Question

The length of the perpendicular from the point \( (1, -2, 5) \) on the line passing through \( (1, 2, 4) \) and parallel to the line given by \( x + y - z = 0 \) and \( x - 2y + 3z - 5 = 0 \) is:

• A. \( \frac{\sqrt{21}}{2} \)
• B. \( \frac{\sqrt{9}}{2} \)
• C. \( \frac{\sqrt{73}}{2} \)
• D. 1

Hint:

To find the perpendicular distance from a point to a line in 3D, use the formula that incorporates the direction ratios of the line and the coordinates of the point.

Answer 1

The Correct Option is A

To find the direction ratios of the given lines, we solve for their intersection and use determinants: \[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 1 & -1
1 & -2 & 3 \end{vmatrix} \] Expanding the determinant, we get: \[ \hat{i} (3 - 2) - \hat{j} (3 - (-1)) + \hat{k} (-2 - 1) = \hat{i} - 4\hat{j} - 3\hat{k} \] So, the direction ratios of the required line are \( (1, -4, -3) \). Equation of the Line: \[ \frac{x-1}{1} = \frac{y-2}{-4} = \frac{z-4}{-3} = \lambda \] Now, let \( P(1, -2, 5) \) be the point from which the perpendicular is drawn. The coordinates of point \( P \) in terms of \( \lambda \) are: \[ (1 + \lambda, 2 - 4\lambda, 4 - 3\lambda) \] Since the perpendicular distance is minimized when the dot product of \( \overrightarrow{PQ} \) with the direction vector is zero, we solve: \[ \overrightarrow{PQ} \cdot (1, -4, -3) = 0 \] Solving for \( \lambda \), we get \( \lambda = \frac{1}{2} \), and the perpendicular point is: \[ \left( \frac{1}{2}, 2, \frac{-5}{2} \right) \] Finally, we compute the perpendicular distance as: \[ PQ = \sqrt{\frac{21}{2}} \] Thus, the correct answer is: \[ \boxed{\frac{\sqrt{21}}{2}} \]