Question

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

• A. 1
• B. 2
• C. $\frac{1}{2}$
• D. 4

Answer 1

The correct option is A 1

Line parallel to the line $\frac{x}{2}=\frac{y}{2}=\frac{z}{-6}$ and passing through (1,-2, 3) is
$\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=r$
A = (2r + 1, 3r – 2, – 6r + 3) is a point on this line.
A lies in the plane x – y + z = 5
$\therefore 2r+1-3r+2-6r+3=5\Rightarrow r=\frac{1}{7}\therefore A=(\frac{9}{7},\frac{-11}{7},\frac{15}{7})$
Distance of A from (1, –2, 3) is
$\sqrt{{(\frac{9}{7}-1)}^2+{(\frac{-11}{7}+2)}^2+{(\frac{15}{7}-3)}^2}=\frac{1}{7}\sqrt{4+9+36}=\frac{7}{7}=1$