Question

If O is the origin and A is the point $(a,b,c)$, then the equation of the plane through A and at right angles to OA is

• A. $a(x-a)-b(y-b)-c(z-c)=0$
• B. $a(x+a)+b(y+b)+c(z+c)=0$
• C. $a(x-a)+b(y-b)+c(z-c)=0$
• D. None of the above

Answer 1

The correct option is C $a(x-a)+b(y-b)+c(z-c)=0$

A plane with normal $ai+bj+ck$ is given by $ax+by+cz+d=0$.

Since it passes through $(a,b,c)$ we have $d=-(a^2+b^2+c^2)$.

Hence, the equation of plane is $ax+by+cz-a^2-b^2-c^2=0$.

By rearranging we get, $a(x-a)+b(y-b)+c(z-c)=0$