Question

If the distances of the point \( (1,2,a) \) from the line \[ \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1} \] along the lines \[ L_1:\ \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b} \quad \text{and} \quad L_2:\ \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c} \] are equal, then \( a+b+c \) is equal to:

• A. \(5\)
• B. \(6\)
• C. \(4\)
• D. \(7\)

Hint:

When distances are measured along lines, compare ratios of direction ratios instead of using perpendicular distance formulas.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Distance "along a line" refers to the distance to a point of intersection between the direction line and the target line.
Step 2: Detailed Explanation:
Target line \(L: \frac{x-1}{1} = \frac{y}{2} = \frac{z-1}{1} = \lambda\). Point on \(L\) is \((\lambda+1, 2\lambda, \lambda+1)\).
Point \(P(1, 2, a)\). Direction from \(P\) to \(L\) is \((\lambda, 2\lambda-2, \lambda+1-a)\).
For \(L_1\), this direction is parallel to \((3, 4, b)\).
\(\frac{\lambda}{3} = \frac{2\lambda-2}{4} \Rightarrow 4\lambda = 6\lambda - 6 \Rightarrow \lambda = 3\).
Direction vector \(= (3, 4, 4-a)\). Thus \(b = 4-a \Rightarrow a+b = 4\).
Distance \(d_1 = \sqrt{3^2 + 4^2 + b^2} = \sqrt{25 + b^2}\).
For \(L_2\), direction is parallel to \((1, 4, c)\).
\(\frac{\lambda}{1} = \frac{2\lambda-2}{4} \Rightarrow 4\lambda = 2\lambda - 2 \Rightarrow \lambda = -1\).
Direction vector \(= (-1, -4, -a)\). Parallel to \((1, 4, c)\) implies \(c = a\) (after matching signs/ratios).
Distance \(d_2 = \sqrt{(-1)^2 + (-4)^2 + (-a)^2} = \sqrt{17 + a^2}\).
Given \(d_1 = d_2 \Rightarrow 25 + b^2 = 17 + a^2 \Rightarrow a^2 - b^2 = 8\).
\((a-b)(a+b) = 8 \Rightarrow (a-b)(4) = 8 \Rightarrow a-b = 2\).
Solving \(a+b=4\) and \(a-b=2\) gives \(a=3, b=1\).
Since \(c=a\), \(c=3\).
Final sum \(a + b + c = 3 + 1 + 3 = 7\).
Step 3: Final Answer:
The sum is 7.