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

The direction vector of the given line is $\vec{b}=\hat{i}+\hat{j}+\hat{k}$.

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

The angle $\theta$ between a line and a plane is given by the formula:

$\sin \theta =\frac{|\vec{b}\cdot \vec{n}|}{|\vec{b}||\vec{n}|}$

Calculating the dot product:

$\vec{b}\cdot \vec{n}=(1)(2)+(1)(-1)+(1)(1)=2-1+1=2$

Calculating the magnitudes of the vectors:

$|\vec{b}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}$

$|\vec{n}|=\sqrt{2^2+(-1{)}^2+1^2}=\sqrt{4+1+1}=\sqrt{6}$

Substituting these values into the formula:

$\sin \theta =\frac{2}{\sqrt{3}\sqrt{6}}=\frac{2}{\sqrt{18}}=\frac{2}{3\sqrt{2}}=\frac{\sqrt{2}}{3}$

$\theta ={\sin }^{-1}\left(\frac{\sqrt{2}}{3}\right)$

Answer: $\theta ={\sin }^{-1}\left(\frac{\sqrt{2}}{3}\right)$