First slide

Straight lines in 3D

Question

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

Easy

A

7
B

5
C

0
D

6

Solution

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 (actual d.c’s _{$\frac{3}{\sqrt{17}}, \frac{2}{\sqrt{17}}, \frac{- 2}{\sqrt{17}}$} )

_{$\Rightarrow (6 - 1) . \frac{3}{\sqrt{17}} + (7 - 2) . \frac{2}{\sqrt{17}} + (7 - 3) \frac{- 2}{\sqrt{17}} = \sqrt{17}$}

_{$∴ ⊥$} 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

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