The acute angle between the planes $2x+y-3z=5$ and $x-4y+2z=7$ is closest to which value?

Step-by-step solution

1. Step 1: Extract normal vectors: ${\mathbf{n}}_1=(2,1,-3)$ for the first plane and ${\mathbf{n}}_2=(1,-4,2)$ for the second.
2. Step 2: Compute dot product ${\mathbf{n}}_1\cdot {\mathbf{n}}_2=2(1)+1(-4)+(-3)(2)=-8$ and take absolute value $8$.
3. Step 3: Compute magnitudes $|{\mathbf{n}}_1|=\sqrt{2^2+1^2+(-3{)}^2}=\sqrt{14}$ and $|{\mathbf{n}}_2|=\sqrt{1^2+(-4{)}^2+2^2}=\sqrt{21}$.
4. Step 4: Cosine of angle $\theta$ is $\cos \theta =\frac{|{\mathbf{n}}_1\cdot {\mathbf{n}}_2|}{|{\mathbf{n}}_1||{\mathbf{n}}_2|}=\frac{8}{\sqrt{14}\sqrt{21}}\approx 0.467$.
5. Step 5: $\theta ={\cos }^{-1}(0.467)\approx 62.2^{\circ }$. The closest option is $60^{\circ }$.

Concept tested: Angle between planes

Published 10 October 2026 · How this solution was prepared and checked