Explanation:
A Given,
\(\mathrm{A}(2,3,5), \mathrm{B}(\alpha, 3,3)\) and \(\mathrm{c}(7,5, \beta)\)
Midpoint of \(\mathrm{BC}, \mathrm{M}=\left(\frac{\alpha+7}{2}, \frac{3+5}{2}, \frac{3+\beta}{2}\right)\)
\(\mathrm{M}=\left(\frac{\alpha+7}{2}, 4, \frac{3+\beta}{2}\right)\)
Direction ration of \(\mathrm{AM}=\left(\frac{\alpha+7}{2}-2,4-3, \frac{3+\beta}{2}-5\right)\)
\(=\left(\frac{\alpha+3}{2}, 1, \frac{\beta-7}{2}\right)\)
Median AM is equally inclined to co - ordinate axes. Therefore, direction ratios of \(x, y, z\) coordinates are the same. Let those coordinate be \(\mathrm{k}\).
So, direction ratios of AM are \((\mathrm{k}, \mathrm{k}, \mathrm{k})\)
Now, equation (i) and (ii) both are direction ratios of AM.
Hence, they must be equal.
\(\frac{\alpha+3}{2}=\mathrm{k}, \mathrm{k}=1, \frac{\beta-7}{2}=\mathrm{k}\)
Using \(\mathrm{k}=1\) obtain from \(\mathrm{y}\) coordinate,
\(\therefore \frac{\alpha+3}{2}=1\)
\(\Rightarrow \alpha=-1\)
Similarly, \(\left(\frac{\beta-7}{2}\right)=1\)
\(\Rightarrow \quad \beta=9\)Now, \(\cos ^{-1}\left(\frac{\alpha}{\beta}\right)=\cos ^{-1}\left(\frac{-1}{9}\right)\)