<p>The coordinates of the foot of the perpendicular from the point \((1, -2, 1)\) on the plane containing the lines, \(\dfrac{x+1}{6} = \dfrac{y-1}{7} = \dfrac{z-3}{8}\) and \(\dfrac{x-1}{3} = \dfrac{y-2}{5} = \dfrac{z-3}{7}\), is</p>
Step-by-Step Solution
Key Concept: Find the plane equation using two lines (via their direction vectors and a common point), then use the perpendicular foot formula: the foot lies on the line through the given point parallel to the plane's normal vector.
Step 1: Extract direction vectors and points from the lines. Line 1: Point P_1 = (-1, 1, 3), direction d_1 = (6, 7, 8) Line 2: Point P_2 = (1, 2, 3), direction d_2 = (3, 5, 7) Step 2: Find the normal to the plane using cross product n = d_1 × d_2. n = |i j k| |6 7 8| = i(49-40) - j(42-24) + k(30-21) = (9, -18, 9) = 9(1, -2, 1) |3 5 7| Normal vector: n = (1, -2, 1) Step 3: Write the plane equation using point P_1(-1, 1, 3). 1(x+1) - 2(y-1) + 1(z-3) = 0 x - 2y + z = 0 Step 4: Find the foot of perpendicular from (1, -2, 1) to the plane. Parametric line through (1, -2, 1) in direction of normal (1, -2, 1): x = 1 + t, y = -2 - 2t, z = 1 + t Step 5: Substitute into plane equation x - 2y + z = 0: (1+t) - 2(-2-2t) + (1+t) = 0 1 + t + 4 + 4t + 1 + t = 0 6t + 6 = 0 → t = -1 Step 6: Find the foot coordinates: x = 1 + (-1) = 0, y = -2 - 2(-1) = 0, z = 1 + (-1) = 0 ∴ Foot of perpendicular is (0, 0, 0)
Correct Answer: A