The harmonic conjugate of the point $(2,3,4)$ with respect to the point $(3,-2,2)$ and $(6,-17,-4)$ is

Question

Aicharya · Accepted Answer

Given, Let $P(2,3,4), A(3,-2,2)$ and $B(6,-17,-4)$ are three points. Let $P$ divides $A B$ in the ratios $\lambda: 1$, then

$ \therefore \quad \begin{aligned} 2 & =\frac{6 \lambda+3}{\lambda+1} \\ 2 \lambda+2 & =6 \lambda+3 \\ 2-3 & =6 \lambda-2 \lambda \\ \lambda & =-\frac{1}{4} \end{aligned} $ Harmonic conjugate $Q$ divides in the ratio $-\lambda: 1$ i.e. $\frac{1}{4}: 1$. $\therefore$ Coordinate of point $Q$. $ \begin{aligned} & =\left[\frac{\frac{1}{4}(6)+3}{\frac{1}{4}+1}, \frac{\frac{1}{4}(-17)-2}{\frac{1}{4}+1}, \frac{\frac{1}{4}(-4)+2}{\frac{1}{4}+1}\right] \\ & =\left(\frac{6+12}{5}, \frac{-17-8}{5}, \frac{-4+8}{5}\right) \\ & =\left(\frac{18}{5},-\frac{25}{5}, \frac{4}{5}\right) \\ & =\left(\frac{18}{5},-5, \frac{4}{5}\right) \end{aligned} $

The harmonic conjugate of the point $(2,3,4)$ with respect to the point $(3,-2,2)$ and $(6,-17,-4)$ is

Solution