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

We are given the point \( P(2, 4, -1) \) and the equation of the line in symmetric form: \[ \frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{-9}. \] Let the parametric form of the line be: \[ x = -5 + t, \quad y = -3 + 4t, \quad z = 6 - 9t, \] where \( t \) is a parameter. Let \( Q(-5, -3, 6) \) be a point on the line and let the direction vector of the line be: \[ \mathbf{v} = \langle 1, 4, -9 \rangle. \] The vector \( \mathbf{PQ} \) from point \( P(2, 4, -1) \) to point \( Q(-5, -3, 6) \) 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 given by: \[ D = \frac{|\mathbf{PQ} \times \mathbf{v}|}{|\mathbf{v}|}. \] First, we compute the cross product \( \mathbf{PQ} \times \mathbf{v} \): \[ \mathbf{PQ} \times \mathbf{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -7 & -7 & 7 \\ 1 & 4 & -9 \end{vmatrix}. \] Expanding this determinant: \[ \mathbf{PQ} \times \mathbf{v} = \hat{i} \begin{vmatrix} -7 & 7 \\ 4 & -9 \end{vmatrix} - \hat{j} \begin{vmatrix} -7 & 7 \\ 1 & -9 \end{vmatrix} + \hat{k} \begin{vmatrix} -7 & -7 \\ 1 & 4 \end{vmatrix}. \] This results in: \[ \mathbf{PQ} \times \mathbf{v} = \hat{i}((-7)(-9) - (7)(4)) - \hat{j}((-7)(-9) - (7)(1)) + \hat{k}((-7)(4) - (-7)(1)). \] Simplifying: \[ \mathbf{PQ} \times \mathbf{v} = \hat{i}(63 - 28) - \hat{j}(63 - 7) + \hat{k}(-28 + 7). \] \[ \mathbf{PQ} \times \mathbf{v} = \hat{i}(35) - \hat{j}(56) + \hat{k}(-21). \] Thus, \[ \mathbf{PQ} \times \mathbf{v} = \langle 35, -56, -21 \rangle. \] Next, we compute the magnitude of \( \mathbf{PQ} \times \mathbf{v} \): \[ |\mathbf{PQ} \times \mathbf{v}| = \sqrt{35^2 + (-56)^2 + (-21)^2} = \sqrt{1225 + 3136 + 441} = \sqrt{4802}. \] Now, we compute the magnitude of \( \mathbf{v} \): \[ |\mathbf{v}| = \sqrt{1^2 + 4^2 + (-9)^2} = \sqrt{1 + 16 + 81} = \sqrt{98}. \] Finally, the distance is: \[ D = \frac{\sqrt{4802}}{\sqrt{98}} = \sqrt{\frac{4802}{98}} = \sqrt{49} = 7. \] Thus, the distance from the point \( P(2, 4, -1) \) to the line is \( \boxed{7} \).