Let ABC be a triangle with A(alpha, 5, beta), | vedclass

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(D) Let $D$ be the midpoint of side $AC$. The coordinates of $D$ are given by:$D = \left( \frac{\alpha + 1}{2}, \frac{5 + 0}{2}, \frac{\beta - 3}{2} \

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(D) Let $D$ be the midpoint of side $AC$. The coordinates of $D$ are given by:$D = \left( \frac{\alpha + 1}{2}, \frac{5 + 0}{2}, \frac{\beta - 3}{2} \right) = \left( \frac{\alpha + 1}{2}, \frac{5}{2}, \frac{\beta - 3}{2} \right)$The vector $\vec{BD}$ is given by:$\vec{BD} = \left( \frac{\alpha + 1}{2} - (-2) \right) \hat{i} + \left( \frac{5}{2} - 1 \right) \hat{j} + \left( \frac{\beta - 3}{2} - 6 \right) \hat{k}$$\vec{BD} = \left( \frac{\alpha + 5}{2} \right) \hat{i} + \left( \frac{3}{2} \right) \hat{j} + \left( \frac{\beta - 15}{2} \right) \hat{k}$Since the median $\vec{BD}$ is equally inclined to the coordinate axes, its direction ratios must be equal in magnitude (or proportional). Thus:$\left| \frac{\alpha + 5}{2} \right| = \left| \frac{3}{2} \right| = \left| \frac{\beta - 15}{2} \right|$Considering the positive case for the direction ratios:$\frac{\alpha + 5}{2} = \frac{3}{2} \implies \alpha + 5 = 3 \implies \alpha = -2$$\frac{\beta - 15}{2} = \frac{3}{2} \implies \beta - 15 = 3 \implies \beta = 18$Therefore, $\alpha + \beta = -2 + 18 = 16$.

Let $ABC$ be a triangle with $A(\alpha, 5, \beta)$, $B(-2, 1, 6)$ and $C(1, 0, -3)$ as its vertices. If the median through $B$ is equally inclined to the coordinate axes, then $\alpha + \beta =$

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