Thread: Angle of intersection of the line and the plane

1. Angle of intersection of the line and the plane

I got this homework question that I do not understand. It is as followed;

Suppose a line intersects a plane at one point. 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 live and plane.

$(x/2) = ((y-1)/1) = ((z+1/-1)$ and x + 2y + 3z - 4 = 0.

I believe the angle of intersection of the line and the plane is the angle the line and plane insect at. There will always be two angles that will always add up to 180 $^o$ and can be modeled mathematically by the formula of $cos(-) = (m1 dot m2)/(|m1||m2|)$

However, to find the angle they intersect at I have no idea because there is only one directional vector. For the formula to work you need two. How do I get the other one? To I find two points that will make the right equation equal to zero and get the second directional vector there?

Also, is my explanation of "angle of intersection of the line and the plane" is correct? Is it detailed enough?

2. This is a bit hard to follow. Let’s start with two intersecting lines
$P+tD~\&~P+sE$. If $D\cdot E\ne 0$ then there is always an acute angle between them: $\theta=\arccos \left( {\dfrac{{\left| {D \cdot E} \right|}}
{{\left\| D \right\|\left\| E \right\|}}} \right)$

How use that same idea but instead let $E=N$ where $N$ is the normal of the plane.
We now have an acute angle, $\theta$, between the line and the normal to the plane.
Then $\frac{\pi}{2}-\theta$ is an acute angle between the plane and the line.

3. I do not understand your explaination. What I said with the plane and line intersecting is that there will be two angles that will add up to 180 degrees because two lines intersecting has to add up to 180 degrees. One of these lines is the entire plane. Therefore, the plane acts like a line which intersects another line, creating an angle of intersection.

For your answer how to find the angle I do not understand. I was looking for the second direction vector, which may be two different points for the right equation that with put in place will equal zero. Take the vectors from that point and then use the given directional vector to find the answer.

Can you give me a different definition so I understand what "angle of intersection of the line and the plane" better?

4. Originally Posted by Barthayn
I do not understand your explaination.
Can you give me a different definition so I understand what "angle of intersection of the line and the plane" better?
Well that is the exact definition you will find in any Vector Geometry textbook.
Find the acute angle between the normal to the plane and the direction of the line.
That is done with the arccosine function using the absolute value of the dot product.
Subtract that angle from $\frac{\pi}{2}$. That is the acute angle between the line and the plane.

5. Originally Posted by Plato
Well that is the exact definition you will find in any Vector Geometry textbook.
Find the acute angle between the normal to the plane and the direction of the line.
That is done with the arccosine function using the absolute value of the dot product.
Subtract that angle from $\frac{\pi}{2}$. That is the acute angle between the line and the plane.
Let N = normal vector and D = directional vector. If the dot product of N an D is not equal to zero there must be a intersection between the line and the plane. The intersecting angle will always either be an acute or right angle. Mathematically, $P+tD~\&~P+sE$. If $D\cdot E\ne 0$ then there is always an acute or right angle between them: $\theta=\arccos \left( {\dfrac{{\left| {D \cdot E} \right|}}
{{\left\| D \right\|\left\| E \right\|}}} \right)$

I included the right angle because the answer to number 2 is 90 degrees, which is a continuous of the first.

To me, it seems like the question still needs more answers for the definition part. What do you think about this and my definition?

6. Hello, Barthayn!

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

$\dfrac{x}{2}\:=\:\dfrac{y-1}{1} \:=\: \dfrac{z+1}{\text{-}1}\;\;\text{ and }\;\;x + 2y + 3z - 4 \:=\: 0.$
Code:
→                   | →
L                   | N
*                   |
*     * - - - - | - - - - - *
*/          |          /
/   *       |         /
/      @ *   |        /
Q * - - - - - - *       /
/              P   .  /
/                     / *
/                     /      *
* - - - - - - - - - - *

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

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

Let $\theta = \angle LPQ.$

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

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

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

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

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