Question
The perpendicular distance of $$P\left( {1,\,2,\,3} \right)$$ from the line $$\frac{{x - 6}}{3} = \frac{{y - 7}}{2} = \frac{{z - 7}}{{ - 2}}$$ is :
A.
$$7$$
B.
$$5$$
C.
$$0$$
D.
$$6$$
Answer :
$$7$$
Solution :
The point $$A\left( {6,\,7,\,7} \right)$$ is on the line . Let the perpendicular from $$P$$ meet the line in $$L$$. Then
$$A{P^2} = {\left( {6 - 1} \right)^2} + {\left( {7 - 2} \right)^2} + {\left( {7 - 3} \right)^2} = 66$$
$$\eqalign{ & {\text{Also }}AL = {\text{ projection of }}AP\,{\text{on line}} \cr & \left( {{\text{actual d}}{\text{.c}}{\text{.'s }}\frac{3}{{\sqrt {17} }},\,\frac{2}{{\sqrt {17} }},\,\frac{{ - 2}}{{\sqrt {17} }}} \right) \cr & \Rightarrow \left( {6 - 1} \right).\frac{3}{{\sqrt {17} }} + \left( {7 - 2} \right).\frac{2}{{\sqrt {17} }} + \left( {7 - 3} \right).\frac{{ - 2}}{{\sqrt {17} }} = \sqrt {17} \cr & \therefore \, \bot \,{\text{distance }}d{\text{ of }}P{\text{ from the line is given by}} \cr & {d^2} = A{P^2} - A{L^2} \cr & \Rightarrow {d^2} = 66 - 17 \cr & \Rightarrow {d^2} = 49 \cr & {\text{So, that }}d = 7 \cr} $$
The point $$A\left( {6,\,7,\,7} \right)$$ is on the line . Let the perpendicular from $$P$$ meet the line in $$L$$. Then
$$A{P^2} = {\left( {6 - 1} \right)^2} + {\left( {7 - 2} \right)^2} + {\left( {7 - 3} \right)^2} = 66$$
$$\eqalign{ & {\text{Also }}AL = {\text{ projection of }}AP\,{\text{on line}} \cr & \left( {{\text{actual d}}{\text{.c}}{\text{.'s }}\frac{3}{{\sqrt {17} }},\,\frac{2}{{\sqrt {17} }},\,\frac{{ - 2}}{{\sqrt {17} }}} \right) \cr & \Rightarrow \left( {6 - 1} \right).\frac{3}{{\sqrt {17} }} + \left( {7 - 2} \right).\frac{2}{{\sqrt {17} }} + \left( {7 - 3} \right).\frac{{ - 2}}{{\sqrt {17} }} = \sqrt {17} \cr & \therefore \, \bot \,{\text{distance }}d{\text{ of }}P{\text{ from the line is given by}} \cr & {d^2} = A{P^2} - A{L^2} \cr & \Rightarrow {d^2} = 66 - 17 \cr & \Rightarrow {d^2} = 49 \cr & {\text{So, that }}d = 7 \cr} $$