3D Geometry
Planes and Lines
Grade 12
Question:
<p><strong>571.</strong> If three planes \(P_1 \equiv 2x + y + z - 1 = 0\), \(P_2 \equiv x - y + z - 2 = 0\) and \(P_3 \equiv \alpha x - y + 3z - 5 = 0\) intersects each other at point \(P\) on \(XOY\) plane and at point \(Q\) on \(YOZ\) plane, where \(O\) is the origin then identify the correct statement(s)?</p>
<p>(a) The value of \(\alpha\) is 4.</p>
<p>(b) Straight line perpendicular to plane \(P_3\) and passing through \(P\) is \(\dfrac{x-1}{4} = \dfrac{y+1}{-1} = \dfrac{z}{3}\).</p>
<p>(c) The length of projection of \(\overrightarrow{PQ}\) on x-axis is 1.</p>
<p>(d) Centroid of the triangle \(OPQ\) is \(\left(\dfrac{1}{3}, \dfrac{-1}{2}, \dfrac{1}{2}\right)\).</p>
Step-by-Step Solution
Key Concept: Points on XOY plane have z=0, and points on YOZ plane have x=0. Use these constraints with the three plane equations to find intersection points P and Q, then use the condition that all three planes pass through both points to determine α.
Step 1: Find point P on XOY plane (z=0). Setting z=0 in the three plane equations: P_1: 2x + y - 1 = 0 P_2: x - y - 2 = 0 From P_2: y = x - 2 Substituting in P_1: 2x + (x - 2) - 1 = 0 → 3x = 3 → x = 1, y = -1 So P = (1, -1, 0) Step 2: Verify P lies on P_3. P_3 at P: α(1) - (-1) + 3(0) - 5 = 0 α + 1 - 5 = 0 → α = 4 Step 3: Find point Q on YOZ plane (x=0). Setting x=0 in the three plane equations: P_1: y + z - 1 = 0 P_2: -y + z - 2 = 0 Adding: 2z - 3 = 0 → z = 3/2, y = -1/2 So Q = (0, -1/2, 3/2) Step 4: Verify Q lies on P_3 with α = 4. P_3 at Q: 4(0) - (-1/2) + 3(3/2) - 5 = 0 0 + 1/2 + 9/2 - 5 = 0 → 5 - 5 = 0 ✓ Conclusion: α = 4, P = (1, -1, 0), Q = (0, -1/2, 3/2) ∴ All statements about these points and α = 4 are correct: A, B, C, D
Correct Answer: A,B,C,D