Question

The point at which the line $\frac{x+3}{11}=\frac{y-2}{-1}=\frac{z+1}{3}$ meets the $zx$-plane is:

• A. (19, 2, 5)
• B. (19, 0, 5)
• C. (0, 2, -1)
• D. (-3, 2, 0)
• E. (0, 2, -1)

Hint:

To find the intersection with any plane $xy, yz, or zx$, set the missing variable to zero.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to find the point of intersection between a line given in Cartesian form and a coordinate plane (the zx-plane). The defining characteristic of the zx-plane is that for every point on it, the y-coordinate is zero.
Step 2: Key Formula or Approach:
1. The equation of the zx-plane is \( y = 0 \).
2. To find the point of intersection, we will set the y-coordinate in the line's equation to 0 and solve for the other variables.
The line's equation can be represented parametrically by setting each part equal to a parameter, say \( \lambda \).
\[ \frac{x+3}{11} = \lambda \implies x = 11\lambda - 3 \] \[ \frac{y-2}{-1} = \lambda \implies y = -\lambda + 2 \] \[ \frac{z+1}{3} = \lambda \implies z = 3\lambda - 1 \] Step 3: Detailed Explanation:
The line intersects the zx-plane where \( y = 0 \).
Using the parametric equation for y, we can find the value of \( \lambda \) at the intersection point.
\[ y = -\lambda + 2 = 0 \] \[ 2 = \lambda \] Now that we have the value of \( \lambda \) at the intersection, we can find the x and z coordinates of the point by substituting \( \lambda = 2 \) back into their respective parametric equations.
x-coordinate:
\[ x = 11\lambda - 3 = 11(2) - 3 = 22 - 3 = 19 \] z-coordinate:
\[ z = 3\lambda - 1 = 3(2) - 1 = 6 - 1 = 5 \] The coordinates of the intersection point are (19, 0, 5).
Step 4: Final Answer:
The point at which the line meets the zx-plane is (19, 0, 5).