Question

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

Hint:

To find the perpendicular distance from a point to a line, use the formula involving the cross product of vectors and the magnitude of the direction vector.

Answer 1

Step 1: The equation of the line is given in parametric form. The point on the line is \( \overrightarrow{r_0} = 7\hat{i} + 7\hat{j} + 6\hat{k} \) and the direction vector is \( \overrightarrow{v} = -2\hat{i} + 2\hat{j} + 3\hat{k} \). Step 2: The formula for the length of the perpendicular from a point \( P(x_1, y_1, z_1) \) to a line is: \[ d = \frac{| (\overrightarrow{r_1} - \overrightarrow{r_0}) \cdot (\overrightarrow{v}) |}{|\overrightarrow{v}|} \] Where \( \overrightarrow{r_1} = 3\hat{i} + 2\hat{j} + 1\hat{k} \) is the point \( P(3, 2, 1) \). Step 3: First, compute \( \overrightarrow{r_1} - \overrightarrow{r_0} \): \[ \overrightarrow{r_1} - \overrightarrow{r_0} = (3\hat{i} + 2\hat{j} + 1\hat{k}) - (7\hat{i} + 7\hat{j} + 6\hat{k}) \] \[ = (-4\hat{i} - 5\hat{j} - 5\hat{k}) \] Step 4: Compute the cross product \( (\overrightarrow{r_1} - \overrightarrow{r_0}) \times \overrightarrow{v} \): \[ (\overrightarrow{r_1} - \overrightarrow{r_0}) \times \overrightarrow{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-4 & -5 & -5
-2 & 2 & 3 \end{vmatrix} \] \[ = \hat{i} \begin{vmatrix} -5 & -5
2 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} -4 & -5
-2 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} -4 & -5
-2 & 2 \end{vmatrix} \] \[ = \hat{i}((-5)(3) - (-5)(2)) - \hat{j}((-4)(3) - (-5)(-2)) + \hat{k}((-4)(2) - (-5)(-2)) \] \[ = \hat{i}(-15 + 10) - \hat{j}(-12 - 10) + \hat{k}(-8 - 10) \] \[ = -5\hat{i} + 22\hat{j} - 18\hat{k} \] Step 5: Find the magnitude of the cross product: \[ |(\overrightarrow{r_1} - \overrightarrow{r_0}) \times \overrightarrow{v}| = \sqrt{(-5)^2 + 22^2 + (-18)^2} \] \[ = \sqrt{25 + 484 + 324} = \sqrt{833} \] Step 6: Compute the magnitude of \( \overrightarrow{v} \): \[ |\overrightarrow{v}| = \sqrt{(-2)^2 + 2^2 + 3^2} = \sqrt{4 + 4 + 9} = \sqrt{17} \] Step 7: Finally, compute the perpendicular distance: \[ d = \frac{\sqrt{833}}{\sqrt{17}} = \sqrt{\frac{833}{17}} = \sqrt{49} = 7 \] Thus, the length of the perpendicular is: \[ \boxed{7} \]