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) 3/4
(b) 4/3
(c) 7/5
(d) 1

Answer 1

(c) 7/5

$$\begin{gathered} \text{Since the plane is perpendicular to the given line, its direction ratios are proportional to 3, 0, 4} \\ \text{So, the required equation of the plane is of the form} \\ 3x+0y+4z+d=0...\left(1\right),\text{where}d\text{is a constant.} \\ \text{Since this plane passes through (1, 1, 1),} \\ 3+0+4+d=0 \\ \Rightarrow d=-7 \\ \text{Substituting this in (1), we get} \\ 3x+0y+4z-7=0...\left(2\right) \\ \text{Perpendicular distance of (2) from the origin} \\ =\frac{\left|3\left(0\right)+0+4\left(0\right)-7\right|}{\sqrt{3^2+0^2+4^2}} \\ =\frac{\left|0+0-7\right|}{\sqrt{25}} \\ =\frac{7}{5}\text{units} \end{gathered}$$