Hello, Barthayn!

Suppose a line $\displaystyle \,L$ intersects a plane at point $\displaystyle \,P.$

Define what is meant by the "angle of intersection of the line and the plane".

Describe a method you can use to determine the angle of intersection of a line and a plane.

Then use your method to calculate the angle of intersecction of the given line and plane.

$\displaystyle \dfrac{x}{2}\:=\:\dfrac{y-1}{1} \:=\: \dfrac{z+1}{\text{-}1}\;\;\text{ and }\;\;x + 2y + 3z - 4 \:=\: 0.$ Code:

→ | →
L | N
* |
* * - - - - | - - - - - *
*/ | /
/ * | /
/ @ * | /
Q * - - - - - - * /
/ P . /
/ / *
/ / *
* - - - - - - - - - - *

We have line $\displaystyle \,L$ with direction vector: .$\displaystyle \vec L \,=\,\langle 2,1,\text{-}1\rangle$

We have a plane with normal vector: .$\displaystyle \vec N \:=\:\langle 1,2,3\rangle$

Let $\displaystyle \theta = \angle LPQ.$

Angle $\displaystyle LPN$ between $\displaystyle \vec N$ and $\displaystyle \vec L$ is given by:

. . $\displaystyle \displaystyle \cos(\angle LPN) \;=\;\frac{\vec N \cdot\vec L}{|\vec N||\vec L|} \;=\;\frac{\langle 1,2,3\rangle \cdot\langle 2,1,\text{-}1\rangle}{\sqrt{1^2+2^2+3^2}\,\sqrt{2^2+1^2+(\tex t{-}1)^2}}$

. . . . . . . . . . . .$\displaystyle \displaystyle =\;\frac{2 + 2 - 3}{\sqrt{14}\,\sqrt{6}} \;=\;\frac{1}{2\sqrt{21}} \;=\;0.109108945 $

. . Hence: .$\displaystyle \angle LPN \:\approx\:83.7^o$

Therefore: .$\displaystyle \theta \;=\; 90^o - 83.7^o \;=\;6.3^o$