Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The direction vector of the line is $\vec{b}=2\hat{i}-3\hat{j}+2\hat{k}$

The normal vector to the plane is $\vec{n}=3\hat{i}-2\hat{j}+5\hat{k}$

The angle $\theta$ between the line and the plane is given by $\sin \theta =\frac{|\vec{b}\cdot \vec{n}|}{|\vec{b}||\vec{n}|}$

$\vec{b}\cdot \vec{n}=(2)(3)+(-3)(-2)+(2)(5)=6+6+10=22$

$|\vec{b}|=\sqrt{2^2+(-3{)}^2+2^2}=\sqrt{4+9+4}=\sqrt{17}$

$|\vec{n}|=\sqrt{3^2+(-2{)}^2+5^2}=\sqrt{9+4+25}=\sqrt{38}$

$\sin \theta =\frac{22}{\sqrt{17}\sqrt{38}}=\frac{22}{\sqrt{646}}$

$\cos \theta =\sqrt{1-{\sin }^2\theta }=\sqrt{1-\frac{484}{646}}=\sqrt{\frac{162}{646}}=\frac{9\sqrt{2}}{\sqrt{646}}$

$\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{22}{9\sqrt{2}}$

$\theta ={\tan }^{-1}\frac{22}{9\sqrt{2}}$

Answer: ${\tan }^{-1}\frac{22}{9\sqrt{2}}$