Question

The square of the distance of the point P(5, 6, 7) from the line $\frac{x-2}{2} = \frac{y-5}{3} = \frac{z-2}{4}$ is equal to:

• A. 3
• B. 5
• C. 6
• D. 8

Answer 1

The Correct Option is C

The problem requires us to find the square of the distance from a point to a line in three-dimensional space. The point is \( P(5, 6, 7) \) and the line is given in symmetric form: \(\frac{x-2}{2} = \frac{y-5}{3} = \frac{z-2}{4}\). Follow these steps to solve the problem:

1. First, identify the direction vector and a point on the line. The equation of the line is given in the symmetric form \(\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}\), where \( (x_0, y_0, z_0) \) is a point on the line and \( (a, b, c) \) is the direction vector.
2. The line can be expressed as \( (x_0, y_0, z_0) = (2, 5, 2) \) and the direction vector \(\vec{d} = (2, 3, 4)\).
3. Use the formula for the distance \( D \) from a point \( (x_1, y_1, z_1) \) to the line defined by a point \( (x_0, y_0, z_0) \) and a direction vector \(\vec{d} = (a, b, c)\): \(D = \frac{\| \vec{d} \times \vec{PQ} \|}{\| \vec{d} \|}\), where \( \vec{PQ} \) is the vector from the line's point \( Q(x_0, y_0, z_0) \) to the point \( P(x_1, y_1, z_1) \).
4. Calculate \( \vec{PQ} = P - Q = (5 - 2, 6 - 5, 7 - 2) = (3, 1, 5) \).
5. Compute the cross product \(\vec{d} \times \vec{PQ}\): \(\vec{d} \times \vec{PQ} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 1 & 5 \end{vmatrix} = \hat{i}(3 \cdot 5 - 4 \cdot 1) - \hat{j}(2 \cdot 5 - 4 \cdot 3) + \hat{k}(2 \cdot 1 - 3 \cdot 3) = \hat{i}(15 - 4) - \hat{j}(10 - 12) + \hat{k}(2 - 9) = 11\hat{i} + 2\hat{j} - 7\hat{k}\).
6. Find the magnitude \(\|\vec{d} \times \vec{PQ}\|\): \(\|\vec{d} \times \vec{PQ}\| = \sqrt{11^2 + 2^2 + (-7)^2} = \sqrt{121 + 4 + 49} = \sqrt{174}\).
7. Calculate the magnitude of the direction vector \(\|\vec{d}\|\): \(\|\vec{d}\| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}\).
8. Substitute these values into the distance formula: \(D = \frac{\sqrt{174}}{\sqrt{29}}\).
9. Square the distance to get the square of the distance: \(D^2 = \frac{174}{29} = 6\).

Thus, the square of the distance of the point from the line is 6.