If A=(5,4,2), B=(6,2,-1), C=(8,-2,-7),then th | vedclass

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Question

(C) The harmonic conjugate of a point $A$ with respect to points $B$ and $C$ is a point $P$ that divides the line segment $BC$ externally in the ratio

Vedclass · Accepted Answer

$(\frac{13}{2}, 1, \frac{-5}{2})$

Vedclass · Answer

(C) The harmonic conjugate of a point $A$ with respect to points $B$ and $C$ is a point $P$ that divides the line segment $BC$ externally in the ratio of the distances $AB$ and $AC$. First,we calculate the distances $AB$ and $AC$: $AB = \sqrt{(6-5)^2 + (2-4)^2 + (-1-2)^2} = \sqrt{1^2 + (-2)^2 + (-3)^2} = \sqrt{1+4+9} = \sqrt{14}$. $AC = \sqrt{(8-5)^2 + (-2-4)^2 + (-7-2)^2} = \sqrt{3^2 + (-6)^2 + (-9)^2} = \sqrt{9+36+81} = \sqrt{126} = 3\sqrt{14}$. The ratio $m:n = AB:AC = \sqrt{14} : 3\sqrt{14} = 1:3$. Since $P$ divides $BC$ externally in the ratio $1:3$,the coordinates of $P$ are given by the section formula for external division: $P = \left(\frac{m x_2 - n x_1}{m-n}, \frac{m y_2 - n y_1}{m-n}, \frac{m z_2 - n z_1}{m-n}\right)$ Substituting $m=1, n=3, B=(6,2,-1), C=(8,-2,-7)$: $x = \frac{1(8) - 3(6)}{1-3} = \frac{8-18}{-2} = \frac{-10}{-2} = 5$. Wait,let us re-evaluate the definition. The harmonic conjugate $P$ of $A$ with respect to $B$ and $C$ means $A$ divides $BC$ internally in ratio $k$ and $P$ divides $BC$ externally in ratio $k$. $A$ divides $BC$ in ratio $k:1$. $5 = \frac{k(8) + 1(6)}{k+1} \implies 5k+5 = 8k+6 \implies 3k = -1 \implies k = -1/3$. Since $k$ is negative,$A$ divides $BC$ externally. Using the property that $P$ is the harmonic conjugate,$P$ divides $BC$ internally in the same ratio $1:3$. $P = \left(\frac{1(8) + 3(6)}{1+3}, \frac{1(-2) + 3(2)}{1+3}, \frac{1(-7) + 3(-1)}{1+3}\right) = \left(\frac{26}{4}, \frac{4}{4}, \frac{-10}{4}\right) = (6.5, 1, -2.5) = (\frac{13}{2}, 1, \frac{-5}{2})$. Thus,the correct option is $C$.

If $A=(5,4,2), B=(6,2,-1), C=(8,-2,-7)$,then the harmonic conjugate of $A$ with respect to $B$ and $C$ is

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