Question

The plane passing through the line L :lx – y + 3(1 – l) z = 1, x + 2y – z = 2 and perpendicular to the plane 3x + 2y + z = 6 is 3x – 8y + 7z = 4. If θ is the acute angle between the line L and the y-axis, then 415 cos2θ is equal to ________.

Answer 1

L :lx – y + 3 (1 – l)z = 1, x + 2y – z = 2
and plane containing the line p : 3x – 8y + 7z = 4
Let $\vec{n}$ be the vector parallel to L.
then
$\vec{n}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\ l & -1 & 3(1-l) \\\ 1 & 2 & -1 \\\ \end{vmatrix}$
$=(6l−5)\hat{i}+(3−2l)\hat{j}+(2l+1)\hat{k}$
∵ R containing L
3(6l – 5) – 8(3 – 2l) + 7 (2l + 1) = 0
18l + 16l + 14l – 15 – 24 + 7 = 0
$∴l=\frac{32}{48}=\frac{2}{3}$
Let θ be the acute angle between L and y-axis
$∴ \cos⁡ \theta=\frac{\frac{5}{3}}{\sqrt{1+\frac{25}{9}+\frac{49}{9}}}=\frac{5}{\sqrt{83}}$
$∴ 415\cos^2⁡ \theta=125$