3D Geometry
Planes and Angles
Grade 12
Question:
<p>Consider the planes \(3x - 6y + 2z + 5 = 0\) and \(4x - 12y + 3z = 3\). The plane \(67x + 162y + 47z + 44 = 0\) bisects the angle between the planes which</p>
<p>(a) contains origin</p>
<p>(b) is acute</p>
<p>(c) is obtuse</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use the angle bisector formula for planes and verify if it contains the origin or forms an acute angle.
Step 1: Rewrite the second plane: \(-4x + 12y - 3z + 3 = 0\) Step 2: The angle bisector formula gives: \[\frac{3x - 6y + 2z + 5}{\sqrt{9 + 36 + 4}} = \frac{-4x + 12y - 3z + 3}{\sqrt{16 + 144 + 9}}\] Step 3: \[\frac{3x - 6y + 2z + 5}{7} = \frac{-4x + 12y - 3z + 3}{13}\] Step 4: \[13(3x - 6y + 2z + 5) = 7(-4x + 12y - 3z + 3)\] Step 5: \[39x - 78y + 26z + 65 = -28x + 84y - 21z + 21\] Step 6: \[67x - 162y + 47z + 44 = 0\] (which is the given plane) Step 7: Check if origin satisfies: \(67(0) - 162(0) + 47(0) + 44 = 44 \neq 0\), so it contains the origin plane. Step 8: Computing \(\cos\theta\) between the planes and finding \(|\tan\theta| < 1\) shows the angle is acute. ∴ Answers are (a) and (b).
Correct Answer: a, b