Question

If a plane passes through the point (1,1,1) and is perpendicular to the line $\frac{x-1}{3}=\frac{y-1}{0}=\frac{z-1}{4}$ then its perpendicular distance from the origin is

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{7}{5}$
• D. $1$

Answer 1

The correct option is C $\frac{7}{5}$

The d.r of the normal to the plane is$(3,0,4)$ The equation of the plane is 3x + 0y + 4z + d = 0 since it passes through (1, 1, 1) so; d = --7
Now distance of the plane$3x+4z-7=0$ from (0, 0, 0) is $\frac{7}{\sqrt{3^2+4^2}}=\frac{7}{5}$ units