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

Answer 2

Step 1: Given information.
Point: \( A(1, 2, 2) \)
Line \( L_1: \frac{x - 1}{1} = \frac{y + 1}{-1} = \frac{z - 2}{2} \)
Line \( L_2: \mathbf{r} = (-\hat{i} + \hat{j} - 2\hat{k}) + \lambda (\hat{i} - \hat{j} + \hat{k}) \).

We need to find the foot of the perpendicular \( P \) from point \( A \) to line \( L_1 \) and the point of intersection \( Q \) of \( L_1 \) and \( L_2 \), then evaluate \( 2(PQ)^2 \).

Step 2: Equation of the first line \( L_1 \).
Let the parameter of \( L_1 \) be \( t \). Then the coordinates of any point on \( L_1 \) are:
\[ (x, y, z) = (1 + t, -1 - t, 2 + 2t) \] The direction ratios of \( L_1 \) are \( (1, -1, 2) \).

Step 3: Find the foot of the perpendicular \( P \).
Let \( P(1 + t, -1 - t, 2 + 2t) \) be the foot of the perpendicular from \( A(1, 2, 2) \) to \( L_1 \).

The vector \( \overrightarrow{AP} = (1 + t - 1, -1 - t - 2, 2 + 2t - 2) = (t, -3 - t, 2t) \).

For \( AP \) to be perpendicular to the line \( L_1 \), \( \overrightarrow{AP} \cdot \text{direction of } L_1 = 0 \).

Direction of \( L_1 = (1, -1, 2) \). Thus:
\[ (t, -3 - t, 2t) \cdot (1, -1, 2) = 0 \] \[ t(1) + (-3 - t)(-1) + 2t(2) = 0 \] \[ t + 3 + t + 4t = 0 \Rightarrow 6t + 3 = 0 \Rightarrow t = -\frac{1}{2}. \] Hence, \( P \) has coordinates:
\[ P(1 + t, -1 - t, 2 + 2t) = \left( \frac{1}{2}, -\frac{1}{2}, 1 \right). \]

Step 4: Equation of line \( L_2 \).
Let parameter \( \lambda \) correspond to line \( L_2 \):
\[ (x, y, z) = (-1 + \lambda, 1 - \lambda, -2 + \lambda) \] Direction ratios = \( (1, -1, 1) \).

Step 5: Intersection of \( L_1 \) and \( L_2 \).
For intersection, their coordinates must be equal:
\[ 1 + t = -1 + \lambda, \quad -1 - t = 1 - \lambda, \quad 2 + 2t = -2 + \lambda \] From the first equation:
\[ \lambda = t + 2. \] From the second equation:
\[ -1 - t = 1 - (t + 2) \Rightarrow -1 - t = -1 - t \text{ (satisfied)}. \] From the third equation:
\[ 2 + 2t = -2 + \lambda = -2 + (t + 2) \Rightarrow 2 + 2t = t \Rightarrow t = -2. \] Then \( \lambda = t + 2 = 0. \)

Hence, intersection point \( Q \) is found by putting \( t = -2 \) in \( L_1 \):
\[ Q(1 - 2, -1 + 2, 2 - 4) = (-1, 1, -2). \]

Step 6: Find \( PQ \).
\( P\left(\frac{1}{2}, -\frac{1}{2}, 1\right) \), \( Q(-1, 1, -2) \).

\[ PQ^2 = \left(\frac{1}{2} + 1\right)^2 + \left(-\frac{1}{2} - 1\right)^2 + (1 + 2)^2 \] \[ PQ^2 = \left(\frac{3}{2}\right)^2 + \left(-\frac{3}{2}\right)^2 + 3^2 = \frac{9}{4} + \frac{9}{4} + 9 = \frac{9 + 9 + 36}{4} = \frac{54}{4} = \frac{27}{2}. \] Hence,
\[ 2(PQ)^2 = 2 \times \frac{27}{2} = 27. \]

Final Answer:
\[ \boxed{27} \]