Question

Find the foot of the perpendicular drawn from point $(2, -1, 5)$ to the line \[ \frac{x - 11}{10} = \frac{y + 2}{-4} = \frac{z + 8}{-11} \] Also, find the length of the perpendicular.

Hint:

To find the foot of a perpendicular from a point to a line, parameterize the line, then set the dot product of the vector from the point to a line point and the line's direction vector to zero.

Answer 1

The given line is in symmetric form: \[ \frac{x - 11}{10} = \frac{y + 2}{-4} = \frac{z + 8}{-11} \] Let this be equal to a parameter $t$. Then any point $P$ on the line can be written as: \[ P(t) = (10t + 11, -4t - 2, -11t - 8) \] Let $A = (2, -1, 5)$ be the given point. The foot of the perpendicular will be the point $P(t)$ such that vector $\vec{AP}$ is perpendicular to the direction vector $\vec{d} = \langle 10, -4, -11 \rangle$. So, $\vec{AP} \cdot \vec{d} = 0$: \[ \vec{AP} = \langle 10t + 11 - 2,\ -4t - 2 + 1,\ -11t - 8 - 5 \rangle = \langle 10t + 9,\ -4t - 1,\ -11t - 13 \rangle \] Taking dot product with $\vec{d}$: \[ (10t + 9)(10) + (-4t - 1)(-4) + (-11t - 13)(-11) = 0 \] Expanding: \[ 100t + 90 + 16t + 4 + 121t + 143 = 0\\ (100 + 16 + 121)t + (90 + 4 + 143) = 0\\ 237t + 237 = 0\\ t = -1 \] Substitute $t = -1$ into $P(t)$: \[ x = 10(-1) + 11 = 1,\quad y = -4(-1) - 2 = 2,\quad z = -11(-1) - 8 = 3 \] So the foot of the perpendicular is $(1, 2, 3)$. Now compute the length of perpendicular $AP$: \[ \vec{AP} = \langle 1 - 2,\ 2 - (-1),\ 3 - 5 \rangle = \langle -1,\ 3,\ -2 \rangle \] \[ |\vec{AP}| = \sqrt{(-1)^2 + 3^2 + (-2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \] % Final Answer Foot of the Perpendicular: $(1, 2, 3)$
Length of the Perpendicular: $\sqrt{14}$ \begin{center} \includegraphics[width=0.55\textwidth]{ig5.png} \end{center}