Question

Distance of a point with position vector a from a plane $râ‹\dots \hat{n}=d$ is

• A. $âˆ\pounds d+aâ‹\dots \hat{n}âˆ\pounds$
• B. $âˆ\pounds d−aâ‹\dots \hat{n}âˆ\pounds$
• C. $âˆ\pounds d+a×\hat{n}âˆ\pounds$
• D. $d+âˆ\pounds a×\hat{n}âˆ\pounds$

Answer 1

Answer: (B)

Consider a point $P$ with position vector a and a plane ${Ï€}_1$ whose equation is $râ‹\dots \hat{n}=d$.

Consider a plane ${Ï€}_2$ through $P$ parallel to the plane ${Ï€}_1$. The unit vector normal to ${Ï€}_2$ is $\hat{n}$. Hence, its equation is $(r−a)â‹\dots \hat{n}=0$ i.e. $râ‹\dots \hat{n}=aâ‹\dots \hat{n}$
Thus, the distance ON' of this plane from the origin is $âˆ\pounds aâ‹\dots \hat{n}âˆ\pounds$. Therefore, the distance $PQ$ from the plane ${Ï€}_1$ is [Fig. (a)] i.e., $ON−ON=âˆ\pounds d−aâ‹\dots \hat{n}âˆ\pounds$
which is the length of the perpendicular from a point to the given plane.
We may establish the similar results for Fig. (b).