Question

Let the line passing through the points P (2, –1, 2) and Q (5, 3, 4) meet the plane x – y + z = 4 at the point R. Then the distance of the point R from the plane x + 2y + 3z + 2 = 0 measured parallel to the line \(\frac{x-7}{2}=\frac{y+3}{2}=\frac{z-2}{1}\) is equal to

• A. \(\sqrt{189}\)
• B. \(\sqrt{31}\)
• C. \(\sqrt{61}\)
• D. 3

Answer 1

The Correct Option is D

We are tasked to find the distance between points \( R \) and \( T \) on the given lines: 

• Line 1: \(\frac{x-5}{3} = \frac{y-3}{4} = \frac{z-4}{2}\)
• Line 2: \(\frac{x+1}{2} = \frac{y+5}{2} = z\)

Step 1: Parametric Equation of Line 1

The parametric coordinates of a point \( R(\lambda) \) on Line 1 are: \[ R(\lambda) = (3\lambda + 5, 4\lambda + 3, 2\lambda + 4). \]

Step 2: Parametric Equation of Line 2

The parametric coordinates of a point \( T(\mu) \) on Line 2 are: \[ T(\mu) = (2\mu - 1, 2\mu - 5, \mu). \]

Step 3: Find \( R \) and \( T \)

Point \( R \):

Substitute \( R(\lambda) \) into the given plane equation: \[ 3x + 4y + 2z = 4. \] Substituting \( R(\lambda) = (3\lambda + 5, 4\lambda + 3, 2\lambda + 4) \): \[ 3(3\lambda + 5) + 4(4\lambda + 3) + 2(2\lambda + 4) = 4. \] Simplify: \[ 9\lambda + 15 + 16\lambda + 12 + 4\lambda + 8 = 4. \] \[ 29\lambda + 35 = 4 \implies \lambda = -2. \] Substituting \( \lambda = -2 \): \[ R = (3(-2) + 5, 4(-2) + 3, 2(-2) + 4) = (-1, -5, 0). \]

Point \( T \):

Substitute \( T(\mu) \) into the given plane equation: \[ 3x + 4y + 2z = 4. \] Substituting \( T(\mu) = (2\mu - 1, 2\mu - 5, \mu) \): \[ 3(2\mu - 1) + 4(2\mu - 5) + 2\mu = 4. \] Simplify: \[ 6\mu - 3 + 8\mu - 20 + 2\mu = 4. \] \[ 16\mu - 23 = 4 \implies \mu = 1. \] Substituting \( \mu = 1 \): \[ T = (2(1) - 1, 2(1) - 5, 1) = (1, -3, 1). \]

Step 4: Find Distance \( RT \)

The coordinates of \( R \) are \((-1, -5, 0)\), and the coordinates of \( T \) are \((1, -3, 1)\).

The distance \( RT \) is: \[ RT = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}. \] Substitute the values: \[ RT = \sqrt{(1 - (-1))^2 + (-3 - (-5))^2 + (1 - 0)^2}. \] Simplify: \[ RT = \sqrt{(1 + 1)^2 + (-3 + 5)^2 + (1 - 0)^2}. \] \[ RT = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3. \]

Final Answer:

The distance \( RT = 3 , \text{units}.\)