Question

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

Answer 1

Answer: 125

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