$\triangle A B C$ is formed by $A(1,8,4), B(0,-11,4)$ and $C(2,-3,1)$. If $D$ is the foot of the…

1. (– 4, 5, 2)
2. (4, 5, – 2)
3. (4, – 5, 2)
4. (4, – 5, – 2)

Solution

The vertices of $\triangle A B C$ are given as $A(1,8,4)$, $B(0,-11,4)$ and $C(2,-3,1)$. Equation of line $B C$, $ \frac{x}{2}=\frac{y+11}{8}=\frac{z-4}{-3}=\lambda \text { (say) } $
As point $D$ is on line $B C$, so coordinates of $D$ are $(2 \lambda, 8 \lambda-11,-3 \lambda+4)$ Since, $A D \perp B C$ $ \begin{aligned} \therefore \quad & A D \cdot B C=0 \\ & (2 \lambda-1,8 \lambda-19,-3 \lambda) \cdot(2,8,-3)=0 \\ \therefore \quad & 2(2 \lambda-1)+8(8 \lambda-19)+(-3)(-3 \lambda)=0 \\ & 4 \lambda-2+64 \lambda-152+9 \lambda=0 \\ & 77 \lambda=154 \\ & \lambda=2 \end{aligned} $ Hence, the coordinates of $D$ are $(4,5,-2)$

Asked in: AP EAMCET 2017 (26 Apr Shift 1)

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