Let $ L_1 $ be the line of intersection of the planes given by the equations
$$
2x + 3y + z = 4 \quad \text{and} \quad x + 2y + z = 5.
$$
Let $ L_2 $ be the line passing through the point $ P(2, -1, 3) $ and parallel to $ L_1 $. Let $ M $ denote the plane given by the equation
$$
2x + y - 2z = 6.
$$
Suppose that the line $ L_2 $ meets the plane $ M $ at the point $ Q $. Let $ R $ be the foot of the perpendicular drawn from $ P $ to the plane $ M $.
Then which of the following statements is (are) TRUE?
Show Hint
- The length of the line segment \( PQ \) is \( 9\sqrt{3} \)
- The length of the line segment \( QR \) is 15
- The area of \( \triangle PQR \) is \( \dfrac{3}{2} \sqrt{234} \)
The acute angle between the line segments \( PQ \) and \( PR \) is \( \cos^{-1} \left(\dfrac{1}{2\sqrt{3}}\right) \)
The Correct Option is A, C
Solution and Explanation
Step 1: Find the equation of line \( L_1 \)
\( L_1 \) is the line of intersection of the planes given by: \[ 2x + 3y + z = 4 \quad (1) \] \[ x + 2y + z = 5 \quad (2) \] To find the line of intersection, subtract equation (2) from equation (1): \[ (2x + 3y + z) - (x + 2y + z) = 4 - 5 \] \[ x + y = -1 \quad \text{or} \quad x = -1 - y \quad (3) \] Substitute \( x = -1 - y \) into equation (2): \[ (-1 - y) + 2y + z = 5 \] \[ -1 - y + 2y + z = 5 \] \[ y + z = 6 \quad \text{or} \quad z = 6 - y \quad (4) \] Now express \( x \) and \( z \) in terms of \( y \): - \( x = -1 - y \)
- \( z = 6 - y \) Let’s use \( y = t \) as the parameter. Then: - \( x = -1 - t \)
- \( y = t \)
- \( z = 6 - t \) To find a point on \( L_1 \), set \( t = 0 \):
- \( x = -1 \), \( y = 0 \), \( z = 6 \)
So, a point on \( L_1 \) is \( (-1, 0, 6) \).
The direction vector of \( L_1 \) can be found by observing the coefficients of \( t \):
- As \( t \) changes, \( x = -1 - t \), \( y = t \), \( z = 6 - t \), so the direction vector is \( (-1, 1, -1) \).
Thus, the parametric equation of \( L_1 \) is: \[ x = -1 - t, \quad y = t, \quad z = 6 - t \] Or in symmetric form: \[ \frac{x + 1}{-1} = \frac{y}{1} = \frac{z - 6}{-1} \] Step 2: Find the equation of line \( L_2 \)
\( L_2 \) passes through the point \( P(2, -1, 3) \) and is parallel to \( L_1 \). Since \( L_2 \) is parallel to \( L_1 \), it has the same direction vector, \( (-1, 1, -1) \). The parametric equation of \( L_2 \) passing through \( P(2, -1, 3) \) with direction vector \( (-1, 1, -1) \) is: \[ x = 2 - s, \quad y = -1 + s, \quad z = 3 - s \] Step 3: Find point \( Q \), the intersection of \( L_2 \) with plane \( M \)
Plane \( M \) is given by: \[ 2x + y - 2z = 6 \quad (5) \] Substitute the parametric equations of \( L_2 \) into the equation of plane \( M \): \[ 2(2 - s) + (-1 + s) - 2(3 - s) = 6 \] \[ 4 - 2s - 1 + s - 6 + 2s = 6 \] \[ (4 - 1 - 6) + (-2s + s + 2s) = 6 \] \[ -3 + s = 6 \] \[ s = 9 \] Now, find the coordinates of \( Q \) by substituting \( s = 9 \) into the equation of \( L_2 \):
- \( x = 2 - 9 = -7 \)
- \( y = -1 + 9 = 8 \)
- \( z = 3 - 9 = -6 \) So, point \( Q \) is \( (-7, 8, -6) \).
Step 4: Find point \( R \), the foot of the perpendicular from \( P \) to plane \( M \)
The normal vector to plane \( M \), \( 2x + y - 2z = 6 \), is \( (2, 1, -2) \). The line from \( P(2, -1, 3) \) perpendicular to plane \( M \) has direction vector \( (2, 1, -2) \). The parametric equation of the line from \( P \) in the direction of the normal is: \[ x = 2 + 2t, \quad y = -1 + t, \quad z = 3 - 2t \] Find where this line intersects plane \( M \): \[ 2(2 + 2t) + (-1 + t) - 2(3 - 2t) = 6 \] \[ 4 + 4t - 1 + t - 6 + 4t = 6 \] \[ (4 - 1 - 6) + (4t + t + 4t) = 6 \] \[ -3 + 9t = 6 \] \[ 9t = 9 \] \[ t = 1 \] Substitute \( t = 1 \):
- \( x = 2 + 2(1) = 4 \)
- \( y = -1 + 1 = 0 \)
- \( z = 3 - 2(1) = 1 \) So, point \( R \) is \( (4, 0, 1) \).
Step 5: Evaluate each option (A) The length of the line segment \( PQ \) is \( 9\sqrt{3} \)
- \( P = (2, -1, 3) \), \( Q = (-7, 8, -6) \)
- Vector \( \overrightarrow{PQ} = Q - P = (-7 - 2, 8 - (-1), -6 - 3) = (-9, 9, -9) \)
- Length of \( PQ = \sqrt{(-9)^2 + 9^2 + (-9)^2} = \sqrt{81 + 81 + 81} = \sqrt{243} = \sqrt{81 \cdot 3} = 9\sqrt{3} \)
Option (A) is true. (B) The length of the line segment \( QR \) is 15
- \( Q = (-7, 8, -6) \), \( R = (4, 0, 1) \)
- Vector \( \overrightarrow{QR} = R - Q = (4 - (-7), 0 - 8, 1 - (-6)) = (11, -8, 7) \)
- Length of \( QR = \sqrt{11^2 + (-8)^2 + 7^2} = \sqrt{121 + 64 + 49} = \sqrt{234} \) Since \( \sqrt{234} \approx 15.297 \), which is not exactly 15, let’s compute \( \sqrt{234} \) more precisely:
- \( 15^2 = 225 \), \( 16^2 = 256 \), so \( \sqrt{234} \) is between 15 and 16, closer to 15 but not exactly 15.
Option (B) is false. (C) The area of \( \triangle PQR \) is \( \frac{3}{2}\sqrt{234} \)
- Vectors \( \overrightarrow{PQ} = (-9, 9, -9) \), \( \overrightarrow{PR} = R - P = (4 - 2, 0 - (-1), 1 - 3) = (2, 1, -2) \).
- Compute the cross product \( \overrightarrow{PQ} \times \overrightarrow{PR} \): \[ \overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\-9 & 9 & -9 \\2 & 1 & -2 \end{vmatrix} \] - \( \mathbf{i} \)-component: \( (9)(-2) - (-9)(1) = -18 + 9 = -9 \)
- \( \mathbf{j} \)-component: \(-[(-9)(-2) - (-9)(2)] = -[18 - (-18)] = -36 \)
- \( \mathbf{k} \)-component: \( (-9)(1) - (9)(2) = -9 - 18 = -27 \)
- So, \( \overrightarrow{PQ} \times \overrightarrow{PR} = (-9, -36, -27) \). - Magnitude: \( \sqrt{(-9)^2 + (-36)^2 + (-27)^2} = \sqrt{81 + 1296 + 729} = \sqrt{2106} = \sqrt{9 \cdot 234} = 3\sqrt{234} \). - Area of \( \triangle PQR = \frac{1}{2} |\overrightarrow{PQ} \times \overrightarrow{PR}| = \frac{1}{2} \cdot 3\sqrt{234} = \frac{3}{2}\sqrt{234} \). Option (C) is true. (D) The acute angle between the line segments \( PQ \) and \( PR \) is \( \cos^{-1}\left(\frac{1}{2\sqrt{3}}\right) \)
- \( \overrightarrow{PQ} = (-9, 9, -9) \), \( \overrightarrow{PR} = (2, 1, -2) \).
- Dot product: \( \overrightarrow{PQ} \cdot \overrightarrow{PR} = (-9)(2) + (9)(1) + (-9)(-2) = -18 + 9 + 18 = 9 \).
- Magnitudes: \( |\overrightarrow{PQ}| = 9\sqrt{3} \), \( |\overrightarrow{PR}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \).
- \( \cos \theta = \frac{\overrightarrow{PQ} \cdot \overrightarrow{PR}}{|\overrightarrow{PQ}| |\overrightarrow{PR}|} = \frac{9}{(9\sqrt{3}) \cdot 3} = \frac{9}{27\sqrt{3}} = \frac{1}{3\sqrt{3}} = \frac{\sqrt{3}}{9} \).
The given angle has \( \cos \theta = \frac{1}{2\sqrt{3}} \). Compare:
- \( \frac{\sqrt{3}}{9} \approx \frac{1.732}{9} \approx 0.192 \)
- \( \frac{1}{2\sqrt{3}} = \frac{1}{2 \cdot 1.732} \approx \frac{1}{3.464} \approx 0.289 \)
These values are not equal, so the angles are different. Option (D) is false.
Final Answer: The true statements are:
- (A) The length of the line segment \( PQ \) is \( 9\sqrt{3} \).
- (C) The area of \( \triangle PQR \) is \( \frac{3}{2}\sqrt{234} \).
Thus, the correct options are (A) and (C).
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