Question

Find the length of the perpendicular drawn from the point \( P(3, 2, 1) \) to the line \( \mathbf{r} = (7\hat{i} + 7\hat{j} + 6\hat{k}) + \lambda (-2\hat{i} + 2\hat{j} + 3\hat{k}) \).

Hint:

Use the cross-product formula for distance from a point to a line; simplify by checking integer results.

Answer 1

Point: \( P(3, 2, 1) \). Line: Point \( \mathbf{a} = (7, 7, 6) \), direction vector \( \mathbf{b} = (-2, 2, 3) \).
Distance formula: \[ d = \frac{|(\mathbf{p} - \mathbf{a}) \cdot (\mathbf{b} \times \mathbf{n})|}{|\mathbf{b}|}, \] or use: \[ d = \frac{|(\mathbf{p} - \mathbf{a}) \times \mathbf{b}|}{|\mathbf{b}|}. \] \[ \mathbf{p} - \mathbf{a} = (3-7, 2-7, 1-6) = (-4, -5, -5). \] \[ (\mathbf{p} - \mathbf{a}) \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-4 & -5 & -5
-2 & 2 & 3 \end{vmatrix} = \hat{i}(-15 - (-10)) - \hat{j}(-12 - 10) + \hat{k}(-8 - 10) = (-5, 22, -18). \] \[ |(\mathbf{p} - \mathbf{a}) \times \mathbf{b}| = \sqrt{(-5)^2 + 22^2 + (-18)^2} = \sqrt{25 + 484 + 324} = \sqrt{833}. \] \[ |\mathbf{b}| = \sqrt{(-2)^2 + 2^2 + 3^2} = \sqrt{4 + 4 + 9} = \sqrt{17}. \] \[ d = \frac{\sqrt{833}}{\sqrt{17}} = \sqrt{\frac{833}{17}} = \sqrt{49} = 7. \] Answer: 7 units.