Find the foot of the perpendicular drawn from | vedclass

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(B) Let $A = (1, 0, 3)$,$B = (4, 7, 1)$,and $C = (3, 5, 3)$. Let $P$ be the foot of the perpendicular from $A$ to the line $BC$.The equation of the li

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $A = (1, 0, 3)$,$B = (4, 7, 1)$,and $C = (3, 5, 3)$. Let $P$ be the foot of the perpendicular from $A$ to the line $BC$.The equation of the line passing through $B$ and $C$ is given by $\vec{r} = \vec{b} + \lambda(\vec{c} - \vec{b})$.Direction ratios of $BC$ are $(3-4, 5-7, 3-1) = (-1, -2, 2)$.So,the line $BC$ is $\frac{x-4}{-1} = \frac{y-7}{-2} = \frac{z-1}{2} = k$.Any point $P$ on the line $BC$ is given by $P = (4-k, 7-2k, 1+2k)$.The vector $\vec{AP} = (4-k-1, 7-2k-0, 1+2k-3) = (3-k, 7-2k, 2k-2)$.Since $AP \perp BC$,the dot product of $\vec{AP}$ and the direction vector of $BC$ (which is $(-1, -2, 2)$) must be zero.$(3-k)(-1) + (7-2k)(-2) + (2k-2)(2) = 0$$-3 + k - 14 + 4k + 4k - 4 = 0$$9k - 21 = 0 \Rightarrow 9k = 21 \Rightarrow k = \frac{21}{9} = \frac{7}{3}$.Substituting $k = \frac{7}{3}$ into the coordinates of $P$:$x = 4 - \frac{7}{3} = \frac{12-7}{3} = \frac{5}{3}$$y = 7 - 2(\frac{7}{3}) = \frac{21-14}{3} = \frac{7}{3}$$z = 1 + 2(\frac{7}{3}) = \frac{3+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 foot of the perpendicular drawn from the point $A(1, 0, 3)$ to the line joining the points $B(4, 7, 1)$ and $C(3, 5, 3)$.

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