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BITSAT 2010
Exam
BITSAT
Year
2010
Subject
Mathematics
Chapter
Three Dimensional Geometry

BITSAT 2010 · Mathematics · Three Dimensional Geometry

The perpendicular distance of P(1,2,3) from the line (...

The perpendicular distance of P(1,2,3)P(1,2,3) from the line (x−63=y−72=z−7−2\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}) is

Options

  1. A7
  2. B5
  3. C0
  4. D6

Correct answer

A. 7

Step-by-step solution

The point A(6,7,7))ison\mathrm{A}(6,7,7)) is on the line.Lettheperpendicularfrom(P)meetthelinein(L).Then(AP2=(6−1)2+(7−2)2+(7−3)2=66)Also(AL=)projectionofAPonlineline. Let the perpendicular from (\mathrm{P}) meet the line in (\mathrm{L}). Then (\mathrm{AP}^2=(6-1)^2+(7-2)^2+(7-3)^2=66) Also (\mathrm{AL}=) projection of AP on line(actual d.c.’s 317,217,−217)⇒(6−1)⋅317+(7−2)⋅217+(7−3)−217=17\begin{aligned}& \left(\text{actual d.c.'s } \frac{3}{\sqrt{17}}, \frac{2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}\right)\\ \Rightarrow & (6-1) \cdot \frac{3}{\sqrt{17}}+(7-2) \cdot \frac{2}{\sqrt{17}}+(7-3) \frac{-2}{\sqrt{17}}\\ & =\sqrt{17} \end{aligned}$$ (\therefore \perp) distance (\mathrm{d}) of (\mathrm{P}) from the line is given by (\mathrm{d}^2=\mathrm{AP}^2-\mathrm{AL}^2=66-17=49 \text{ so that } \mathrm{d}=7)
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