Question:
If the equation of a plane passing through \(A(1,p,2)\), \(B(3,2,4)\) and parallel to the z axis, is \(3x-2y-q = 0\), then \(\ldots\)
If the equation of a plane passing through \(A(1,p,2)\), \(B(3,2,4)\) and parallel to the z axis, is \(3x-2y-q = 0\), then \(\ldots\)
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A plane parallel to the z-axis has no z term. Use the two points to find q and p.
Updated On: Oct 1, 2026
- \(p = -1, q = 5\)
- \(p = 1, q = -5\)
- \(p = -2, q = -5\)
- \(p = 2, q = -5\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
A plane parallel to the z-axis has a normal perpendicular to the z-axis, so its equation has no \(z\) term. That fits the given form \(3x - 2y - q = 0\).
Step 2: Key Formula or Approach:
Each point on the plane must satisfy \(3x - 2y - q = 0\).
Step 3: Detailed Explanation:
Point \(B(3, 2, 4)\):
\[ 3(3) - 2(2) - q = 0 \Rightarrow 9 - 4 = q \Rightarrow q = 5 \]
Point \(A(1, p, 2)\):
\[ 3(1) - 2p - 5 = 0 \Rightarrow -2p = 2 \Rightarrow p = -1 \]
So \(p = -1\) and \(q = 5\). Options (B), (C) and (D) all have \(q = -5\), which would give a plane \(3x - 2y + 5 = 0\), and B does not lie on it: \(9 - 4 + 5 = 10 \ne 0\).
Final Answer:
\(p = -1\) and \(q = 5\), option (A). \[ \boxed{p=-1,\ q=5 \text{ (A)}} \]
A plane parallel to the z-axis has a normal perpendicular to the z-axis, so its equation has no \(z\) term. That fits the given form \(3x - 2y - q = 0\).
Step 2: Key Formula or Approach:
Each point on the plane must satisfy \(3x - 2y - q = 0\).
Step 3: Detailed Explanation:
Point \(B(3, 2, 4)\):
\[ 3(3) - 2(2) - q = 0 \Rightarrow 9 - 4 = q \Rightarrow q = 5 \]
Point \(A(1, p, 2)\):
\[ 3(1) - 2p - 5 = 0 \Rightarrow -2p = 2 \Rightarrow p = -1 \]
So \(p = -1\) and \(q = 5\). Options (B), (C) and (D) all have \(q = -5\), which would give a plane \(3x - 2y + 5 = 0\), and B does not lie on it: \(9 - 4 + 5 = 10 \ne 0\).
Final Answer:
\(p = -1\) and \(q = 5\), option (A). \[ \boxed{p=-1,\ q=5 \text{ (A)}} \]
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