Thread: Angle between line and plane

1. Angle between line and plane

Hi if i was needing to find the angle between the plane
P:r = (0,0,1) + lambda(1,1,0) + mu(0,1,2)
and the line
L: z=0, y=4x+1

I think the equation of the line would be (1, 5 0)
and i believe that the plane is bounded by the direction vectors (1,1,0) and (0,1,2)

I was told be a friend to try and use the trig identity of cos(pi/2 -x) = sin x

any help or just a push in the right direction for this would be greatly appreciated

2. Re: Angle between line and plane

The angle between a line and a plane is found by finding angle between the line and normal to the plane. Since one would know both the vectors we can fond the angle by finding the dot product of the vectors.

3. Re: Angle between line and plane

Originally Posted by ducked
Hi if i was needing to find the angle between the plane
P:r = (0,0,1) + lambda(1,1,0) + mu(0,1,2)

and the line
L: z=0, y=4x+1

I think the equation of the line would be (1, 5 0)
You mean a vector in the direction of the line. You are mistaken. Perhaps it would be easier if the line were written as
x= t, y= 4t+ 1, z= 0.

[quote]and i believe that the plane is bounded by the direction vectors (1,1,0) and (0,1,2) [quote]
"Bounded" is the wrong word here- a plane has no "boundaries". (1, 1, 0) and (0, 1, 2) are vectors [b]in the plane and so their cross product is perpedicular to the plane.

I was told be a friend to try and use the trig identity of cos(pi/2 -x) = sin x

any help or just a push in the right direction for this would be greatly appreciated
As ibdutt said, the angle between a line and a plane is the angle between a vector in the direction of the line and a vector perpendicular to the plane. And that can be calculated
using $u\cdot v= |u||v|cos(\theta)$. There is nothing here that involves changing from sine to cosine or vice versa.

4. Re: Angle between line and plane

Thanks very much for you help

5. Re: Angle between line and plane

Originally Posted by ducked
Hi if i was needing to find the angle between the plane
P:r = (0,0,1) + lambda(1,1,0) + mu(0,1,2)
and the line L: z=0, y=4x+1
I have a different take on this question.
In most cases when speaking of the angle between a line and a plane, usually that is understood to be an acute angle (the line is not parallel to the plane).

Notation: $R=$
$\Pi:N\cdot \overrightarrow {PX} =0$ is a plane and $\ell: Q+tD$ is a line such that $N\cdot D\ne 0$.

Thus the acute angle between $\Pi~\&~\ell$ is $\frac{ \pi}{2} - \arccos \left( {\frac{{\left| {N \cdot D} \right|}}{{\left\| N \right\|\left\| D \right\|}}} \right)$

6. Re: Angle between line and plane

You mean a vector in the direction of the line. You are mistaken. Perhaps it would be easier if the line were written as
x= t, y= 4t+ 1, z= 0.
Im confused, know how to find the equation of the line ?

7. Re: Angle between line and plane

You were given the line as "L: z=0, y=4x+1". Taking x= t as parameter, y= 4x+1= 4t+1 so that x= t, y= 4t+1, z= 0.

8. Re: Angle between line and plane

Originally Posted by ducked
Im confused, know how to find the equation of the line ?
Suppose that $a\cdot b\cdot c\ne 0$ then
$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$

$\ell(t)==+t$

$\\ x=x_0+at \\ y=y_0+bt \\ z=z_0+ct \right$

Are all ways of writing the equation of a line through $(x_0,y_0,z_0)$ with direction $ai+bj+ck$.

9. Re: Angle between line and plane

I found the normal to the plane to be (2,-2,1), The direction of the line to be (1,4,0) and thus the angle to be equal to arccos(-6/(3(sqrt(17)))) Can anyone confirm if this is correct?

10. Re: Angle between line and plane

yeah i have the same as you algerber, but im not sure if we have the right answer

11. Re: Angle between line and plane

Originally Posted by algerber
I found the normal to the plane to be (2,-2,1), The direction of the line to be (1,4,0) and thus the angle to be equal to arccos(-6/(3(sqrt(17)))) Can anyone confirm if this is correct?
Originally Posted by ducked
yeah i have the same as you algerber, but im not sure if we have the right answer
The answer depends upon how one defines the angle between a line and a plane.
The answer above is not the answer used in most vector geometry textbooks.
$\arccos(-6/(3(\sqrt{17})))$ is the obtuse angle between the plane's normal and the given line's direction.

But that does not seen natural to use as the meaning for the angle between a line and a plane.

So how does your textbook/lecturenotes define that term?

12. Re: Angle between line and plane

In our course we generally look for the acute angle , so would I be best to go theta equal to 119.0171
Then subtract 90 to get in first Quadrant . Than go 90-29.02 to get a final angle of 60.98 ?

13. Re: Angle between line and plane

Originally Posted by ducked
In our course we generally look for the acute angle , so would I be best to go theta equal to 119.0171
Then subtract 90 to get in first Quadrant . Than go 90-29.02 to get a final angle of 60.98 ?
But again, that is the acute between the normal and the line,
That is not the usual meaning between a line and a plane.

The normal is perpendicular to the plane. Does it seem natural to take the angle between the line and the normal as the angle between the line and the plane?

But what does your text material say about it? Surely it is defined (explained) somewhere in your notes.

14. Re: Angle between line and plane

Yes when examining your point it does make intuitive sense , I have been searching through my notes for some form of definition but nothing is there to define it. So the 119.017 I have calculated is the angle between the normal to the plane and line. I understand why I must find the angle between the line and the plane but I am not sure how to utilize the formula shown in your response #5. Could you elaborate how to use this.

15. Re: Angle between line and plane

Originally Posted by ducked
Yes when examining your point it does make intuitive sense , I have been searching through my notes for some form of definition but nothing is there to define it. So the 119.017 I have calculated is the angle between the normal to the plane and line. I understand why I must find the angle between the line and the plane but I am not sure how to utilize the formula shown in your response #5. Could you elaborate how to use this.
I really do not remember the details of this question,
Do you understand the operations in the formula $\frac{ \pi}{2} - \arccos \left( {\frac{{\left| {N \cdot D} \right|}}{{\left\| N \right\|\left\| D \right\|}}} \right)~?$

If you do, then $N$ is the normal of the given plane and $D$ is the direction of the given line.

Just apply. You should get a value between $0\text{ and }\frac{\pi}{2}~.$

BTW I absolutely refuse to use degrees.

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