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

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Aicharya · Accepted Answer

Since $\mathrm{AD}$ is the median $ 	herefore \quad \mathrm{D}=\left(\frac{\lambda-1}{2}, 4, \frac{\mu+2}{2}ight) $ Now, dR's of AD is $ a=\left(\frac{\lambda-1}{2}-2ight)=\frac{\lambda-5}{2} $ $ b=4-3=1, c=\frac{\mu+2}{2}-5=\frac{\mu-8}{2} $ Also, $a, b, c$ are $\mathrm{dR}$ 's $	herefore a=k l, b=k m, c=k n$ where $l=m=n$ and $l^2+m^2+n^2=1$ $ \Rightarrow l=m=n=\frac{1}{\sqrt{3}} $ Now, $a=1, b=1$ and $c=1$ $\Rightarrow \lambda=7$ and $\mu=10$

Let $\mathrm{ABC}$ be a triangle with vertices at points $\mathrm{A}$ $(2,3,5)$, B $(-1,3,2)$ and…

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