Question

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:

• A. \(2\)
• B. \(5\)
• C. \(7\)
• D. \(9\)

Answer 1

The Correct Option is B

Concept: For two lines in symmetric form, we introduce parameters and equate coordinates at the intersection point. Since the intersection point lies on the \(xy\)-plane, its \(z\)-coordinate must be zero.
Step 1:Introduce parameters for both lines.} Let \[ \frac{x+1}{3}=\frac{y+a}{5}=\frac{z+b+1}{7}=\lambda \] Then \[ x=3\lambda-1,\quad y=5\lambda-a,\quad z=7\lambda-b-1 \] For the second line, let \[ \frac{x-2}{1}=\frac{y-b}{4}=\frac{z-2a}{7}=\mu \] Thus \[ x=\mu+2,\quad y=4\mu+b,\quad z=7\mu+2a \]
Step 2:Equate coordinates.} At the intersection point, \[ 3\lambda-1=\mu+2 \] \[ 5\lambda-a=4\mu+b \] \[ 7\lambda-b-1=7\mu+2a \]
Step 3:Use the \(xy\)-plane condition.} Since the point lies on the \(xy\)-plane, \[ z=0 \] Thus \[ 7\lambda-b-1=0 \] \[ 7\mu+2a=0 \]
Step 4:Solve the equations.} From \[ 7\lambda=b+1 \] \[ \lambda=\frac{b+1}{7} \] and \[ \mu=-\frac{2a}{7} \] Substitute into \[ 3\lambda-1=\mu+2 \] \[ 3\left(\frac{b+1}{7}\right)-1=-\frac{2a}{7}+2 \] Solving gives \[ a+b=5 \]