3D Geometry
Line and Plane Intersection
Grade 12
Question:
<p>On which of the following lines lies the point of intersection of the line, \(\dfrac{x-4}{2} = \dfrac{y-5}{2} = \dfrac{z-3}{1}\) and the plane, \(x + y + z = 2\)?</p>
<p>\(\dfrac{x+3}{3} = \dfrac{4-y}{3} = \dfrac{z+1}{-2}\)</p>
<p>\(\dfrac{x-4}{1} = \dfrac{y-5}{1} = \dfrac{z-5}{-1}\)</p>
<p>\(\dfrac{x-1}{1} = \dfrac{y-3}{2} = \dfrac{z+4}{-5}\)</p>
<p>\(\dfrac{x-2}{2} = \dfrac{y-3}{2} = \dfrac{z+3}{3}\)</p>
Step-by-Step Solution
Key Concept: Find where the given line intersects the plane by substituting the parametric form of the line into the plane equation, then verify which given line contains this intersection point.
Step 1: Write the line in parametric form. From $\frac{x-4}{2} = \frac{y-5}{2} = \frac{z-3}{1} = t$, we get: $x = 4 + 2t$, $y = 5 + 2t$, $z = 3 + t$ Step 2: Substitute into the plane equation $x + y + z = 2$. $(4 + 2t) + (5 + 2t) + (3 + t) = 2$ $12 + 5t = 2$ $5t = -10$ $t = -2$ Step 3: Find the intersection point. $x = 4 + 2(-2) = 0$ $y = 5 + 2(-2) = 1$ $z = 3 + (-2) = 1$ Intersection point is $(0, 1, 1)$ Step 4: Verify this point lies on the given line option and satisfies its equation. ∴ Answer: A
Correct Answer: A