Question

lf $\Theta$ is the angle between two lines whose $d$.rs are $1,-2,1$ and $4,3,2$ then $sec(\frac{\theta }{2})+cosec(\frac{\theta }{2})=$

• A. $\sqrt{2}$
• B. $\infty$
• C. $2\sqrt{2}$
• D. $\frac{1}{2\sqrt{2}}$

Answer 1

The correct option is C $2\sqrt{2}$

The direction ratios are $1,-2,1$ and $4,3,2$

Angle between two lines with direction ratios $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$\cos \theta =\frac{x_1{x}_2+y_1{y}_2+z_1{z}_2}{\sqrt{(x_1{)}^2+(y_1{)}^2+(z_1{)}^2}\sqrt{(x_2{)}^2+(y_2{)}^2+(z_2{)}^2}}$

$⟹\cos \theta =\frac{1\times 4+(-2)\left(3\right)+1\times 2}{\sqrt{1^2+(-2{)}^2+1^2}\sqrt{4^2+3^2+2^2}}$

$=\frac{4-6+2}{\sqrt{6}\sqrt{29}}=0$

$\therefore \theta =co{s}^{-1}\left(0\right)=\frac{\pi }{2}={90}^o$

Then, $\sec \left(\theta /2\right)+\text{cosec}\left(\theta /2\right)=\sec {45}^o+\text{cosec}{45}^o=\sqrt{2}+\sqrt{2}=2\sqrt{2}$
$\therefore \sec \theta +\text{cosec}\theta =2\sqrt{2}$

Hence, option C is correct.