$A(2,3,5), B(\alpha, 3,3)$ and $C(7,5, \beta)$ are the vertices of a triangle. If the median through $A$ is equally inclined with the co-ordinate axes, then $\cos ^{-1}\left(\frac{\alpha}{\beta}\right)=$

Question

$A(2,3,5), B(\alpha, 3,3)$ and $C(7,5, \beta)$ are the vertices of a triangle. If the median through $A$ is equally inclined with the co-ordinate axes, then $\cos ^{-1}\left(\frac{\alpha}{\beta}\right)=$, $\cos ^{-1}\left(\frac{-1}{9}\right)$, $\frac{\pi}{2}$, $\frac{\pi}{3}$, $\cos ^{-1}\left(\frac{2}{5}\right)$

MARKS App · Accepted Answer

Given, points $A(2,3,5), B(\alpha, 3,3)$ and $C(7,5, \beta)$ $\therefore$ Mid-point of $B C$ is $D\left(\frac{\alpha+7}{2}, 4, \frac{3+\beta}{2}\right)$ $\because$ Direction ratios of line joining points $A(2,3,5)$ and $D\left(\frac{\alpha+7}{2}, 4, \frac{3+\beta}{2}\right)$ is $\left(\frac{\alpha+3}{2}, 1, \frac{\beta-7}{2}\right)$. $\because$ The line segment $A D$ is equally inclined with the co-ordinate axes, so $\begin{aligned} & \frac{\alpha+3}{2}=1=\frac{\beta-7}{2} \ & \Rightarrow \quad \alpha=-1 \text { and } \beta=9 \ & \therefore \cos ^{-1}\left(\frac{\alpha}{\beta}\right)=\cos ^{-1}\left(-\frac{1}{9}\right) \end{aligned}$ Hence, option (1) is correct.

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