Question

The angle between the straight lines $x−1=\frac{2y+3}{3}=\frac{z+5}{2}$ and $x−3r+2;y=−2r−1;z=2,$ where $r$ is a parameter, is

• A. $\frac{Ï€}{4}$
• B. ${\cos }^{−1}\left(\frac{−3}{\sqrt{182}}\right)$
• C. ${\sin }^{−1}\left(\frac{−3}{\sqrt{182}}\right)$
• D. $\frac{Ï€}{2}$
• E. 0

Answer 1

Answer: (D)

Given equation of lines are,
$x−1=\frac{2y+3}{3}=\frac{z+5}{2}$
and $x=3r+2;y=−2r−1;z=2$
$⇒\frac{x−1}{1}=\frac{y+\frac{3}{2}}{\frac{3}{2}}=\frac{z+5}{2}$
and $\frac{x−2}{3}=r;\frac{y+1}{−2}=r;\frac{z−2}{0}=r$
$⇒\frac{x−1}{1}=\frac{y+\frac{3}{2}}{\frac{3}{2}}=\frac{z+5}{2}$
and $\frac{x−2}{3}=\frac{y+1}{−2}=\frac{z−2}{0}$
DR's of lines Ist and IInd lines are $\left(1,\frac{3}{2},2\right)$ and $(3,−2,0)$
$∴$ The angle between two straight lines is
$\cos θ=\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$=\frac{1×3−\frac{3}{2}×2+2×0}{\sqrt{1^2+{\left(\frac{3}{2}\right)}^2+2^2}\sqrt{3^2+2^2+0^2}}$
$=\frac{3−3+0}{\sqrt{1+\frac{9}{4}+4}\sqrt{9+4}}$
$=\frac{0}{\sqrt{\frac{29}{4}}\sqrt{13}}$
$⇒\cos θ=0$
$⇒θ=\frac{π}{2}$