Question

Find the coordinates of the foot of the perpendicular drawn from the point (5, 4, 2) to the line $\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}.$ Hence, or otherwise, deduce the length of the perpendicular.

Answer 1

$$\begin{gathered} \text{Let}\mathit{\text{M}}\text{be the foot of the perpendicular of the point}\mathit{\text{P}}\text{(5, 4, 2) on the line}\frac{x+1}{2}\text{=}\frac{y-3}{3}\text{=}\frac{z-1}{-1} \\ \mathrm{Therefore},\mathrm{its}\mathrm{equation}\mathrm{is} \\ \frac{\mathrm{x}+1}{2}\text{=}\frac{\mathrm{y}-3}{3}\text{=}\frac{\mathrm{z}-1}{-1}\text{= r} \\ \text{Then,}\mathit{\text{M}}\text{is in the form}\left(2r-1,3r+3,-r+1\right) \\ \text{Direction ratios of MP are}2r-1-5,3r+3-4,-r+1-2\text{or}2r-6,3r-1,-r-1. \\ \text{Since MP is perpendicular to the given line (2, 3, -1),} \\ 2\left(2r-6\right)+3\left(3r-1\right)-1\left(-r-1\right)=0\text{(Because}{a}_1{a}_2+b_1{b}_2+c_1{c}_2=0\text{)} \\ \Rightarrow 4r-12+9r-3+r+1=0 \\ \Rightarrow 14r-14=0 \\ \Rightarrow r=1 \\ \text{So,}M=\left(2r-1,3r+3,-r+1\right)=\left(2\left(1\right)-1,3\left(1\right)+3,-1+1\right)=\left(1,6,0\right) \\ \text{Length of the perperndicular,}MP=\sqrt{{\left(1-5\right)}^2+{\left(6-4\right)}^2+{\left(0-2\right)}^2}=\sqrt{16+4+4}=\sqrt{24}=2\sqrt{6}\text{units} \end{gathered}$$