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: Normalise the denominators
Divide the y-terms: $3y-1=3\left(y-\frac13\right)$ and $1-2y=-2\left(y-\frac12\right)$.

Step 2: Dot product
The direction vectors are $(-3,\frac{2\lambda +1}{3},-1)$ and $(2\mu ,-3,-7)$; their dot product is $-6\mu -(2\lambda +1)+7$.

Step 3: Solve
Setting it to zero: $6\mu +2\lambda =6$, hence $3\mu +\lambda =3$.

Final Answer:
The perpendicularity condition reduces to $3\mu +\lambda =3$. \[ \boxed{\text{(B) }3} \]