Let ABC be a triangle with vertices at points | vedclass

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(C) Let $D$ be the midpoint of $BC$. The coordinates of $D$ are given by:$D = \left( \frac{\lambda - 1}{2}, \frac{5 + 3}{2}, \frac{\mu + 2}{2} \right)

Vedclass · Accepted Answer

$(7, 10)$

Vedclass · Answer

(C) Let $D$ be the midpoint of $BC$. The coordinates of $D$ are given by:$D = \left( \frac{\lambda - 1}{2}, \frac{5 + 3}{2}, \frac{\mu + 2}{2} \right) = \left( \frac{\lambda - 1}{2}, 4, \frac{\mu + 2}{2} \right)$The direction ratios of the median $AD$ are:$a = \frac{\lambda - 1}{2} - 2 = \frac{\lambda - 5}{2}$$b = 4 - 3 = 1$$c = \frac{\mu + 2}{2} - 5 = \frac{\mu - 8}{2}$Since the median $AD$ is equally inclined with the axes,its direction cosines $l, m, n$ are equal,i.e.,$|l| = |m| = |n|$. Since $l^2 + m^2 + n^2 = 1$,we have $l = m = n = \pm \frac{1}{\sqrt{3}}$.Thus,the direction ratios $a, b, c$ must be proportional to $1, 1, 1$. Since $b = 1$,we must have $a = 1$ and $c = 1$.Setting $a = 1$:$\frac{\lambda - 5}{2} = 1 \Rightarrow \lambda - 5 = 2 \Rightarrow \lambda = 7$Setting $c = 1$:$\frac{\mu - 8}{2} = 1 \Rightarrow \mu - 8 = 2 \Rightarrow \mu = 10$Therefore,$(\lambda, \mu) = (7, 10)$.

Let $ABC$ be a triangle with vertices at points $A(2, 3, 5)$,$B(-1, 3, 2)$ and $C(\lambda, 5, \mu)$ in three-dimensional space. If the median through $A$ is equally inclined with the axes,then $(\lambda, \mu)$ is equal to

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