UPSC NDA Math Model Paper 9

Given, triangle whose vertices are at A = (3, -1, 2) $=\vec{3i}-\vec{j}+\vec{2k}$ $B = (1, -1, -3) =\vec{i}-\vec{j}-\vec{3k}$ and $C = (4, -3, 1) = \vec{4i}-\vec{3j}+\vec{k}$ Let A, B , and C be the vertices of the △ABC , then the area of that triangle $=\frac{1}{2}\times\left|\vec{AB}\times\vec{AC}\right|$ $\vec{AB}=(\vec{i}-\vec{j}-\vec{3k})-(\vec{3i}-\vec{j}+\vec{2k})$ $\Rightarrow\vec{AB}=-2\vec{i}+0\vec{k}-5\vec{k}$ $\vec{AC}=(\vec{4i}-\vec{3j}+\vec{k})-(\vec{3i}-\vec{j}+\vec{2k})$ $\Rightarrow\vec{AC}=\vec{i}-2\vec{j}-\vec{k}$ Now, $\vec{AB}\times\vec{AC}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -2 & 0 & -5 \\ 1 & -2 & -1 \end{vmatrix}$ $\Rightarrow\vec{AB}\times\vec{AC}=\vec{i}(-10)-\vec{j}(2+5)+\vec{k}(4-0)$ $\Rightarrow\vec{AB}\times\vec{AC}=-10\vec{i}-7\vec{j}+4\vec{k}$ $\Rightarrow\left|\vec{AB}\times\vec{AC}\right|=\sqrt{(-10)^{2}+(-7)^{2}+(4)^{2}}$ $\Rightarrow\left|\vec{AB}\times\vec{AC}\right|=\sqrt{165}$ The area of that triangle $=\frac{1}{2}\times\left|\vec{AB}\times\vec{AC}\right|$ $\Rightarrow$ The area of that triangle $=\frac{1}{2}\times\sqrt{165}$ Hence, the area of a triangle whose vertices are at (3, -1, 2), (1, -1, -3) and (4, -3, 1) is $\sqrt{165}/2$