Find the acute angle (in degrees, rounded to the nearest integer) between the planes $x-2y+3z=4$ and $2x+y-z=0$.

Step-by-step solution

1. Identify normal vectors: ${\mathbf{n}}_1=(1,-2,3)$ and ${\mathbf{n}}_2=(2,1,-1)$.
2. Compute dot product: ${\mathbf{n}}_1\cdot {\mathbf{n}}_2=1\cdot 2+(-2)\cdot 1+3\cdot (-1)=-3$.
3. Magnitudes: $||{\mathbf{n}}_1||=\sqrt{1^2+(-2{)}^2+3^2}=\sqrt{14}$, $||{\mathbf{n}}_2||=\sqrt{2^2+1^2+(-1{)}^2}=\sqrt{6}$.
4. Compute cosine of angle between normals: $cos\theta =\frac{|{\mathbf{n}}_1\cdot {\mathbf{n}}_2|}{||{\mathbf{n}}_1||||{\mathbf{n}}_2||}=\frac{3}{\sqrt{14}\sqrt{6}}=\frac{3}{\sqrt{84}}$.
5. Find $\theta ={\cos }^{-1}(\frac{3}{\sqrt{84}})\approx 71°$.
6. Round to nearest integer: $71°$.
7. Choose the option that matches this value.

Concept tested: Angle between planes

Published 17 September 2026 · How this solution was prepared and checked