Find the acute angle (in degrees, rounded to the nearest integer) between the line $x=-1+2t$, $y=3-t$, $z=4+3t$ and the plane $2x+y-2z=5$.

Step-by-step solution

1. Extract the direction vector of the line: $\mathbf{d}=⟨2,-1,3⟩$.
2. Identify the normal vector of the plane: $\mathbf{n}=⟨2,1,-2⟩$.
3. Compute the angle $\varphi$ between $\mathbf{d}$ and $\mathbf{n}$ using $\cos \varphi =\frac{|\mathbf{d}\cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$.
4. Calculate the dot product: $\mathbf{d}\cdot \mathbf{n}=2\cdot 2+(-1)\cdot 1+3\cdot (-2)=4-1-6=-3$, so $|\mathbf{d}\cdot \mathbf{n}|=3$.
5. Find the magnitudes: $|\mathbf{d}|=\sqrt{2^2+(-1{)}^2+3^2}=\sqrt{14}$ and $|\mathbf{n}|=\sqrt{2^2+1^2+(-2{)}^2}=\sqrt{9}=3$.
6. Thus $\cos \varphi =\frac{3}{\sqrt{14}\times 3}=\frac{1}{\sqrt{14}}$.
7. The angle between the line and the plane is $\alpha =90^{\circ }-\varphi$; equivalently $\sin \alpha =\cos \varphi =\frac{1}{\sqrt{14}}$.
8. Compute $\alpha =\arcsin \left(\frac{1}{\sqrt{14}}\right)\approx 15.5^{\circ }$.
9. Round to the nearest integer: $\alpha \approx 16^{\circ }$.

Concept tested: Angle between a line and a plane

Published 14 September 2026 · How this solution was prepared and checked