The harmonic conjugate of the point P(2,3,4) | vedclass

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Question

(A) Let the points be $A(3,-2,2)$ and $B(6,-17,-4)$. Let $P(2,3,4)$ divide the line segment $AB$ in the ratio $\lambda : 1$.Using the section formula

Vedclass · Accepted Answer

$\left(\frac{18}{5},-5, \frac{4}{5}\right)$

Vedclass · Answer

(A) Let the points be $A(3,-2,2)$ and $B(6,-17,-4)$. Let $P(2,3,4)$ divide the line segment $AB$ in the ratio $\lambda : 1$.Using the section formula for the $x$-coordinate:$2 = \frac{6\lambda + 3}{\lambda + 1}$$2\lambda + 2 = 6\lambda + 3$$-4\lambda = 1 \implies \lambda = -\frac{1}{4}$.The harmonic conjugate $Q$ of $P$ with respect to $A$ and $B$ divides $AB$ externally in the same ratio $\lambda : 1$,which is equivalent to dividing $AB$ internally in the ratio $-\lambda : 1$.Thus,$Q$ divides $AB$ in the ratio $\frac{1}{4} : 1$,or $1 : 4$.Using the section formula for $Q$ with ratio $m:n = 1:4$:$Q = \left( \frac{1(6) + 4(3)}{1+4}, \frac{1(-17) + 4(-2)}{1+4}, \frac{1(-4) + 4(2)}{1+4} \right)$$Q = \left( \frac{6+12}{5}, \frac{-17-8}{5}, \frac{-4+8}{5} \right)$$Q = \left( \frac{18}{5}, -\frac{25}{5}, \frac{4}{5} \right) = \left( \frac{18}{5}, -5, \frac{4}{5} \right)$.

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

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