Question

If the lines \(\frac{3-x}{2}=\frac{5 y-2}{3 \lambda+1}=5-\mathrm{z}\) and \(\frac{x+2}{-1}=\frac{1-3 y}{7}=\frac{4-z}{2 \mu}\) are at right angles, then \(7 \lambda-10 \mu=\)

• A. \(143\)
• B. \(\frac{23}{3}\)
• C. \(137\)
• D. \(\frac{137}{5}\)

Hint:

Write both lines in standard symmetric form, read the direction ratios and set their dot product to zero.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Two lines are perpendicular when the dot product of their direction ratios is zero.

Step 2: Line 1
\(\dfrac{x-3}{-2}=\dfrac{y-\frac{2}{5}}{\frac{3\lambda+1}{5}}=\dfrac{z-5}{-1}\). Direction ratios: \(\left(-2,\dfrac{3\lambda+1}{5},-1\right)\).

Step 3: Line 2
\(\dfrac{x+2}{-1}=\dfrac{y-\frac{1}{3}}{-\frac{7}{3}}=\dfrac{z-4}{-2\mu}\). Direction ratios: \(\left(-1,-\dfrac{7}{3},-2\mu\right)\).

Step 4: Calculation
\(2-\dfrac{7(3\lambda+1)}{15}+2\mu=0\). Multiplying by 15: \(30-21\lambda-7+30\mu=0\), so \(21\lambda-30\mu=23\).
Divide by 3: \(7\lambda-10\mu=\dfrac{23}{3}\).

Final Answer:
The value of \(7\lambda-10\mu\) is \(\frac{23}{3}\). \[ \boxed{\text{(B) }\dfrac{23}{3}} \]