ABC is a triangle with vertices A(2, 3, 5),B( | vedclass

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

Vedclass · Accepted Answer

$1348$

Vedclass · Answer

(B) Let $D$ be the midpoint of $BC$. The coordinates of $D$ are $\left( \frac{\lambda - 1}{2}, \frac{5 + 3}{2}, \frac{\mu + 2}{2} \right) = \left( \frac{\lambda - 1}{2}, 4, \frac{\mu + 2}{2} \right)$.The direction ratios of the median $AD$ are $\left( \frac{\lambda - 1}{2} - 2, 4 - 3, \frac{\mu + 2}{2} - 5 \right) = \left( \frac{\lambda - 5}{2}, 1, \frac{\mu - 8}{2} \right)$.Since the median $AD$ is equally inclined to the coordinate axes,its direction ratios must be equal: $\frac{\lambda - 5}{2} = 1 = \frac{\mu - 8}{2}$.From $\frac{\lambda - 5}{2} = 1$,we get $\lambda - 5 = 2$,so $\lambda = 7$.From $\frac{\mu - 8}{2} = 1$,we get $\mu - 8 = 2$,so $\mu = 10$.Now,calculate $(\lambda^3 + \mu^3 + 5) = 7^3 + 10^3 + 5 = 343 + 1000 + 5 = 1348$.

$ABC$ is a triangle with vertices $A(2, 3, 5)$,$B(-1, 3, 2)$,and $C(\lambda, 5, \mu)$. If the median through $A$ is equally inclined to the coordinate axes,then the value of $(\lambda^3 + \mu^3 + 5)$ is equal to:

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