Question

The harmonic conjugate of $(2,3,4)$ w.r.t the points $(3,-2,2)$ and $(6,-17,-4)$ is:

• A. $(\frac{-1}{2},\frac{1}{3},\frac{-1}{4})$
• B. $(\frac{1}{2},\frac{1}{3},\frac{1}{4})$
• C. $(\frac{18}{7},2,\frac{1}{5})$
• D. $(\frac{18}{5},-5,\frac{4}{5})$

Answer 1

The correct option is D $(\frac{18}{5},-5,\frac{4}{5})$

Given: Let $P(2,3,4),A(3,-2,2),B(6,-17,-4)$
Let $P$ divides $AB$ in the ratio $k:1$, then
Hence,$2=\frac{6k+3}{k+1}$
$2k+2=6k+3$
$4k=-1$
$k=\frac{-1}{4}$
Therefore harmonic conjugate $Q$ must divide $AB$ in the $-k:1$
$\Rightarrow -k:1=-\left(\frac{-1}{4}\right):1=\frac{1}{4}:1$
Hence, coordinates of $Q$ are
$(\frac{\frac{1}{4}\times 6+1\times 3}{\frac{1}{4}+1},\frac{\frac{1}{4}\times -17+1\times -2}{\frac{1}{4}+1},\frac{\frac{1}{4}\times -4+1\times 2}{\frac{1}{4}+1})$
$\Rightarrow Q(\frac{18}{5},\frac{-25}{5},\frac{4}{5})$
$\Rightarrow Q(\frac{18}{5},-5,\frac{4}{5})$