Question

The distance of the point \( Q(0, 2, -2) \) from the line passing through the point \( P(5, -4, 3) \) and perpendicular to the lines \[ \vec{r} = (-3\hat{i} + 2\hat{k}) + \lambda (2\hat{i} + 3\hat{j} + 5\hat{k}), \quad \lambda \in \mathbb{R} \] and \[ \vec{r} = (\hat{i} - 2\hat{j} + \hat{k}) + \mu (-\hat{i} + 3\hat{j} + 2\hat{k}), \quad \mu \in \mathbb{R} \] is

• A. \( \sqrt{86} \)
• B. \( \sqrt{20} \)
• C. \( \sqrt{54} \)
• D. \( \sqrt{74} \)

Answer 1

The Correct Option is D

A vector in the direction of the required line can be obtained by the cross product of:

\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ -1 & 3 & 2 \end{vmatrix} = -9\hat{i} - 9\hat{j} + 9\hat{k} \]

Required line:

\[ \vec{r} = (5\hat{i} - 4\hat{j} + 3\hat{k}) + \lambda (-9\hat{i} - 9\hat{j} + 9\hat{k}) \] \[ \vec{r} = (5\hat{i} - 4\hat{j} + 3\hat{k}) + \lambda (1\hat{i} + \hat{j} - \hat{k}) \]

Now, the distance of \( (0, 2, -2) \) is:

\[ \text{P.V. of } P = (5 + \lambda)\hat{i} + (-4 + \lambda)\hat{j} + (3 - \lambda)\hat{k} \] \[ \vec{AP} = (5 + \lambda)\hat{i} + (-6 + \lambda)\hat{j} + (5 - \lambda)\hat{k} \] \[ \vec{AP} \cdot (\hat{i} + \hat{j} - \hat{k}) = 0 \] \[ 5 + \lambda - 6 + \lambda - 5 + \lambda = 0 \quad \implies \quad \lambda = 2 \]

\[ |\vec{AP}| = \sqrt{(5 + \lambda)^2 + (-6 + \lambda)^2 + (5 - \lambda)^2} \] \[ |\vec{AP}| = \sqrt{49 + 16 + 9} = \sqrt{74} \]