Question

Let the line passing through the points \( (-1, 2, 1) \) and parallel to the line \[ \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4} \] intersect the line \[ \frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1} \] at the point P. Then the distance of P from the point Q(4, -5, 1) is:

• A. 5
• B. 10
• C. \( 5\sqrt{6} \)
• D. \( 5\sqrt{5} \)

Hint:

To find the distance between two points in 3D, use the distance formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \). Always check your solution by verifying the parametric equations of the lines.

Answer 1

The Correct Option is D

To solve this problem, we need to find the point of intersection of two lines and then calculate the distance between this intersection point and a given point \( Q(4, -5, 1) \). 

1. First, identify the equations of both lines.
   • The first line passes through the point \( (-1, 2, 1) \) and is parallel to the line given by the direction ratios \(\langle 2, 3, 4 \rangle\). Therefore, the parametric equation of the line is: \(x = -1 + 2t, \, y = 2 + 3t, \, z = 1 + 4t\)
   • The second line is given by the ratios: \(\frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1} = \lambda\)
   • Thus, the parametric equation of the second line is: \(x = 3\lambda - 2, \, y = 2\lambda + 3, \, z = \lambda + 4\)
2. The next step is to find the intersection, which requires solving the equations:
   • \(-1 + 2t = 3\lambda - 2\)
   • \(2 + 3t = 2\lambda + 3\)
   • \(1 + 4t = \lambda + 4\)
3. Solving these equations simultaneously:
   1. From the first equation: 
      \(2t + 1 = 3\lambda\)
      Thus, \(\lambda = \frac{2t + 1}{3}\)
   2. From the second equation: 
      \(3t - 1 = 2\lambda\)
      Thus, \(\lambda = \frac{3t - 1}{2}\)
   3. Equating both expressions for \(\lambda\): 
      \(\frac{2t + 1}{3} = \frac{3t - 1}{2}\) 
      Cross-multiplying gives \(4t + 2 = 9t - 3\) 
      Solving for \(t\), we get \(5t = 5\) so \(t = 1\).
   4. Substitute \(t = 1\) into \(\lambda = \frac{2t + 1}{3}\) or \(\lambda = \frac{3t - 1}{2}\) to find \(\lambda\). 
      Using the first: \(\lambda = \frac{2(1) + 1}{3} = 1\) (check consistency with second gives same result).
4. Calculate coordinates of intersection point \( P \) using either parametric equation with \(= 1\) or \(\lambda =\).
   • For the first line: \(x = 1, y = 5, z = 5\)
5. Now, compute the distance between points \( P(1, 5, 5) \) and \( Q(4, -5, 1) \) using the distance formula: 
   \(d = \sqrt{(4 - 1)^2 + (-5 - 5)^2 + (1 - 5)^2}\)
   = \(\sqrt{3^2 + (-10)^2 + (-4)^2} = \sqrt{9 + 100 + 16} = \sqrt{125} = 5\sqrt{5}\).

Therefore, the distance of point \( P \) from \( Q \) is \(5\sqrt{5}\).

Answer 2

Step 1: The parametric equations of the line passing through the points \( (-1, 2, 1) \) and parallel to the line \( \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4} \) are: \[ \frac{x - (-1)}{2} = \frac{y - 2}{3} = \frac{z - 1}{4} = t. \] Thus, the parametric equations are: \[ x = -1 + 2t, \quad y = 2 + 3t, \quad z = 1 + 4t. \] Step 2: The parametric equations of the line \( \frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1} \) are: \[ \frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1} = s. \] Thus, the parametric equations are: \[ x = -2 + 3s, \quad y = 3 + 2s, \quad z = 4 + s. \] Step 3: To find the point of intersection \( P \), equate the parametric equations of the two lines: \[ -1 + 2t = -2 + 3s, \quad 2 + 3t = 3 + 2s, \quad 1 + 4t = 4 + s. \] Step 4: Solve the system of equations for \( t \) and \( s \). From the first equation: \[ -1 + 2t = -2 + 3s \implies 2t - 3s = -1. \] From the second equation: \[ 2 + 3t = 3 + 2s \implies 3t - 2s = 1. \] From the third equation: \[ 1 + 4t = 4 + s \implies 4t - s = 3. \] Step 5: Solve the system of equations: 1. \( 2t - 3s = -1 \) 2. \( 3t - 2s = 1 \) 3. \( 4t - s = 3 \) From equation (3), solve for \( s \): \[ s = 4t - 3. \] Substitute this into equations (1) and (2): From equation (1): \[ 2t - 3(4t - 3) = -1 \implies 2t - 12t + 9 = -1 \implies -10t = -10 \implies t = 1. \] Substitute \( t = 1 \) into the equation for \( s \): \[ s = 4(1) - 3 = 1. \] Step 6: Substitute \( t = 1 \) and \( s = 1 \) into the parametric equations of the lines to find the coordinates of the intersection point \( P \): \[ x = -1 + 2(1) = 1, \quad y = 2 + 3(1) = 5, \quad z = 1 + 4(1) = 5. \] Thus, \( P(1, 5, 5) \). 
Step 7: Now, calculate the distance from \( P(1, 5, 5) \) to the point \( Q(4, -5, 1) \). The distance formula is: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}. \] Substitute the coordinates of \( P \) and \( Q \): \[ d = \sqrt{(4 - 1)^2 + (-5 - 5)^2 + (1 - 5)^2} = \sqrt{3^2 + (-10)^2 + (-4)^2} = \sqrt{9 + 100 + 16} = \sqrt{125} = 5\sqrt{5}. \] Thus, the distance from \( P \) to \( Q \) is \( 5\sqrt{5} \).