triangle ABC is formed by A(1, 8, 4),B(0, -11 | vedclass

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(D) Let $D(h, k, l)$ be the foot of the perpendicular $AD$ on $BC$. The direction ratios of line $BC$ are $(2-0, -3-(-11), 1-4) = (2, 8, -3)$.The equa

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$(4, 5, -2)$

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(D) Let $D(h, k, l)$ be the foot of the perpendicular $AD$ on $BC$. The direction ratios of line $BC$ are $(2-0, -3-(-11), 1-4) = (2, 8, -3)$.The equation of line $BC$ passing through $C(2, -3, 1)$ is $\frac{x-2}{2} = \frac{y+3}{8} = \frac{z-1}{-3} = \lambda$.Any point on $BC$ is $D(2\lambda+2, 8\lambda-3, -3\lambda+1)$.The vector $\vec{AD} = (2\lambda+2-1, 8\lambda-3-8, -3\lambda+1-4) = (2\lambda+1, 8\lambda-11, -3\lambda-3)$.Since $AD \perp BC$,the dot product of $\vec{AD}$ and the direction vector of $BC$ $(2, 8, -3)$ is zero:$2(2\lambda+1) + 8(8\lambda-11) - 3(-3\lambda-3) = 0$$4\lambda + 2 + 64\lambda - 88 + 9\lambda + 9 = 0$$77\lambda - 77 = 0 \Rightarrow \lambda = 1$.Substituting $\lambda = 1$ in the coordinates of $D$:$h = 2(1)+2 = 4$$k = 8(1)-3 = 5$$l = -3(1)+1 = -2$Thus,the coordinates of $D$ are $(4, 5, -2)$.

$\triangle ABC$ is formed by $A(1, 8, 4)$,$B(0, -11, 4)$,and $C(2, -3, 1)$. If $D$ is the foot of the perpendicular drawn from $A$ to $BC$,then the coordinates of $D$ are

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