Question:
The point at which the line $\frac{x+3}{11}=\frac{y-2}{-1}=\frac{z+1}{3}$ meets the $zx$-plane is:
The point at which the line $\frac{x+3}{11}=\frac{y-2}{-1}=\frac{z+1}{3}$ meets the $zx$-plane is:
Show Hint
To find the intersection with any plane $xy, yz, or zx$, set the missing variable to zero.
Updated On: Apr 28, 2026
- (19, 2, 5)
- (19, 0, 5)
- (0, 2, -1)
- (-3, 2, 0)
- (0, 2, -1)
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The Correct Option is B
Solution and Explanation
Step 1: Concept
In the $zx$-plane, the $y$-coordinate must be zero $(y = 0)$.
Step 2: Analysis
Set $y = 0$ in the line equation: $\frac{y-2}{-1} = \frac{0-2}{-1} = 2$. Now equate other parts to 2: $\frac{x+3}{11} = 2 \implies x+3 = 22 \implies x = 19$. $\frac{z+1}{3} = 2 \implies z+1 = 6 \implies z = 5$.
Step 3: Conclusion
The point is (19, 0, 5). Final Answer: (B)
In the $zx$-plane, the $y$-coordinate must be zero $(y = 0)$.
Step 2: Analysis
Set $y = 0$ in the line equation: $\frac{y-2}{-1} = \frac{0-2}{-1} = 2$. Now equate other parts to 2: $\frac{x+3}{11} = 2 \implies x+3 = 22 \implies x = 19$. $\frac{z+1}{3} = 2 \implies z+1 = 6 \implies z = 5$.
Step 3: Conclusion
The point is (19, 0, 5). Final Answer: (B)
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