A(2,3,5), B(alpha, 3,3) and C(7,5, beta) are | vedclass

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(A) Given,vertices of the triangle are $A(2,3,5), B(\alpha, 3,3)$ and $C(7,5, \beta)$.Let $D$ be the midpoint of $BC$. The coordinates of $D$ are $\le

Vedclass · Accepted Answer

$\cos^{-1}\left(-\frac{1}{9}\right)$

Vedclass · Answer

(A) Given,vertices of the triangle are $A(2,3,5), B(\alpha, 3,3)$ and $C(7,5, \beta)$.Let $D$ be the midpoint of $BC$. The coordinates of $D$ are $\left(\frac{\alpha+7}{2}, \frac{3+5}{2}, \frac{3+\beta}{2}\right) = \left(\frac{\alpha+7}{2}, 4, \frac{3+\beta}{2}\right)$.The direction ratios of the median $AD$ are $\left(\frac{\alpha+7}{2} - 2, 4 - 3, \frac{3+\beta}{2} - 5\right) = \left(\frac{\alpha+3}{2}, 1, \frac{\beta-7}{2}\right)$.Since the median $AD$ is equally inclined to the coordinate axes,its direction ratios must be proportional to $(1, 1, 1)$.Thus,$\frac{\alpha+3}{2} = 1$ and $\frac{\beta-7}{2} = 1$.Solving these,we get $\alpha+3 = 2 \Rightarrow \alpha = -1$ and $\beta-7 = 2 \Rightarrow \beta = 9$.Therefore,$\cos^{-1}\left(\frac{\alpha}{\beta}\right) = \cos^{-1}\left(-\frac{1}{9}\right)$.

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

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