Question:
A straight line through the point (1,-1,0) meets the line $\frac{x-1}{1}=\frac{y+1}{1}=\frac{z-1}{-1}$ at right angle. Its equation is ________.
A straight line through the point (1,-1,0) meets the line $\frac{x-1}{1}=\frac{y+1}{1}=\frac{z-1}{-1}$ at right angle. Its equation is ________.
Show Hint
Use the dot product of direction vectors for perpendicular lines.
Updated On: Jun 26, 2026
- $\frac{x-1}{1}=\frac{y+1}{1}=\frac{z}{2}$
- $\frac{x-1}{1}=\frac{y-1}{1}=\frac{z}{4}$
- $\frac{x-1}{-1}=\frac{y-1}{1}=\frac{z}{6}$
- $\frac{x-1}{-1}=\frac{y-1}{-1}=\frac{z}{3}$
- $\frac{x-1}{1}=\frac{y+1}{1}=\frac{z}{-2}$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The line passes through $P(1,-1,0)$ and some point $Q$ on the given line. $\vec{PQ}$ is perpendicular to the given line's direction $(1, 1, -1)$.
Step 2: Meaning
General point $Q = (k+1, k-1, -k+1)$. $\vec{PQ} = (k, k, -k+1)$.
Step 3: Analysis
$\vec{PQ}\cdot(1, 1, -1) = 0 \implies k + k + k - 1 = 0 \implies 3k = 1 \implies k = 1/3$. Direction of line is $(1/3, 1/3, 2/3)$, proportional to $(1, 1, 2)$.
Step 4: Conclusion
The equation is $\frac{x-1}{1}=\frac{y+1}{1}=\frac{z}{2}$. Final Answer: (A)
The line passes through $P(1,-1,0)$ and some point $Q$ on the given line. $\vec{PQ}$ is perpendicular to the given line's direction $(1, 1, -1)$.
Step 2: Meaning
General point $Q = (k+1, k-1, -k+1)$. $\vec{PQ} = (k, k, -k+1)$.
Step 3: Analysis
$\vec{PQ}\cdot(1, 1, -1) = 0 \implies k + k + k - 1 = 0 \implies 3k = 1 \implies k = 1/3$. Direction of line is $(1/3, 1/3, 2/3)$, proportional to $(1, 1, 2)$.
Step 4: Conclusion
The equation is $\frac{x-1}{1}=\frac{y+1}{1}=\frac{z}{2}$. Final Answer: (A)
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