The coordinates of the foot of the perpendicu | vedclass

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(B) Let the line $AB$ pass through $A(4, 7, 1)$ and $B(3, 5, 3)$. The direction ratios of line $AB$ are $(3-4, 5-7, 3-1) = (-1, -2, 2)$.The equation o

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$\left( \frac{5}{3}, \frac{7}{3}, \frac{17}{3} \right)$

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(B) Let the line $AB$ pass through $A(4, 7, 1)$ and $B(3, 5, 3)$. The direction ratios of line $AB$ are $(3-4, 5-7, 3-1) = (-1, -2, 2)$.The equation of line $AB$ is $\frac{x-4}{-1} = \frac{y-7}{-2} = \frac{z-1}{2} = k$.Any point $D$ on the line $AB$ is given by $D = (-k+4, -2k+7, 2k+1)$.Since $PD$ is perpendicular to $AB$,the direction ratios of $PD$ are $((-k+4)-1, (-2k+7)-0, (2k+1)-3) = (-k+3, -2k+7, 2k-2)$.Since $PD \perp AB$,the dot product of their direction ratios is zero:$(-1)(-k+3) + (-2)(-2k+7) + (2)(2k-2) = 0$$k-3 + 4k-14 + 4k-4 = 0$$9k - 21 = 0 \implies k = \frac{21}{9} = \frac{7}{3}$.Substituting $k = \frac{7}{3}$ into the coordinates of $D$:$x = -\frac{7}{3} + 4 = \frac{5}{3}$$y = -2(\frac{7}{3}) + 7 = -\frac{14}{3} + \frac{21}{3} = \frac{7}{3}$$z = 2(\frac{7}{3}) + 1 = \frac{14}{3} + \frac{3}{3} = \frac{17}{3}$.Thus,the foot of the perpendicular is $\left( \frac{5}{3}, \frac{7}{3}, \frac{17}{3} \right)$.

The coordinates of the foot of the perpendicular drawn from point $P(1, 0, 3)$ to the line joining points $A(4, 7, 1)$ and $B(3, 5, 3)$ are:

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