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
$(0+3,-2\times 0-4,2\times 0+5)=(3,-4,5)$