<p>A plane which passes through the point (3, 2, 0) and the line \(\dfrac{x-4}{1} = \dfrac{y-7}{5} = \dfrac{z-4}{4}\) is</p>
Step-by-Step Solution
Key Concept: A plane passing through a point and a line must contain both the given point and the entire line. This means the plane contains the line's direction vector and the vector from any point on the line to the given point.
Step 1: Identify the given information. Point: P(3, 2, 0) Line: passes through A(4, 7, 4) with direction vector d = (1, 5, 4) Step 2: Find vector AP from a point on the line to the given point. AP = (3-4, 2-7, 0-4) = (-1, -5, -4) Step 3: The normal vector to the plane is perpendicular to both d and AP . n = d × AP = (1, 5, 4) × (-1, -5, -4) n = | i j k | = i (5·(-4) - 4·(-5)) - j (1·(-4) - 4·(-1)) + k (1·(-5) - 5·(-1)) |1 5 4| |-1 -5 -4| n = i (-20+20) - j (-4+4) + k (-5+5) = (0, 0, 0) Step 4: Since the cross product is zero, the point P lies on the line itself. Recalculate using point A(4,7,4) on the line. AP = (-1, -5, -4) is parallel to d = (1, 5, 4), confirming P is on the line. Step 5: Take another point on the line, say B(5, 12, 8), and find n = d × BP where BP = (-2, -10, -8). Since BP is parallel to d, any plane through the line works. The plane equation is of the form a(x-4) + b(y-7) + c(z-4) = 0 where (a,b,c)·(1,5,4) = 0. ∴ Answer: A
Correct Answer: A