Question

The angle between the two lines $\frac{2-x}{1}=\frac{y}{2}=\frac{z+3}{1}$ and $\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-5}{2}$ is

• A. ${{0}^{o}}$
• B. ${{90}^{o}}$
• C. ${{45}^{o}}$
• D. None of these

Answer 1

Answer: (B)

${{90}^{o}}$

Answer 2

Given lines can be rewritten as
$\frac{x-2}{-1}=\frac{y}{2}=\frac{z+3}{1}$
and $\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-5}{2}$
Here, $({{a}_{1}},{{b}_{1}},{{c}_{1}})=(1,2,1)$
and $({{a}_{2}},{{b}_{2}},{{c}_{2}})=(4,1,2)$
$\therefore$ $\cos \theta =\frac{{{a}_{1}}{{a}_{2}}+{{b}_{2}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
$=\frac{-1\times 4+2\times 1+1\times 2}{\sqrt{{{(-1)}^{2}}+{{2}^{2}}+{{1}^{2}}}\sqrt{{{4}^{2}}+{{1}^{2}}+{{2}^{2}}}}$
$=\frac{-4+2+2}{\sqrt{6}\,\sqrt{21}}=0$
$\Rightarrow$ $\theta ={{90}^{o}}$