<p>The line \(\dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-2}{4}\) meets the plane \(x + 2y + 3z = 15\) at a point P. The distance of P from the origin is:</p>
Step-by-Step Solution
Key Concept: Find the intersection point P by substituting the parametric form of the line into the plane equation, then calculate the distance from origin using the distance formula.
Step 1: Convert the line to parametric form. Let <code>t</code> be the parameter: x = 1 + 2t, y = -1 + 3t, z = 2 + 4t Step 2: Substitute into the plane equation x + 2y + 3z = 15: (1 + 2t) + 2(-1 + 3t) + 3(2 + 4t) = 15 1 + 2t - 2 + 6t + 6 + 12t = 15 5 + 20t = 15 20t = 10 t = 1/2 Step 3: Find coordinates of point P: x = 1 + 2(1/2) = 2 y = -1 + 3(1/2) = 1/2 z = 2 + 4(1/2) = 4 Step 4: Calculate distance from origin using OP = √(x^2 + y^2 + z^2): OP = √(4 + 1/4 + 16) = √(20.25) = √(81/4) = 9/2 = 4.5 ∴ Answer: 4.5
Correct Answer: 4.5