Discussion Forum

The distance of the point ( 1, -2, 3) from the plane x - y + z =5, measured parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z-1}{-6}$ is

Answer:

Explanation:

Equation of the line through (1, -2, 3)

parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z-1}{-6}$ is

$\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=r$ (say)           ...(1)

Then any point on (1) is (2r + 1, 3r - 2, -6r+3)

If this point lies on the plane x - y + z = 5
then

(2r+1)-(3r-2)+(-6r+3)= 5 ⇒ r = $\frac{1}{7}$

Hence the point is $(\frac{9}{7},-\frac{11}{7},\frac{15}{7})$

Distance between (1,-2, 3 ) and $(\frac{9}{7},-\frac{11}{7},\frac{15}{7})$

$=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=\sqrt{\frac{49}{49}}=1$