<p>The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines \(\vec{r} = (\hat{i} + \hat{j}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})\) and \(\vec{r} = (\hat{i} + \hat{j}) + \mu(-\hat{i} + \hat{j} - 2\hat{k})\) is</p>
Step-by-Step Solution
Key Concept: Find the plane equation by using two direction vectors from the given lines (which share a common point), then calculate the perpendicular distance from the given point to this plane using the standard distance formula.
Step 1: Identify the common point and direction vectors of both lines. Both lines pass through point P = (1, 1, 0). Direction vector of line 1: d_1 = (1, 2, -1) Direction vector of line 2: d_2 = (-1, 1, -2) Step 2: Find the normal vector to the plane using cross product. n = d_1 × d_2 = | i j k | / |1 2 -1| / |-1 1 -2| n = i (2(-2) - (-1)(1)) - j (1(-2) - (-1)(-1)) + k (1(1) - 2(-1)) n = i (-4 + 1) - j (-2 - 1) + k (1 + 2) n = (-3, 3, 3) or simplified n = (-1, 1, 1) Step 3: Write the plane equation using point (1, 1, 0) and normal (-1, 1, 1). -1(x - 1) + 1(y - 1) + 1(z - 0) = 0 -x + 1 + y - 1 + z = 0 Plane: -x + y + z = 0 or x - y - z = 0 Step 4: Find perpendicular distance from point (2, 1, 4) to plane x - y - z = 0. Distance = |1(2) - 1(1) - 1(4)| / √(1^2 + (-1)^2 + (-1)^2) Distance = |2 - 1 - 4| / √3 = |-3| / √3 = 3/√3 = √3 ∴ Answer: C (√3)
Correct Answer: C