Question

Find the distance of the point (1,-2,3) from the plane x-y+z=5 measure parallel to the line whose direction cosines are proportional to 2,3,-6
.

Answer 1

Answer: 1

Step-by-step explanation:

The given plane line is x - y+z =5.....(i)

$\frac{x}{2} = \frac{y}{3} = \frac{z}{-6}$

So , the direction of ratio of the line from the point (2,3,-6) to the given plane will same as that of given line.

Let the line from point (1, -2, 3) meets at Q on the plane

Equation of line passing through (2,3,-6) and having (1, -2, 3)

$\frac{x-1}{2} =\frac{y+2}{3} = \frac{z-3}{-6} =$λ

x = 2λ +1 , y =3λ -2 , z = -6λ+3

Coordinates of Q are (2λ +1 , 3λ -2 , -6λ+3)

Q lies in the plane , as it is point of intersection of line and plane by (i)

2λ +1 -3λ +2 -6λ+3 = 5

-7λ =- 1

λ =1/7

Put λ =1/7 in Q (2λ +1 , 3λ -2 , -6λ+3)

Coordinates of Q are  ($\frac{9}{7} ,\frac{-11}{7}, \frac{15}{7}$ )

By Distance formula = $\sqrt{(9/7-1)^{2}+(-11/7+2)^{2}+(15/7-3)^{2} }$

                                 = $\sqrt{\frac{49}{49} }$

                                 = 1