Question:medium
If the point of intersection of the lines
\[
\frac{x+1}{3}=\frac{y+a}{5}=\frac{z+b+1}{7}
\]
\[
\frac{x-2}{1}=\frac{y-b}{4}=\frac{z-2a}{7}
\]
lies on the \(xy\)-plane, then the value of \(a+b\) is:
If the point of intersection of the lines
\[
\frac{x+1}{3}=\frac{y+a}{5}=\frac{z+b+1}{7}
\]
\[
\frac{x-2}{1}=\frac{y-b}{4}=\frac{z-2a}{7}
\]
lies on the \(xy\)-plane, then the value of \(a+b\) is:
Updated On: Apr 10, 2026
- \(2\)
- \(5\)
- \(7\)
- \(9\)
Show Solution
The Correct Option is B
Solution and Explanation
Was this answer helpful?
0
Top Questions on 3D Geometry
- The lines \( \frac{1 - x}{2} = \frac{y - 1}{3} = \frac{z}{1} \) and \( \frac{2x - 3}{2p} = \frac{y}{-1} = \frac{z - 4}{7} \) are perpendicular to each other for \( p \) equal to:
- CBSE Class XII - 2024
- Mathematics
- 3D Geometry
- Find the distance between the line \( \frac{x}{2} = \frac{2y - 6}{4} = \frac{1 - z}{-1} \) and another line parallel to it passing through the point \( (4, 0, -5) \).
- CBSE Class XII - 2024
- Mathematics
- 3D Geometry
- If the lines \( \frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2} \) and \( \frac{x - 1}{3k} = \frac{y - 1}{1} = \frac{z - 6}{-7} \) are perpendicular to each other, find the value of \( k \) and hence write the vector equation of a line perpendicular to these two lines passing through the point \( (3, -4, 7) \).
- CBSE Class XII - 2024
- Mathematics
- 3D Geometry
- If \( d_1 \) is the shortest distance between the lines \(\frac{x+1}{2} = \frac{y-1}{-12} = \frac{z-1}{-7} \quad \text{and} \quad \frac{x-1}{7} = \frac{y+8}{2} = \frac{z-4}{5},\) and \( d_2 \) is the shortest distance between the lines \(\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-6}{-3} \quad \text{and} \quad \frac{x+2}{1} = \frac{y+2}{2} = \frac{z-1}{6},\) then the value of \( \frac{32\sqrt{3}d_1}{d_2} \) is:
- JEE Main - 2024
- Mathematics
- 3D Geometry
- Want to practice more? Try solving extra ecology questions todayView All Questions
