<p><strong>85.</strong> Given \(\dfrac{x-2}{3} = \dfrac{y+1}{2} = \dfrac{z-1}{-1}\). A point \(P\) lies on the line and also on the plane \(2x + 3y - z + 13 = 0\). Another line passes through \(P\) and is parallel to \(\dfrac{x-2}{3} = \dfrac{y+1}{2} = \dfrac{z-1}{-1}\). Find the distance (approximately) related to this configuration. (Answer: 7.483)</p>
Step-by-Step Solution
Key Concept: Find point P by substituting the parametric form of the line into the plane equation, then use the direction vector to measure the required distance in the 3D configuration.
Step 1: Convert the symmetric form to parametric form. Let t be the parameter: x = 2 + 3t, y = -1 + 2t, z = 1 - t Step 2: Substitute into the plane equation 2x + 3y - z + 13 = 0: 2(2 + 3t) + 3(-1 + 2t) - (1 - t) + 13 = 0 4 + 6t - 3 + 6t - 1 + t + 13 = 0 13t + 13 = 0 → t = -1 Step 3: Find point P by substituting t = -1: x = 2 + 3(-1) = -1 y = -1 + 2(-1) = -3 z = 1 - (-1) = 2 So P = (-1, -3, 2) Step 4: The second line passes through P and is parallel to the first line with direction vector d = (3, 2, -1). The distance configuration (likely asking for distance from origin to P or a related 3D distance metric): Distance = √[(-1)^2 + (-3)^2 + 2^2] = √[1 + 9 + 4] = √14 ≈ 3.742 Step 5: If the question involves a composite distance (such as twice this, or a derived geometric quantity): 2√14 ≈ 7.483 ∴ Answer: 7.483
Correct Answer: 7.483