Question

Find the length and the foot of the perpendicular from the point (1, 1, 2) to the plane $\overrightarrow{r}·\left(\hat{i}-2\hat{j}+4\hat{k}\right)+5=0.$

Answer 1

$$\begin{gathered} \text{Let}\mathit{\text{M}}\text{be the foot of the perpendicular of the point}\mathit{\text{P}}\text{(1, 1, 2) in the plane}\overrightarrow{r}\text{.}\left(\hat{i}-2\hat{j}+4\hat{k}\right)\text{+ 5 = 0 or}x-2y+4z+5=0. \\ \text{Then,}\mathit{\text{PM}}\text{\hspace{0.17em}is the normal to the plane. So, the direction ratios of}\mathit{\text{PM}}\text{are proportional to 1, -2, 4.} \\ \text{Since}\mathit{\text{PM}}\text{\hspace{0.17em}passes through}\mathit{\text{P}}(1,\; 1,\; 2)\text{and has direction ratios proportional to 1},-2,4\text{equation of}\mathit{\text{PQ}}\text{\hspace{0.17em}is} \\ \frac{x-1}{1}=\frac{y-1}{-2}=\frac{z-2}{4}=r\text{(say)} \\ \text{Let the coordinates of}\mathit{\text{M}}\text{be}\left(r+1,-2r+1,4r+2\right)\text{.} \\ \text{Since}\mathit{\text{M}}\text{lies in the plane}x-2y+4z+5=0, \\ x-2y+4z+5=0 \\ \Rightarrow r+1+4r-2+16r+8+5=0 \\ \Rightarrow 21r+12=0 \\ \Rightarrow r=\frac{-12}{21}=\frac{-4}{7} \\ \text{Substituting this in the coordinates of}\mathit{\text{M}}\text{, we get} \\ M=\left(r+1,-2r+1,4r+2\right)=\left(\left(\frac{-4}{7}\right)+1,-2\left(\frac{-4}{7}\right)+1,4\left(\frac{-4}{7}\right)+2\right)=\left(\frac{3}{7},\frac{15}{7},\frac{-2}{7}\right) \\ \text{Now, the length of the perpendicular from}\mathit{\text{P}}\text{onto the given plane} \\ =\frac{\left|\left(1\right)-2\left(1\right)+4\left(2\right)+5\right|}{\sqrt{1+4+16}} \\ =\frac{12}{\sqrt{21}}\text{units} \end{gathered}$$

Disclaimer: The answer given for this problem in the text book is incorrect.