<p>Let \(P(3, 2, 6)\) be a point in space and \(Q\) be a point on the line \(\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(-3\hat{i} + \hat{j} + 5\hat{k})\). Then the value of \(\mu\) for which the vector \(\overrightarrow{PQ}\) is parallel to the plane \(x - 4y + 3z = 1\) is</p>
Step-by-Step Solution
Key Concept: A vector is parallel to a plane if and only if it is perpendicular to the plane's normal vector; use the condition dot product = 0 to find μ.
Step 1: Find point Q on the line. Any point on the line is Q = (1 - 3μ, -1 + μ, 2 + 5μ) Step 2: Find vector PQ. Since P = (3, 2, 6), we have:
PQ = (1 - 3μ - 3, -1 + μ - 2, 2 + 5μ - 6) = (-2 - 3μ, -3 + μ, -4 + 5μ) Step 3: The normal vector to the plane x - 4y + 3z = 1 is n = (1, -4, 3) Step 4: For PQ to be parallel to the plane, PQ must be perpendicular to the normal vector:
PQ · n = 0
(-2 - 3μ)(1) + (-3 + μ)(-4) + (-4 + 5μ)(3) = 0 Step 5: Expand:
-2 - 3μ + 12 - 4μ - 12 + 15μ = 0
-2 + 12 - 12 + (-3 - 4 + 15)μ = 0
-2 + 8μ = 0
8μ = 2
μ = 1/4 ∴ Answer: 1/4
Correct Answer: 1/4