Find the coordinates of the foot of the perpe | vedclass

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(B) Let the line passing through $A(4, 7, 1)$ and $B(3, 5, 3)$ be $L$. The direction ratios of line $AB$ are $(3-4, 5-7, 3-1) = (-1, -2, 2)$.The equat

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

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(B) Let the line passing through $A(4, 7, 1)$ and $B(3, 5, 3)$ be $L$. The direction ratios of line $AB$ are $(3-4, 5-7, 3-1) = (-1, -2, 2)$.The equation of line $AB$ in parametric form is $x = 4 - r, y = 7 - 2r, z = 1 + 2r$,where $r$ is a parameter.Let $Q$ be the foot of the perpendicular from $P(1, 0, 3)$ to line $AB$. Then $Q$ has coordinates $(4-r, 7-2r, 1+2r)$.The vector $\vec{PQ} = (4-r-1, 7-2r-0, 1+2r-3) = (3-r, 7-2r, 2r-2)$.Since $PQ \perp AB$,the dot product of $\vec{PQ}$ and the direction vector of $AB$,$\vec{v} = (-1, -2, 2)$,must be zero:$(3-r)(-1) + (7-2r)(-2) + (2r-2)(2) = 0$$-3 + r - 14 + 4r + 4r - 4 = 0$$9r - 21 = 0 \Rightarrow r = \frac{21}{9} = \frac{7}{3}$.Substituting $r = \frac{7}{3}$ into the coordinates of $Q$:$x = 4 - \frac{7}{3} = \frac{5}{3}$$y = 7 - 2(\frac{7}{3}) = 7 - \frac{14}{3} = \frac{7}{3}$$z = 1 + 2(\frac{7}{3}) = 1 + \frac{14}{3} = \frac{17}{3}$.Thus,the foot of the perpendicular is $\left( \frac{5}{3}, \frac{7}{3}, \frac{17}{3} \right)$.

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

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