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 \(\ldots\)

• A. \(p = -1, q = 5\)
• B. \(p = 1, q = -5\)
• C. \(p = -2, q = -5\)
• D. \(p = 2, q = -5\)

Hint:

A plane parallel to the z-axis has no z term. Use the two points to find q and p.

Answer 1

The Correct Option is A

Step 1: Use the normal vector:
The normal is $(3,-2,0)$, perpendicular to the direction vector $\overrightarrow{AB} = (2, 2-p, 2)$.

Step 2: Dot product is zero:
$3(2) - 2(2-p) + 0 = 0$, so $6 - 4 + 2p = 0$, so $p = -1$.

Step 3: Find q from point B:
$3\cdot3 - 2\cdot2 = 5 = q$.

Step 4: Check A:
$3(1) - 2(-1) - 5 = 3 + 2 - 5 = 0$. A lies on the plane.

Final Answer:
$p = -1$, $q = 5$, option (A). \[ \boxed{p=-1,\ q=5 \text{ (A)}} \]