Question

A plane passes through a fixed point $(a,b,c)$. The locus of the foot of perpendicular to it from the origin is sphere of radius

• A. $\sqrt{a^2+b^2+c^2}$
• B. $\frac{\sqrt{a^2+b^2+c^2}}{2}$
• C. $a^2+b^2+c^2$
• D. $\frac{a^2+b^2+c^2}{2}$

Answer 1

The correct option is B $\frac{\sqrt{a^2+b^2+c^2}}{2}$

An equation of plane can be constructed by equation,
$n.(r-r_1)=0$ where $r$ is any position vector on the place $r_1$ is fixed position vector in a plane and $n$ is direction ratios of normal to the plane.
Therefore $A(x-a)+B(x-b)+C(x-c)=0$
Normal from origin to the point (h,k,l) on sphere will have direction ratios $h-0,k-0,l-0$.
$(h,k,l)$ are the coordinates of foot of perpendicular so they ll lie on the plane hence satisfy the equation.
$\left(h\right)(h-a)+k(k-b)+l(l-c)=0$
$h^2+k^2+l^2-ah-bk-cl=0$
On replacing $h,k,l$ by $x,y,z$, we get equation of a sphere with radius $\frac{\sqrt{a^2+b^2+c^2}}{2}$
Hence, option B is the correct option.