Question

Find the distance of the point $P(2, 4, -1)$ from the line \[ \frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{-9}. \]

Hint:

To find the distance from a point to a line, use the formula: \[ D = \frac{|\mathbf{PQ} \times \mathbf{v}|}{|\mathbf{v}|}, \] where \( \mathbf{PQ} \) is the vector from the point to a point on the line and \( \mathbf{v} \) is the direction vector of the line.

Answer 1

Given point \( P(2, 4, -1) \) and line \( \frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{-9} \). The line's parametric form is \( x = -5 + t, y = -3 + 4t, z = 6 - 9t \). A point on the line is \( Q(-5, -3, 6) \) and the direction vector is \( \mathbf{v} = \langle 1, 4, -9 \rangle \). The vector from \( P \) to \( Q \) is \( \mathbf{PQ} = \langle -5 - 2, -3 - 4, 6 - (-1) \rangle = \langle -7, -7, 7 \rangle \). The distance \( D \) from a point to a line is \( D = \frac{|\mathbf{PQ} \times \mathbf{v}|}{|\mathbf{v}|} \). The cross product is \( \mathbf{PQ} \times \mathbf{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -7 & -7 & 7 \\ 1 & 4 & -9 \end{vmatrix} = \hat{i}(63 - 28) - \hat{j}(63 - 7) + \hat{k}(-28 + 7) = 35\hat{i} - 56\hat{j} - 21\hat{k} = \langle 35, -56, -21 \rangle \). The magnitude of \( \mathbf{PQ} \times \mathbf{v} \) is \( |\mathbf{PQ} \times \mathbf{v}| = \sqrt{35^2 + (-56)^2 + (-21)^2} = \sqrt{1225 + 3136 + 441} = \sqrt{4802} \). The magnitude of \( \mathbf{v} \) is \( |\mathbf{v}| = \sqrt{1^2 + 4^2 + (-9)^2} = \sqrt{1 + 16 + 81} = \sqrt{98} \). The distance is \( D = \frac{\sqrt{4802}}{\sqrt{98}} = \sqrt{\frac{4802}{98}} = \sqrt{49} = 7 \). The distance from point \( P(2, 4, -1) \) to the line is \( \boxed{7} \).