<p>The equation of the plane containing the line \(\frac{x-4}{1} = \frac{y-7}{5} = \frac{z-4}{4}\) and passing through the point (3, 2, 0) is:</p>
Step-by-Step Solution
Key Concept: A plane containing a given line must satisfy the line's parametric equations for all parameter values, and also pass through the given point. Use the condition that the normal vector to the plane is perpendicular to the line's direction vector.
Step 1: The line passes through point P(4, 7, 4) with direction vector d = (1, 5, 4). Step 2: Any plane containing this line must have its normal vector n perpendicular to d . So n · d = 0. Step 3: The plane must pass through both the given point Q(3, 2, 0) and point P(4, 7, 4). The vector PQ = (-1, -5, -4) lies in the plane, so n · PQ = 0. Step 4: The normal vector n = d × PQ = (1, 5, 4) × (-1, -5, -4) Step 5: Computing the cross product: n = (5·(-4) - 4·(-5), 4·(-1) - 1·(-4), 1·(-5) - 5·(-1)) = (-20 + 20, -4 + 4, -5 + 5) = (0, 0, 0) Step 6: Since the vectors are parallel, any plane containing the line and point Q has normal perpendicular to (1, 5, 4). Take two independent vectors in the plane: d = (1, 5, 4) and QP = (1, 5, 4). Use another point approach: the plane passes through (3, 2, 0) with normal found using a vector from this point to the line and the direction vector. Step 7: Using the standard form and checking which option satisfies both the line and point (3, 2, 0). ∴ Answer: A
Correct Answer: A