Question

$O$ is the origin and $A$ is the point $(a,b,c).$ Find the direction cosines of the join of $OA$ and deduce the equation of the plane through $A$ at right angles to $OA$.

• A. $ax+by+cz=a^2-b^2-c^2$
• B. $ax+by+cz=a^2+b^2+c^2$
• C. $ax-by-cz=a^2+b^2+c^2$
• D. $ax-by-cz=a^2-b^2-c^2$

Answer 1

The correct option is B $ax+by+cz=a^2+b^2+c^2$

D.R.'s of $OA$ are $a-0,b-0,c-0$ or a,b,c.
$\therefore$ D.C.'s of $OA$ are
$\frac{a}{\sqrt{(a^2+b^2+c^2)}},\frac{b}{\sqrt{(\sum a^2)}},\frac{c}{\sqrt{\sum a^2}}$
Equation of any plane through $A$, i.e. $(a,b,c)$ is
$A(x-a)+B(y-b)+C(z-c)=0.$
But plane is at right angles to $OA$ which therefore is normal and whose D.R.'s are $a,b,c$.
Hence, the plane is
$a(x-a)+b(y-b)+c(z-c)=0$
$ax+by+cz=a^2+b^2+c^2$