214}}{2}$

Given,

$A=(1,2,3),B=(-2,1,-4),C=(3,4,-2)$

$\Rightarrow AB=(-2-1)i+(1-2)j+(-4-3)k=-3i-j-7k$

$\Rightarrow AC=(3-1)i+(4-2)j+(-2-3)k=2i+2j-5k$

$AB\times AC=\left[\begin{array}{ccc}i& j& k\\ -3& -1& -7\\ 2& 2& -5\end{array}\right]$

$=i(5+14)-j(15+14)+k(-6+2)$

$=19i-29j-4k$

Area of triangle $=\frac{1}{2}(19i-29j-4k)$

$=\frac{1}{2}\sqrt{361+841+16}$

$=\frac{\sqrt{1214}}{2}$