Question

If $A(-1,4,-3)$ is one end and $B$ is other end of diameter of the sphere $x^2+y^2+z^2-3x-2y+2z+15=0$, then $B$ is

• A. $(4,-2,1)$
• B. $(-4,-2,1)$
• C. $(4,-3,1)$
• D. $(3,-4,1)$

Answer 1

Answer: (A)

$(4,-2,1)$

Here centre of the sphere is $(\frac{3}{2},1,-1)$
Let other end of the diameter $B$ is $(a,b,c)$
Since $AB$ is diameter of the sphere $\Rightarrow$ Centre will be mid point of $AB$.
$\Rightarrow \frac{-1+a}{2}=\frac{3}{2}\Rightarrow a=4$
$\Rightarrow \frac{4+b}{2}=1\Rightarrow b=-2$
and $\frac{-3+c}{2}=-1\Rightarrow c=1$
Hence, $B\equiv (4,-2,1)$