Question

An angle between the plane, $x+y+z=5$ and the line of intersection of the planes, $3x+4y+z-1=0$ and $5x+8y+2z+14=0$, is :

• A. ${\sin }^{-1}\left(\sqrt{\frac{3}{17}}\right)$
• B. ${\cos }^{-1}\left(\sqrt{\frac{3}{17}}\right)$
• C. ${\cos }^{-1}\left(\frac{3}{\sqrt{17}}\right)$
• D. ${\sin }^{-1}\left(\frac{3}{\sqrt{17}}\right)$

Answer 1

The correct option is A ${\sin }^{-1}\left(\sqrt{\frac{3}{17}}\right)$

Given planes:
$3x+4y+z-1=0$ and $5x+8y+2z+14=0$
The normal to the planes will be,
$\overrightarrow{n_1}=3\hat{i}+4\hat{j}+\hat{k}\overrightarrow{n_2}=5\hat{i}+8\hat{j}+2\hat{k}$
The direction ratios of the line of intersection will be,
$\overrightarrow{n_1}\times \overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 4& 1\\ 5& 8& 2\end{array}\right|\Rightarrow \overrightarrow{n_1}\times \overrightarrow{n_2}=-\hat{j}+4\hat{k}$
Normal vector to plane $x+y+z=5$
$\overrightarrow{n_3}=\hat{i}+\hat{j}+\hat{k}$
The angle between plane and the line of intersection will be,
$\sin \theta =\frac{\overrightarrow{n_3}\cdot (\overrightarrow{n_1}\times \overrightarrow{n_2})}{|\overrightarrow{n_3}||\overrightarrow{n_1}\times \overrightarrow{n_2}|}\Rightarrow \sin \theta =\frac{-1+4}{\sqrt{3}\sqrt{17}}\Rightarrow \theta ={\sin }^{-1}\sqrt{\frac{3}{17}}$