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$\triangle \mathrm{ABC}$ is formed by $\mathrm{A}(1,8,4), \mathrm{B}(0,-11,4)$ and $C(2,-3,1)$. If $D$ is the foot of the perpendicular drawn from $\mathrm{A}$ to $\mathrm{BC}$, then the coordinates of $\mathrm{D}$ are
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Verified Answer
The correct answer is:
$(4,5,-2)$
Let $\mathrm{D}(h, k, l)$ be foot of Perpendicular $\mathrm{AD}$ on $\mathrm{BC}$. Equation of line $\mathrm{BC}$ is
$$
\frac{x-2}{2}=\frac{y+3}{8}=\frac{z-1}{-3}=\lambda \text { (say) }
$$
Since equation passes through $\mathrm{D}(h, k, l)$
Hence $\frac{h-2}{2}=\lambda, \frac{k+3}{8}=\lambda, \frac{l-1}{-3}=\lambda$
$$
\Rightarrow h=2 \lambda+2, \quad k=8 \lambda-3, \quad l=-3 \lambda+1
$$
Since $\mathrm{AD} \perp \mathrm{BC}$ Hence
$$
\begin{aligned}
& \begin{aligned}
& \Rightarrow[(2 \lambda+2)-1] \cdot(2-0)+ {[(8 \lambda-3)-8] \cdot(-3+11) } \\
&+[(-3 \lambda+1)-4] \cdot(1-4)=0 \\
& \Rightarrow \lambda=1
\end{aligned}
\end{aligned}
$$
Hence $h=4, \quad k=5, \quad l=-2$
$$
\mathrm{D}(h, k, l)=\mathrm{D}(4,5,-2)
$$
$$
\frac{x-2}{2}=\frac{y+3}{8}=\frac{z-1}{-3}=\lambda \text { (say) }
$$
Since equation passes through $\mathrm{D}(h, k, l)$
Hence $\frac{h-2}{2}=\lambda, \frac{k+3}{8}=\lambda, \frac{l-1}{-3}=\lambda$
$$
\Rightarrow h=2 \lambda+2, \quad k=8 \lambda-3, \quad l=-3 \lambda+1
$$
Since $\mathrm{AD} \perp \mathrm{BC}$ Hence
$$
\begin{aligned}
& \begin{aligned}
& \Rightarrow[(2 \lambda+2)-1] \cdot(2-0)+ {[(8 \lambda-3)-8] \cdot(-3+11) } \\
&+[(-3 \lambda+1)-4] \cdot(1-4)=0 \\
& \Rightarrow \lambda=1
\end{aligned}
\end{aligned}
$$
Hence $h=4, \quad k=5, \quad l=-2$
$$
\mathrm{D}(h, k, l)=\mathrm{D}(4,5,-2)
$$
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