Three Dimensional Geometry Question 10
Question: The perpendicular distance of P (1, 2, 3) form the lie $ \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2} $ is
Options:
A) 7
B) 5
C) 0
D) 6
Show Answer
Answer:
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 $ AP^{2}={{(6-1)}^{2}}+{{(7-2)}^{2}}+{{(7-3)}^{2}}=66 $ Also AL=Projection of AP on line $ ( actual,d.c.’s\frac{3}{\sqrt{17}},\frac{2}{\sqrt{17}},\frac{-2}{\sqrt{17}} ) $
$ \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}=AP^{2}-AL^{2}=66-17=49 $ So that d=7
BETA
