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

Step 1: Understanding the Question:
We need to find the shortest distance from a given point to a given line in 3D space, and then square that distance.

Step 2: Key Formula or Approach:
Let the line be given by \(\mathbf{r} = \mathbf{a} + t\mathbf{d}\), where \(\mathbf{a}\) is a point on the line and \(\mathbf{d}\) is the direction vector of the line. Let the given point be \(P\) with position vector \(\mathbf{p}\).
The distance from \(P\) to the line is given by:
\[ \text{Distance} = \frac{|\vec{AP} \times \mathbf{d}|}{|\mathbf{d}|} \] where \(\vec{AP} = \mathbf{p} - \mathbf{a}\). We are asked for the square of this distance.

Step 3: Detailed Explanation:
From the equation of the line: \[ \frac{x-2}{2} = \frac{y-5}{3} = \frac{z-2}{4} \] A point on the line is \(A = (2, 5, 2)\), so: \[ \mathbf{a} = 2\mathbf{i} + 5\mathbf{j} + 2\mathbf{k} \] Direction vector: \[ \mathbf{d} = 2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k} \] Given point: \[ P = (5, 6, 7), \quad \mathbf{p} = 5\mathbf{i} + 6\mathbf{j} + 7\mathbf{k} \] Vector \(\vec{AP}\): \[ \vec{AP} = \mathbf{p} - \mathbf{a} = 3\mathbf{i} + 1\mathbf{j} + 5\mathbf{k} \] Cross product: \[ \vec{AP} \times \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & 5 \\ 2 & 3 & 4 \end{vmatrix} \] \[ = \mathbf{i}(1 \cdot 4 - 5 \cdot 3) - \mathbf{j}(3 \cdot 4 - 5 \cdot 2) + \mathbf{k}(3 \cdot 3 - 1 \cdot 2) \] \[ = \mathbf{i}(4 - 15) - \mathbf{j}(12 - 10) + \mathbf{k}(9 - 2) \] \[ = -11\mathbf{i} - 2\mathbf{j} + 7\mathbf{k} \] Magnitude: \[ |\vec{AP} \times \mathbf{d}| = \sqrt{(-11)^2 + (-2)^2 + 7^2} = \sqrt{174} \] \[ |\mathbf{d}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{29} \] Distance: \[ \frac{\sqrt{174}}{\sqrt{29}} = \sqrt{\frac{174}{29}} = \sqrt{6} \] Square of distance: \[ (\sqrt{6})^2 = 6 \]

Step 4: Final Answer:
The square of the distance is \(6\).