<p>The sum of the intercepts on the coordinate axes of the plane passing through the point (–2, –2, 2) and containing the line joining the points (1, –1, 2) and (1, 1, 1), is</p>
Step-by-Step Solution
Key Concept: Find the plane equation by using a point on the plane and the direction vector of the line it contains. The intercepts are found by setting two variables to zero sequentially in the plane equation.
Step 1: Find the direction vector of the line joining (1, –1, 2) and (1, 1, 1). Direction vector: d = (1−1, 1−(−1), 1−2) = (0, 2, −1) Step 2: The plane passes through (−2, −2, 2) and contains the line, so it passes through (1, −1, 2) as well. Vector from (1, −1, 2) to (−2, −2, 2): v = (−3, −1, 0) Step 3: The normal vector to the plane is n = d × v . n = (0, 2, −1) × (−3, −1, 0) = (−1, 3, 6) Step 4: Plane equation using point (1, −1, 2) and normal (−1, 3, 6): −1(x − 1) + 3(y + 1) + 6(z − 2) = 0 −x + 1 + 3y + 3 + 6z − 12 = 0 −x + 3y + 6z − 8 = 0 or x − 3y − 6z + 8 = 0 Step 5: Find intercepts by setting two variables to zero: x-intercept (y=0, z=0): x + 8 = 0 → x = −8 y-intercept (x=0, z=0): −3y + 8 = 0 → y = 8/3 z-intercept (x=0, y=0): −6z + 8 = 0 → z = 4/3 Step 6: Sum of intercepts = −8 + 8/3 + 4/3 = −8 + 12/3 = −8 + 4 = −4 ∴ Answer: D
Correct Answer: D