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

Step 1: Understanding the Concept:
A point lying on the \(xy\)-plane has its \(z\)-coordinate equal to zero. If the intersection point lies on this plane, we can set \(z=0\) in the parametric forms of both lines and solve for the unknown constants \(a\) and \(b\).
Step 2: Key Formula or Approach:
1. Intersection condition: Set both lines equal to parameters \(\lambda\) and \(\mu\).
2. \(xy\)-plane condition: \(z = 0\).
Step 3: Detailed Explanation:
Let the intersection point be \(P(x, y, z)\). Since it is on the \(xy\)-plane, \(z = 0\).
For Line 1: \(\frac{x+1}{3} = \frac{y+a}{5} = \frac{0+b+1}{7}\).
From this, \(\frac{x+1}{3} = \frac{b+1}{7} \implies x = \frac{3b+3}{7} - 1 = \frac{3b-4}{7}\).
And \(\frac{y+a}{5} = \frac{b+1}{7} \implies y = \frac{5b+5}{7} - a = \frac{5b+5-7a}{7}\).
For Line 2: \(\frac{x-2}{1} = \frac{y-b}{4} = \frac{0-2a}{7}\).
From this, \(x-2 = -\frac{2a}{7} \implies x = 2 - \frac{2a}{7} = \frac{14-2a}{7}\).
And \(\frac{y-b}{4} = -\frac{2a}{7} \implies y = b - \frac{8a}{7} = \frac{7b-8a}{7}\).
Equating the \(x\)-coordinates:
\(\frac{3b-4}{7} = \frac{14-2a}{7} \implies 2a + 3b = 18 \quad \dots \text{(Eq. 1)}\).
Equating the \(y\)-coordinates:
\(\frac{5b+5-7a}{7} = \frac{7b-8a}{7} \implies 5b + 5 - 7a = 7b - 8a \implies a - 2b = -5 \implies 2a - 4b = -10 \quad \dots \text{(Eq. 2)}\).
Subtracting Eq. 2 from Eq. 1:
\((2a + 3b) - (2a - 4b) = 18 - (-10) \implies 7b = 28 \implies b = 4\).
Substitute \(b=4\) into \(a - 2b = -5\):
\(a - 8 = -5 \implies a = 3\).
Finally, \(a + b = 3 + 4 = 7\).
Step 4: Final Answer:
The value of \(a+b\) is 7.