Question

The point of intersection of the lines \(\frac{x-3}{2} = \frac{y-2}{2} = \frac{z-6}{1}\) and \(\frac{x-2}{3} = \frac{y-4}{2} = \frac{z-1}{3}\) is:

• A. \( (3,4,3) \)
• B. \( (7,6,6) \)
• C. \( (4,3,3) \)
• D. \( (10,11,10) \)
• E. \( (11,10,10) \)

Hint:

When finding the intersection of lines given in symmetric form, convert them to parametric form and solve the system of equations that results from setting the components equal.

Answer 1

Answer: (E)

First, express the lines in parametric form: \[ {Line 1: } x = 3 + 2t, \, y = 2 + 2t, \, z = 6 + t \] \[ {Line 2: } x = 2 + 3s, \, y = 4 + 2s, \, z = 1 + 3s \] To find the intersection, equate the parametric equations and solve for \(t\) and \(s\): \[ 3 + 2t = 2 + 3s \] \[ 2 + 2t = 4 + 2s \] \[ 6 + t = 1 + 3s \] From the second equation: \[ 2t - 2s = 2 \implies t - s = 1 \] From the third equation: \[ t - 3s = -5 \] Solving these equations: \[ t - s = 1 \] \[ t - 3s = -5 \] Subtract the first from the second: \[ 2s = 6 \implies s = 3 \] \[ t = 4 \] Substitute \(t = 4\) into the equations for Line 1: \[ x = 3 + 2 \times 4 = 11 \] \[ y = 2 + 2 \times 4 = 10 \] \[ z = 6 + 4 = 10 \] The point of intersection is \((11, 10, 10)\).