Find the acute angle between the line $x=5+3t,y=-2+4t,z=1-2t$ and the plane $2x-y+3z=7$.

Step-by-step solution

1. Determine the direction vector of the line: $\mathbf{v}=(3,4,-2)$.
2. Determine the normal vector of the plane: $\mathbf{n}=(2,-1,3)$.
3. Compute the dot product: $\mathbf{v}cdotmathbfn=3cdot2+4(-1)+(-2)cdot3=-4$, take absolute value $4$.
4. Find magnitudes: $|\mathbf{v}|=\sqrt{3^2+4^2+(-2{)}^2}=\sqrt{29}$, $|\mathbf{n}|=\sqrt{2^2+(-1{)}^2+3^2}=\sqrt{14}$.
5. Compute $cosalpha=\frac{4}{\sqrt{29}\sqrt{14}}=\frac{4}{\sqrt{406}}$, giving $alphaapprox78.6^{c}irc$ as the angle between $\mathbf{v}$ and $\mathbf{n}$.
6. The acute angle between the line and the plane is $90^{c}irc-alphaapprox90^{c}irc-78.6^{c}irc=11.4^{c}irc$.
7. Rounded to the nearest integer, the required acute angle is $11^{c}irc$.

Concept tested: Angle between line and plane

Published 10 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked