The foot of the perpendicular drawn from the | vedclass

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(D) Let the points be $A(0, -11, 4)$ and $B(2, -3, 1)$. The equation of the line passing through $A$ and $B$ is given by $\frac{x-0}{2-0} = \frac{y-(-

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$(4, 5, -2)$

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(D) Let the points be $A(0, -11, 4)$ and $B(2, -3, 1)$. The equation of the line passing through $A$ and $B$ is given by $\frac{x-0}{2-0} = \frac{y-(-11)}{-3-(-11)} = \frac{z-4}{1-4} = \lambda$. This simplifies to $\frac{x}{2} = \frac{y+11}{8} = \frac{z-4}{-3} = \lambda$. So,any point $P$ on this line is $(2\lambda, 8\lambda-11, -3\lambda+4)$. Let $Q$ be the point $(1, 8, 4)$. The direction ratios of the line $PQ$ are $(2\lambda-1, 8\lambda-11-8, -3\lambda+4-4) = (2\lambda-1, 8\lambda-19, -3\lambda)$. Since $PQ$ is perpendicular to the line,the dot product of the direction ratios of $PQ$ and the line $(2, 8, -3)$ must be zero: $2(2\lambda-1) + 8(8\lambda-19) - 3(-3\lambda) = 0$. $4\lambda - 2 + 64\lambda - 152 + 9\lambda = 0$. $77\lambda - 154 = 0 \Rightarrow \lambda = 2$. Substituting $\lambda = 2$ into the coordinates of $P$: $x = 2(2) = 4$,$y = 8(2)-11 = 5$,$z = -3(2)+4 = -2$. Thus,the foot of the perpendicular is $(4, 5, -2)$.

The foot of the perpendicular drawn from the point $(1, 8, 4)$ on the line joining the points $(0, -11, 4)$ and $(2, -3, 1)$ is

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