Question

If the lines \(\frac{1-x}{3}=\frac{3y-1}{2\lambda +1}=4-z\) and \(\frac{x+2}{2\mu }=\frac{1-2y}{6}=\frac{5-z}{7}\) are perpendicular, then \(3\mu +\lambda =\ldots\)

• A. \(-4\)
• B. \(3\)
• C. \(8\)
• D. \(6\)

Hint:

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

Answer 1

The Correct Option is B

Step 1: Direction of line 1
\(\frac{x-1}{-3}=\frac{y-\frac13}{\frac{2\lambda +1}{3}}=\frac{z-4}{-1}\), so the direction ratios are \(\left(-3,\frac{2\lambda +1}{3},-1\right)\).

Step 2: Direction of line 2
\(\frac{x+2}{2\mu }=\frac{y-\frac12}{-3}=\frac{z-5}{-7}\), so the direction ratios are \((2\mu ,-3,-7)\).

Step 3: Perpendicular condition
\(-3(2\mu )+\frac{2\lambda +1}{3}(-3)+(-1)(-7)=0\).

Step 4: Simplify
\(-6\mu -2\lambda -1+7=0\Rightarrow 6\mu +2\lambda =6\Rightarrow 3\mu +\lambda =3\).

Final Answer:
The value of \(3\mu +\lambda\) is 3. \[ \boxed{\text{(B) }3} \]