Question

Let P be the foot of the perpendicular from the point \( (1, 2, 2) \) on the line \[ \frac{x-1}{1} = \frac{y + 1}{-1} = \frac{z - 2}{2} \] Let the line \( \mathbf{r} = (-\hat{i} + \hat{j} - 2\hat{k}) + \lambda (\hat{i} - \hat{j} + \hat{k})\), \( \lambda \in \mathbb{R} \), intersect the line \(L\) at \(Q\). Then \( 2(PQ)^2 \) is equal to:

• A. 27
• B. 25
• C. 29
• D. 19

Hint:

For finding the foot of a perpendicular in 3D geometry, equating parametric forms of the given lines is the key step.

Answer 1

The Correct Option is A

Step 1: Finding coordinates of points. The general point on line \(L\) is: \[ Q(1 + \mu, -1 - \mu, 2 + 2\mu) \] The given point \(P = (1, 2, 2)\).

Step 2: Finding perpendicular condition. We use the condition that the line joining \(P\) and \(Q\) must be perpendicular to the direction vector of the line: \[ \mathbf{AP} \cdot \mathbf{d} = 0 \] Where \( \mathbf{AP} = (1 + \mu - 1, -1 - \mu - 2, 2 + 2\mu - 2) = (\mu, -3 - \mu, 2\mu) \) And direction vector \( \mathbf{d} = (1, -1, 2) \) Dot product condition: \[ \mu + (-3 - \mu)(-1) + 2\mu \times 2 = 0 \] Simplifying: \[ \mu + 3 + \mu + 4\mu = 0 \] \[ 6\mu = -3 \implies \mu = -\frac{1}{2} \] Substituting \( \mu = -\frac{1}{2} \) in the parametric form: \[ Q = \left(1 - \frac{1}{2}, -1 + \frac{1}{2}, 2 - 1\right) = \left( \frac{1}{2}, -\frac{1}{2}, 1 \right) \]

Step 3: Finding distance \( PQ \) \[ PQ = \sqrt{ \left( 1 - \frac{1}{2} \right)^2 + \left( 2 - \left(-\frac{1}{2}\right) \right)^2 + (2 - 1)^2 } \] \[ PQ = \sqrt{ \left( \frac{1}{2} \right)^2 + \left( \frac{5}{2} \right)^2 + (1)^2 } \] \[ PQ = \sqrt{ \frac{1}{4} + \frac{25}{4} + 1 } = \sqrt{ \frac{1 + 25 + 4}{4} } \] \[ PQ = \sqrt{\frac{30}{4}} = \frac{\sqrt{30}}{2} \]

Step 4: Final Calculation \[ 2(PQ)^2 = 2 \times \left( \frac{30}{4} \right) = 2 \times 7.5 = 27 \]