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

Concept: The distance of a point from a line measured along another line is proportional to the ratio of direction cosines of the two lines.
Step 1: Direction ratios Direction ratios of the given line: \[ (1,2,1) \] Direction ratios of \( L_1 \): \[ (3,4,b) \] Direction ratios of \( L_2 \): \[ (1,4,c) \]
Step 2: Condition for equal distances For equal distances measured along the two lines: \[ \frac{a-1}{b} = \frac{a-1}{c} \Rightarrow b = c \]
Step 3: Use coplanarity condition Since all lines pass through \( (1,2,a) \), direction ratios must satisfy proportionality: \[ \frac{3}{1} = \frac{4}{2} = \frac{b}{1} \Rightarrow b = 3 \] Thus, \[ b = c = 3 \]
Step 4: Find \( a \) From the given line, \[ \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1} \Rightarrow a = 1 \]
Step 5: Calculate the required sum \[ a+b+c = 1+3+1 = 5 \]