question_answer

The distance of point A(-2, 3, 1) from the line PQ through P(-3, 5, 2), which make equal angles with the axes is -

A) \[2/\sqrt{3}\]           

B)        \[\sqrt{14/3}\]

C) \[16/\sqrt{3}\]           

D)        \[5/\sqrt{3}\]

Correct Answer: B

Solution :

[b] \[A\left( -2,3,1 \right)\] Equation of line PQ \[\Rightarrow \frac{x+3}{1}=\frac{y-5}{1}=\frac{z-2}{1}=\lambda \] \[\Rightarrow Q\left\{ \begin{align}   & {{x}_{1}}=x=\lambda -3 \\  & {{y}_{1}}=y=\lambda +5 \\  & {{z}_{1}}=z=\lambda +2 \\ \end{align} \right\}\] \[AQ\bot \] line PQ \[\Rightarrow \left\{ \begin{align}   & 1(\lambda -3+2)+1\left( \lambda +5-3 \right)+1\left( \lambda +2-1 \right)=0 \\  & 3\lambda +2=0\,\,\,\,\,\,\,\,\,\Rightarrow \lambda =\frac{-2}{3} \\ \end{align} \right.\] \[Q({{x}_{1}},{{y}_{1}},{{z}_{1}})=\left( \frac{-11}{3},\frac{13}{3},\frac{4}{3} \right)\] \[AQ=\sqrt{{{\left( \frac{-11}{3}+2 \right)}^{2}}+{{\left( \frac{13}{3}-3 \right)}^{2}}+{{\left( \frac{4}{3}-1 \right)}^{2}}}\] \[=\sqrt{\frac{25}{9}+\frac{16}{9}+\frac{1}{9}}=\sqrt{\frac{42}{9}}=\sqrt{\frac{14}{3}}\]