Question

If the angle between the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-3}{\lambda }$ and plane x+2y + 3z = 4 , is ${\cos }^{-1}\left(\sqrt{\frac{5}{14}}\right)$ find the value of $\lambda$.

Answer 1

Given equation of line is $\frac{x}{1}=\frac{y-1}{2}=\frac{z-3}{\lambda }$

Equation of plane is $x+2y+3z=4$

And also angle between them is ${\text{cos}}^{-1}(\sqrt{\frac{5}{14}})={\text{sin}}^{-1}(\frac{3}{\sqrt{14}})$

So the d.r's of line and plane will be $(1,2,\lambda )$ and $(1,2,3)$ respectively

$⟹{\text{sin}}^{-1}(\frac{3}{\sqrt{14}})={\text{sin}}^{-1}(\frac{1+4+3\lambda }{\sqrt{1+4+{\lambda }^2}\sqrt{1+4+9}})$

$⟹\frac{3}{\sqrt{14}}=\frac{(5+3\lambda )}{\sqrt{14}\sqrt{(5+{\lambda }^2)}}$

$⟹9(5+{\lambda }^2)=(5+3\lambda {)}^2$

$⟹45+9{\lambda }^2=25+30\lambda +9{\lambda }^2$

$⟹\lambda =\frac{2}{3}$