The $△\mathrm{ABC}$is defined by the vertices $\mathrm{A}(1,-2,2)\mathrm{B}(1,4,0)$ and $\mathrm{C}(-4,1,1)$ . Let M be the foot of the altitude drawn from the vertex B to side A C .Then $\stackrel{-}{\mathrm{BM}}=$

 $\mathrm{A}(1,-2,2)\mathrm{B}(1,4,0)$ and $\mathrm{C}(-4,1,1)$

$\overline{AB}=6\hat{j}-2\hat{k},\overline{AC}=-5\hat{i}+3\hat{j}-\hat{k}$

Here $\stackrel{-}{\mathrm{MB}}=$ component vector of $\stackrel{-}{\mathrm{AB}}$ perpendicular$\overline{AC}$ $=\stackrel{-}{\mathrm{AB}}-\frac{(\stackrel{-}{\mathrm{AB}}\cdot \stackrel{-}{\mathrm{AC}})\stackrel{-}{\mathrm{AC}}}{{\left|\overline{\mathrm{AC}}\right|}^2}$ 

$$\begin{gathered} \overline{MB}=6\hat{j}-2\hat{k}-\frac{\left(0+18+2\right)\left(-5\hat{i}+3\hat{j}-\hat{k}\right)}{25+9+1} \\ \Rightarrow \overline{MB}=6\hat{j}-2\hat{k}-\frac{4\left(-5\hat{i}+3\hat{j}-\hat{k}\right)}{7}=\frac{7\left(6\hat{j}-2\hat{k}\right)-4\left(-5\hat{i}+3\hat{j}-\hat{k}\right)}{7} \\ \Rightarrow \overline{MB}=\frac{20\hat{i}+30\hat{j}-10\hat{k}}{7}\Rightarrow \overline{BM}=\left(\frac{-20}{7},\frac{-30}{7},\frac{10}{7}\right) \end{gathered}$$and hence find