question_answer

The perpendicular distance of P (1, 2, 3) form the lie \[\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}\] is

A) 7

B) 5

C) 0

D) 6

Correct Answer: A

Solution :

[a] The point \[A(6,7,7)\] is on the line. Let the perpendicular from P meet the line in L. then \[A{{P}^{2}}={{(6-1)}^{2}}+{{(7-2)}^{2}}+{{(7-3)}^{2}}=66\] Also AL=Projection of AP on line \[\left( actual\,\,\,d.c.'s\frac{3}{\sqrt{17}},\frac{2}{\sqrt{17}},\frac{-2}{\sqrt{17}} \right)\] \[\Rightarrow (6-1).\frac{3}{\sqrt{17}}+(7-2.)\frac{2}{\sqrt{17}}+(7-3)\frac{-2}{\sqrt{17}}=\sqrt{17}\] \[\therefore \,\,\,\bot \] Distance d of P from the line is given by \[{{d}^{2}}=A{{P}^{2}}-A{{L}^{2}}=66-17=49\] So that d=7