If the foot of the perpendicular drawn from t | vedclass

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(B) Let $A = (1, 0, 3)$,$B = (\alpha, 7, 1)$,and $P = \left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$.Since $P$ is the foot of the perpendicular

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$4$

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(B) Let $A = (1, 0, 3)$,$B = (\alpha, 7, 1)$,and $P = \left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$.Since $P$ is the foot of the perpendicular from $A$ to the line passing through $B$,the vector $\vec{AP}$ must be perpendicular to the vector $\vec{BP}$.First,find the direction ratios of $\vec{AP}$:$\vec{AP} = \left(\frac{5}{3} - 1, \frac{7}{3} - 0, \frac{17}{3} - 3\right) = \left(\frac{2}{3}, \frac{7}{3}, \frac{8}{3}\right)$.Next,find the direction ratios of $\vec{BP}$:$\vec{BP} = \left(\frac{5}{3} - \alpha, \frac{7}{3} - 7, \frac{17}{3} - 1\right) = \left(\frac{5}{3} - \alpha, -\frac{14}{3}, \frac{14}{3}\right)$.Since $\vec{AP} \perp \vec{BP}$,their dot product must be zero:$\vec{AP} \cdot \vec{BP} = 0$$\left(\frac{2}{3}\right)\left(\frac{5}{3} - \alpha\right) + \left(\frac{7}{3}\right)\left(-\frac{14}{3}\right) + \left(\frac{8}{3}\right)\left(\frac{14}{3}\right) = 0$$\frac{2}{3} \left(\frac{5}{3} - \alpha\right) - \frac{98}{9} + \frac{112}{9} = 0$$\frac{2}{3} \left(\frac{5}{3} - \alpha\right) + \frac{14}{9} = 0$Multiply by $9$:$6 \left(\frac{5}{3} - \alpha\right) + 14 = 0$$10 - 6\alpha + 14 = 0$$24 = 6\alpha$$\alpha = 4$.

If the foot of the perpendicular drawn from the point $A(1, 0, 3)$ on a line passing through $B(\alpha, 7, 1)$ is $P\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$,then $\alpha$ is equal to:

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