Question

The point of intersection of the straight lines $r =(3 \hat{ i }-4 \hat{ j }+5 \hat{ k })+\lambda(-\hat{ i }-2 \hat{ j }+2 \hat{ k })$ and
$\frac{3-x}{-1}=\frac{y+4}{2}=\frac{z-5}{7}$ is

• A. $(- 3, - 4, - 5)$
• B. $(- 3, 4, 5)$
• C. $(- 3, 4, - 5)$
• D. $(- 3, - 4, 5)$
• E. $(3, - 4, 5)$

Answer 1

Answer: (E)

Given equation of straight lines are
$r =(3 \hat{ i }-4 \hat{ j }+5 \hat{ k })+\lambda(-\hat{ i }-2 \hat{ j }+2 \hat{ k })$
or $\frac{x-3}{-1}=\frac{y+4}{2}=\frac{z-5}{2}=\lambda \,..(i)$
and $\frac{3-x}{-1}=\frac{y+4}{2}=\frac{z-5}{7}=\mu$
or, $\frac{x-3}{1}=\frac{y+4}{2}=\frac{z-5}{7}=\mu \,.....(ii)$
From Eq. (i), coordinates are
$(-\lambda+3,-2 \lambda-4,2 \lambda+5)$
From Eq. (ii) coordinates are
$(\mu+3,2 \mu-4,7 \mu+5)$
For intersecting the given lines,
$-\lambda+3 =\mu+3$
$\Rightarrow \mu+\lambda =0\,.....(iii)$
$-2 \lambda-4=2 \mu-4 $
$ \Rightarrow 2 \mu+2 \lambda=0 \,.....(iv)$
$ 2 \lambda+5=7 \mu+5$
$\Rightarrow 7 \mu-2 \lambda=0 \,......(v)$
From Eqs. (iii), (vi), (v)
$\mu=0, \lambda=0$
Hence, intersecting points are
$(0+3,-2 \times 0-4,2 \times 0+5)=(3,-4,5)$