Let the line l:x=(1-y)/(-2)=(z-3)/( ), R meet the plane P:x+2y+3z=4 at the point ( , , ).

Question

Let the line $l:x=\frac{1-y}{-2}=\frac{z-3}{\lambda },\lambda \in R$ meet the plane $P:x+2y+3z=4$ at the point $(\alpha ,\beta ,\gamma )$. If the angle between the line $l$ and the plane $P$ is ${\cos }^{-1}(\sqrt{\frac{5}{14}})$, then $\alpha +2\beta +6\gamma $ is equal to \_\_\_\_\_\_\_\_\_\_\_.

Answer: 11

Accepted Answer

$L:\frac{x-0}{1}=\frac{y-1}{2}=\frac{z-3}{\lambda }$

$P:x+2y+3z=4$

Vector parallel to line : $\langle 1,2,\lambda angle =\vec{b}$

Normal vector to plane $P:\cdot {\sqrt{{1}^{2}+{2}^{2}+{\lambda }^{2}}\cdot \sqrt{{1}^{2}+{2}^{2}+{3}^{2}}}$

$\Rightarrow \frac{3}{\sqrt{14}}=\frac{1+4+3\lambda }{\sqrt{{\lambda }^{2}+5}\sqrt{14}}\Rightarrow \lambda =\frac{2}{3}$

${L}_{1}\equiv \frac{x-0}{3}=\frac{y-1}{6}=\frac{z-3}{2}=\mu $

Any point on line : $(3\mu ,6\mu +1,2\mu +3)$

It lies on P

$\begin{gathered} 	herefore 3\mu +12\mu +2+6\mu +9=4 \ \Rightarrow \mu =\frac{-1}{3} \end{gathered}$

Hence, $\alpha =3\mu =-1,\beta =6\mu +1=-1,\gamma =2\mu +3=\frac{7}{3}$

Now, $\alpha +2\beta +6\gamma =11$

Solutions · 1