Question

The value of $\lambda$ for which the lines $\frac{1-x}{3}=\frac{y-2}{2\lambda}=\frac{z-3}{2}$ and $\frac{x-1}{3\lambda}=\frac{y-1}{1}=\frac{6-z}{7}$

• A. -1
• B. -2
• C. 1
• D. 2

Answer 1

Answer: (D)

2

Answer 2

The two lines
$\frac{1 - x}{3} = \frac{y-2}{2\lambda} =\frac{z-3}{2}$
and $\frac{x-1}{3\lambda}=\frac{y-1}{1}=\frac{6-z}{7}$ are perpendicular if
$-3 \times 3\lambda+ 2\lambda \times1 + 7 \times 2$ = 0
$\Rightarrow$ $-9 -1 + 2-1 + 14 = u \Rightarrow - 7\lambda$ = - 14
$\Rightarrow$ $\lambda = 2$