Question

The perpendicular distance of the line \( \frac{x - 1}{2} = \frac{y + 2}{-1} = \frac{z + 3}{2} \) from the point \( P(2, -10, 1) \) is:

• A. \( 4\sqrt{3} \)
• B. \( 5\sqrt{2} \)
• C. \( 3\sqrt{5} \)
• D. \( 6 \)

Hint:

For finding the perpendicular distance from a point to a line in three dimensions: - Convert the given symmetric form of the line to its direction ratios. - Use the distance formula \( d = \frac{| \mathbf{a} \cdot (\mathbf{r_0} - \mathbf{r_1}) |}{|\mathbf{a}|} \) where \( \mathbf{a} \) is the direction vector of the line.

Answer 1

The Correct Option is A

The given line is in symmetric form. To find the perpendicular distance from the point \( P(2, -10, 1) \) to the line, we use the formula for the distance from a point to a line in space: \[ d = \frac{| \mathbf{a} \cdot (\mathbf{r_0} - \mathbf{r_1}) |}{|\mathbf{a}|} \] where \( \mathbf{a} \) is the direction vector of the line, \( \mathbf{r_0} \) is the position vector of the point, and \( \mathbf{r_1} \) is a point on the line. The direction vector \( \mathbf{a} \) is \( \langle 2, -1, 2 \rangle \), and \( \mathbf{r_1} = (1, -2, -3) \). The vector \( \mathbf{r_0} - \mathbf{r_1} = \langle 2 - 1, -10 + 2, 1 + 3 \rangle = \langle 1, -8, 4 \rangle \). Now, applying the formula for distance: \[ d = \frac{| \langle 2, -1, 2 \rangle \cdot \langle 1, -8, 4 \rangle |}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{| 2(1) + (-1)(-8) + 2(4) |}{\sqrt{4 + 1 + 4}} = \frac{| 2 + 8 + 8 |}{3} = \frac{18}{3} = 6. \] Thus, the perpendicular distance is \( 6 \).