Question

Find the distance of the point 3i - 2j + k from the plane 3x + y - z + 2 = 0 measured parallel to the line x-3/2 = y+2/-3 = z-1/1.

Answer 1

Solution:

As, you can see point (3,-2,1) lies on the line,

$\frac{x-3}{2}=\frac{y-(-2)}{-3}=\frac{z-1}{1}$---(1)

So, we have to find Perpendicular distance of point from the plane.

Distance of Point (3,2,-1) which lies on the line, can be found by finding a line perpendicular to this line passing through the point (3,2,-1).

Let this Perpendicular line has direction ratios a, b, and c.

Equation of this line can be written as

$\frac{x-3}{a}=\frac{y-(-2)}{b}=\frac{z-1}{c}$---(2)

Equation of plane is, 3 x + y-z+2=0

Direction ratios of this line and Direction ratios of plane will be proportional.

$\frac{a}{3}=\frac{b}{1}=\frac{c}{-1}=k$

So, Line (2) is equivalent to

$\frac{x-3}{3}=\frac{y-(-2)}{1}=\frac{z-1}{-1}$

Any point a distance P from the above line can be found as,

x=3 P +3, y= P -2, Z=-P +1

Substituting the values of , x , y and z in equation of plane, 3 x + y- z+2=0

3 (3 P +3) + P -2+ P -1+2=0

9 P +8+2 P =0

11 P= -8

$P=\frac{-8}{11}$

So, Point which is perpendicular to (3,-2,1) which lies on the plane is

 $(\frac{-24}{11}+3=\frac{9}{11}, \frac{-8}{11}-2=\frac{-30}{11},\frac{8}{11}+1=\frac{19}{11}$

Distance between (3,-2,1) and $(\frac{9}{11},\frac{-30}{11}. \frac{19}{11})$ is equal to$\sqrt{(\frac{24}{11})^2+(\frac{8}{11})^2+(\frac{8}{11})^2}=\frac{\sqrt{704}}{11}=\frac{26.5329}{11}=2.4$

Distance between two points , (A,B,C) and (P,Q,R ) is given by

$=\sqrt{(A-P)^2+(B-Q)^2+(C-R)^2}$