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

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(B) The vertices of $	riangle ABC$ are $A(1, 8, 4)$,$B(0, -11, 4)$,and $C(2, -3, 1)$.First,find the equation of the line $BC$ passing through $B(0, -

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$(4, 5, – 2)$

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(B) The vertices of $	riangle ABC$ are $A(1, 8, 4)$,$B(0, -11, 4)$,and $C(2, -3, 1)$.First,find the equation of the line $BC$ passing through $B(0, -11, 4)$ and $C(2, -3, 1)$.The direction ratios of $BC$ are $(2-0, -3-(-11), 1-4) = (2, 8, -3)$.The equation of line $BC$ is $\frac{x-0}{2} = \frac{y+11}{8} = \frac{z-4}{-3} = \lambda$.Any point $D$ on the line $BC$ can be represented as $(2\lambda, 8\lambda - 11, -3\lambda + 4)$.Since $AD \perp BC$,the dot product of vector $\vec{AD}$ and the direction vector of $BC$ must be zero.Vector $\vec{AD} = (2\lambda - 1, 8\lambda - 11 - 8, -3\lambda + 4 - 4) = (2\lambda - 1, 8\lambda - 19, -3\lambda)$.The direction vector of $BC$ is $\vec{v} = (2, 8, -3)$.Setting $\vec{AD} \cdot \vec{v} = 0$:$2(2\lambda - 1) + 8(8\lambda - 19) + (-3)(-3\lambda) = 0$$4\lambda - 2 + 64\lambda - 152 + 9\lambda = 0$$77\lambda - 154 = 0$$77\lambda = 154 \implies \lambda = 2$.Substituting $\lambda = 2$ into the coordinates of $D$:$D = (2(2), 8(2) - 11, -3(2) + 4) = (4, 16 - 11, -6 + 4) = (4, 5, -2)$.

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

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