3D Geometry
Perpendicularity of lines
Grade 12
Question:
<p>The equations of lines are: <br/> \(x - ay - b = 0,\; cy - z + d = 0\) and \(x - a'y - b' = 0,\; c'y - z + d' = 0\). <br/> If these two lines are perpendicular, then which of the following is correct?</p>
<p>\(aa' + 1 + cc' = 0\)</p>
<p>\(aa' + 1 - cc' = 0\)</p>
<p>\(aa' - 1 + cc' = 0\)</p>
<p>\(aa' - 1 - cc' = 0\)</p>
Step-by-Step Solution
Key Concept: Two lines in 3D are perpendicular when their direction vectors have a dot product of zero. Convert the symmetric plane equations into parametric form to find direction vectors, then apply the perpendicularity condition.
Step 1: Express each line as intersection of two planes. Line 1: x - ay - b = 0 and cy - z + d = 0 Line 2: x - a'y - b' = 0 and c'y - z + d' = 0 Step 2: Find direction vector of Line 1 using cross product of normal vectors. Normal to plane 1: n_1 = (1, -a, 0), Normal to plane 2: n_1' = (0, c, -1) Direction vector d_1 = n_1 × n_1' = (a, 1, c) Step 3: Find direction vector of Line 2 similarly. Normal to plane 3: n_2 = (1, -a', 0), Normal to plane 4: n_2' = (0, c', -1) Direction vector d_2 = n_2 × n_2' = (a', 1, c') Step 4: Apply perpendicularity condition: d_1 · d_2 = 0 (a)(a') + (1)(1) + (c)(c') = 0 ∴ Answer: aa' + cc' + 1 = 0
Correct Answer: A