Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the vectors parallel to the plane be ${\vec{v}}_1=3\hat{i}+2\hat{j}-\hat{k}$ and ${\vec{v}}_2=\hat{i}-2\hat{j}-2\hat{k}$.

The normal vector $\vec{n}$ to the plane is given by the cross product of ${\vec{v}}_1$ and ${\vec{v}}_2$:

$\vec{n}={\vec{v}}_1\times {\vec{v}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 2& -1\\ 1& -2& -2\end{array}\right|$

$\vec{n}=\hat{i}(-4-2)-\hat{j}(-6-(-1))+\hat{k}(-6-2)=-6\hat{i}+5\hat{j}-8\hat{k}$

The normal vector to the $xy$-plane is $\hat{k}$.

The acute angle $\theta$ between the two planes is the acute angle between their normal vectors, given by:

$\cos \theta =\frac{|\vec{n}\cdot \hat{k}|}{|\vec{n}||\hat{k}|}$

$\cos \theta =\frac{|(-6\hat{i}+5\hat{j}-8\hat{k})\cdot \hat{k}|}{\sqrt{(-6{)}^2+5^2+(-8{)}^2}\sqrt{1^2}}$

$\cos \theta =\frac{|-8|}{\sqrt{36+25+64}}=\frac{8}{\sqrt{125}}=\frac{8}{5\sqrt{5}}$

$\theta ={\cos }^{-1}\left(\frac{8}{5\sqrt{5}}\right)$

Answer: ${\cos }^{-1}\left(\frac{8}{5\sqrt{5}}\right)$