Question

A straight line drawn from the point P(1,3, 2), parallel to the line \(\frac{x-2}{1}=\frac{y-4}{2}=\frac{z-6}{1}\), intersects the plane L₁ : x - y + 3z = 6 at the point Q. Another straight line which passes through Q and is perpendicular to the plane L₁ intersects the plane L₂ : 2x - y + z = -4 at the point R. Then which of the following statements is (are) TRUE ?

• A. The length of the line segment PQ is \(\sqrt6\)
• B. The coordinates of R are (1, 6, 3)
• C. The centroid of the triangle PQR is \((\frac{4}{3},\frac{14}{3},\frac{5}{3})\)
• D. The perimeter of the triangle PQR is \(\sqrt2+\sqrt6+\sqrt{11}\)

Answer 1

The Correct Option is A, C

1. Equation of Line Parallel to $ \frac{x - 2}{1} = \frac{y - 4}{2} = \frac{z - 6}{1} $ Through Point $P(1, 3, 2)$:
We are given the equation of a line and the point $P(1, 3, 2)$. The equation of the line parallel to the given one can be written as:

$ \frac{x - 1}{1} = \frac{y - 3}{2} = \frac{z - 2}{1} = \lambda $ (Let)

Substituting $ \lambda = 1 $ in the above equation, we get:

$ Q(2, 5, 3) $

2. Equation of Line Through $ Q(2, 5, 3) $ Perpendicular to $ L_1 $:
The equation of the line through $ Q(2, 5, 3) $ perpendicular to $ L_1 $ is given by:

$ \frac{x - 2}{1} = \frac{y - 5}{-1} = \frac{z - 3}{3} = \mu $ (Let)

3. Putting Any Point $ (\mu + 2, -\mu + 5, 3\mu + 3) $ in $ L_2 $:
We substitute the coordinates of the point in $ L_2 $ and simplify the calculations.

4. Point $ R(1, 6, 0) $:
After simplification, we find that point $ R(1, 6, 0) $ lies on $ L_2 $. Now, let's calculate the distances:

5. Distance Between Points $ P $ and $ Q $:
We use the distance formula to find $ PQ $:

$ PQ = \sqrt{(2 - 1)^2 + (5 - 3)^2 + (3 - 2)^2} = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6} $

6. Distance Between Points $ Q $ and $ R $:
Similarly, we calculate $ QR $:

$ QR = \sqrt{(1 - 2)^2 + (6 - 5)^2 + (0 - 3)^2} = \sqrt{(-1)^2 + (1)^2 + (-3)^2} = \sqrt{1 + 1 + 9} = \sqrt{11} $

7. Distance Between Points $ P $ and $ R $:
Next, we calculate $ PR $:

$ PR = \sqrt{(1 - 1)^2 + (6 - 3)^2 + (0 - 2)^2} = \sqrt{0^2 + 3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} $

8. Final Calculation:
The total distance is the sum of $ PQ $, $ QR $, and $ PR $:

$ PQ + QR + PR = \sqrt{6} + \sqrt{11} + \sqrt{13} $

Final Answer:

• (A) $ PQ = \sqrt{6} $
• (B) $ R(1, 6, 0) $
• (C) Centroid: $ \left( \frac{4}{3}, \frac{14}{3}, \frac{5}{3} \right) $
• (D) $ PQ + QR + PR = \sqrt{6} + \sqrt{11} + \sqrt{13} $

Answer 2

To solve the problem, we analyze the points \(P\), \(Q\), and \(R\) step-by-step using the given information.

Given:
Point \(P = (1, 3, 2)\)
Line parallel to \(\frac{x-2}{1} = \frac{y-4}{2} = \frac{z-6}{1}\) passing through \(P\)
Plane \(L_1: x - y + 3z = 6\)
Plane \(L_2: 2x - y + z = -4\)

1. Find point \(Q\):
The line through \(P\) parallel to the given line has direction vector \(\vec{d} = (1, 2, 1)\). Parametric equations of the line through \(P\):
\[ x = 1 + t, \quad y = 3 + 2t, \quad z = 2 + t \]

Find \(t\) such that point \(Q\) lies on plane \(L_1\):
\[ (1 + t) - (3 + 2t) + 3(2 + t) = 6 \] \[ 1 + t - 3 - 2t + 6 + 3t = 6 \] \[ (1 - 3 + 6) + (t - 2t + 3t) = 6 \] \[ 4 + 2t = 6 \implies 2t = 2 \implies t = 1 \]

Coordinates of \(Q\):
\[ x = 1 + 1 = 2, \quad y = 3 + 2 \times 1 = 5, \quad z = 2 + 1 = 3 \] So, \(Q = (2, 5, 3)\).

2. Find point \(R\):
The line through \(Q\) perpendicular to plane \(L_1\) has direction vector equal to the normal vector of \(L_1\), \(\vec{n}_1 = (1, -1, 3)\). Parametric form:

\[ x = 2 + s, \quad y = 5 - s, \quad z = 3 + 3s \]

Find \(s\) such that point \(R\) lies on plane \(L_2\):

\[ 2(2 + s) - (5 - s) + (3 + 3s) = -4 \] \[ 4 + 2s - 5 + s + 3 + 3s = -4 \] \[ (4 - 5 + 3) + (2s + s + 3s) = -4 \] \[ 2 + 6s = -4 \implies 6s = -6 \implies s = -1 \]

Coordinates of \(R\):
\[ x = 2 - 1 = 1, \quad y = 5 + 1 = 6, \quad z = 3 - 3 = 0 \] So, \(R = (1, 6, 0)\).

3. Check the statements:

• Length \(PQ\):
  \[ PQ = \sqrt{(2 - 1)^2 + (5 - 3)^2 + (3 - 2)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \] Statement 1 is TRUE.
• Coordinates of \(R\):
  Calculated \(R = (1, 6, 0)\), given option is \((1, 6, 3)\), so Statement 2 is FALSE.
• Centroid of \(\triangle PQR\):
  \[ G = \left(\frac{1 + 2 + 1}{3}, \frac{3 + 5 + 6}{3}, \frac{2 + 3 + 0}{3}\right) = \left(\frac{4}{3}, \frac{14}{3}, \frac{5}{3}\right) \] Statement 3 is TRUE.
• Perimeter of \(\triangle PQR\):
  Calculate distances:
  \[ PQ = \sqrt{6} \quad \text{(already calculated)} \] \[ QR = \sqrt{(1 - 2)^2 + (6 - 5)^2 + (0 - 3)^2} = \sqrt{1 + 1 + 9} = \sqrt{11} \] \[ PR = \sqrt{(1 - 1)^2 + (6 - 3)^2 + (0 - 2)^2} = \sqrt{0 + 9 + 4} = \sqrt{13} \] Sum: \[ \sqrt{6} + \sqrt{11} + \sqrt{13} \neq \sqrt{2} + \sqrt{6} + \sqrt{11} \] Statement 4 is FALSE.

Final Answer:
Statements 1 and 3 are TRUE.