The coordinates of the foot of the perpendicu | vedclass

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(D) Let the given point be $P(2, -1, 5)$ and the line be $\vec{r} = (11 \hat{i} - 2 \hat{j} - 8 \hat{k}) + \lambda(10 \hat{i} - 4 \hat{j} - 11 \hat{k}

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$(1, 2, 3)$

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(D) Let the given point be $P(2, -1, 5)$ and the line be $\vec{r} = (11 \hat{i} - 2 \hat{j} - 8 \hat{k}) + \lambda(10 \hat{i} - 4 \hat{j} - 11 \hat{k})$.Any point $M$ on the line can be represented as $(10\lambda + 11, -4\lambda - 2, -11\lambda - 8)$.The direction ratios of the line $PM$ are $(10\lambda + 11 - 2, -4\lambda - 2 - (-1), -11\lambda - 8 - 5) = (10\lambda + 9, -4\lambda - 1, -11\lambda - 13)$.Since $PM$ is perpendicular to the given line,the dot product of the direction ratios of $PM$ and the line vector $(10, -4, -11)$ must be zero:$10(10\lambda + 9) - 4(-4\lambda - 1) - 11(-11\lambda - 13) = 0$.$100\lambda + 90 + 16\lambda + 4 + 121\lambda + 143 = 0$.$237\lambda + 237 = 0 \Rightarrow \lambda = -1$.Substituting $\lambda = -1$ into the coordinates of $M$:$M = (10(-1) + 11, -4(-1) - 2, -11(-1) - 8) = (1, 2, 3)$.

The coordinates of the foot of the perpendicular drawn from the point $2 \hat{i} - \hat{j} + 5 \hat{k}$ to the line $\vec{r} = (11 \hat{i} - 2 \hat{j} - 8 \hat{k}) + \lambda(10 \hat{i} - 4 \hat{j} - 11 \hat{k})$ are

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