The angle between the planes $x-2y+3z=4$ and $2x+y-z=0$ is closest to which of the following values?\n(a) $50^{c}irc$\n(b) $70^{c}irc$\n(c) $71^{c}irc$\n(d) $80^{c}irc$

Step-by-step solution

1. Identify the normal vectors: $\overrightarrow{n_1}=(1,-2,3)$ and $\overrightarrow{n_2}=(2,1,-1)$.
2. Compute the dot product $\overrightarrow{n_1}cdotvec{n}_2=1(2)+(-2)(1)+3(-1)=-3$; take its absolute value $3$.
3. Calculate the magnitudes: $|\overrightarrow{n_1}|=\sqrt{1+4+9}=\sqrt{14}$ and $|\overrightarrow{n_2}|=\sqrt{4+1+1}=\sqrt{6}$.
4. Find $cos\theta =\frac{3}{\sqrt{14}\sqrt{6}}=\frac{3}{\sqrt{84}}=\frac{3}{2\sqrt{21}}$.
5. Determine $\theta$ by $\arccos (\frac{3}{2\sqrt{21}})\approx 70.9^{c}irc$.
6. Round to the nearest integer to get $71^{c}irc$, which matches option (c).

Concept tested: Angle between two planes

Published 6 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked