Question

The plane passing through the line $L : \ell x$ $y +3(1-\ell) \quad z =1, \quad x +2 y - z =2$ and perpendicular to the plane $3 x+2 y+z=6$ is $3 x-$ $8 y+7 z=4$. If $\theta$ is the acute angle between the line $L$ and the $y$-axis, then $415 \cos ^2 \theta$ is equal to _____

Answer 1

$ \overrightarrow{ n }_1=\ell \hat{ i }-\hat{ j }+3(1-\ell) \hat{ k } $
$ \overrightarrow{ n }_2=\hat{ i }+2 \hat{ j }-\hat{ k }$
Dircction ratio of linc $=\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ \ell & -1 & 3(1-\ell) \\ 1 & 2 & -1\end{vmatrix}$
$=(6 \ell-5) \hat{ i }+(3-2 \ell) \hat{ j }+(2 \ell+1) \hat{ k }$
$3 x -8 y +7 z =4$ will contain the line $(6 \ell-5) \hat{ i }+(3-2 \ell) \hat{ j }+(2 \ell+1) \hat{ k }$
Normal of $3 x -8 y +7 z =4$ will be perpendicular to the line
$ =3(6 \ell-5)+(3-2 \ell)(-8)+7(2 \ell+1)=0$
$ \Rightarrow \ell=\frac{2}{3}$
$\therefore$ direction ratio of line $\left(-1, \frac{5}{3}, \frac{7}{3}\right)$
Angle with y axis
$\cos \theta=\frac{5 / 3}{\sqrt{1+\frac{25}{9}+\frac{49}{9}}} $
$ \cos \theta=\frac{5}{\sqrt{83}} $
$ \therefore 415 \cos ^2 \theta=\frac{25}{83} \times 415=125$