\(A(2,3,5), B(\alpha, 3,3)\) and \(C(7,5, \beta)\) are the vertices of a triangle. If the median through…
\(A(2,3,5), B(\alpha, 3,3)\) 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 \(\cos ^{-1}\left(\frac{\alpha}{\beta}\right)=\)
\(\cos ^{-1}\left(\frac{-1}{9}\right)\)
\(\frac{\pi}{2}\)
\(\frac{\pi}{3}\)
\(\cos ^{-1}\left(\frac{2}{5}\right)\)
Solution
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.