# Homework Help: The angle between a line and a plane

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1. Nov 27, 2014

### mafagafo

1. The problem statement, all variables and given/known data
In Exercises 45-46, show that the plane and line with the given equations intersect, and then find the acute angle of intersection between them.
45. The plane given by x + y + 2z = 0 and the line given by
x = 2 + t
y = 1 - 2t
z = 3 + t

Verbatim from Poole - Linear Algebra: A Modern Introduction.
2. Relevant equations
$$\cos{\theta}=\frac{\vec{u}\cdot\vec{v}}{||u||\cdot{}||v||}$$

3. The attempt at a solution
The normal vector of the plane is n = [1, 1, 2].
The direction vector of the line is d = [1, -2, 1].
The acute angle between the line and the plane should be the complement of
$$\cos^{-1}{\frac{\vec{n}\cdot\vec{d}}{||n||\cdot{}||d||}}$$
$$\frac{\vec{n}\cdot\vec{d}}{||n||\cdot{}||d||}=\frac{1}{6}$$
The complement is, approximately, 9.59406822686046 degrees.

The book says it is close to 80.4 degrees. (Not the complement, but theta on my calculations.)

What am I doing wrong?

2. Nov 27, 2014

### ehild

3. Nov 27, 2014

### mafagafo

It seems that many reviewers also found errors in this book. I know that this is not the best place to ask, but could you point me a better textbook on the subject? I don't think I am willing to put up with more hundreds of pages and a all the mistakes that come with them.

4. Nov 27, 2014

### LCKurtz

Just to add a comment, since $\hat n\cdot \hat d$ came out positive, you know that both of those vectors point to the same side of the plane. That is why you take the complement of the $\cos^{-1}$. Had the dot product come out negative, what would you have done?

5. Nov 27, 2014

### mafagafo

I would subtract $\pi{}/2$ from $\cos^{-1}$. (Based on what I am "rendering" in my brain.)

BTW, would you like to suggest a book?

6. Dec 2, 2014

### mafagafo

I did not misunderstand the term acute angle. But the calculation gives me the cosine of the angle between the vector and the normal of the plane, right? So if I want the acute angle between the vector and the plane I should get the complement of that angle.

7. Dec 2, 2014

### ehild

@Mark,
The angle of a plane and a straight line is equal to the angle the line encloses with its projection onto the plane. See: http://www.netcomuk.co.uk/~jenolive/vect14.html
cos-1(1/6) is the angle between the normal of the plane and the straight line. The angle between the plane and the line is 90°-cos-1(1/6), that is sin-1(1/6)

8. Dec 2, 2014

### mafagafo

Could you comment the answer I gave LCKurtz, he never came back.

9. Dec 2, 2014

### Staff: Mentor

You are right, and I apologize for misunderstanding what you're trying to do. I mistakenly thought you were trying to find the angle between the plane's normal and the line rather than the angle between the plane itself and the line.

• Linear Algebra and Differential Equations, Charles G. Cullen
• An Introduction to Linear Algebra with Applications, Steven Roman
• Linear Algebra, P. G. Kumpel and J.A. Thorpe
• Elementary Linear Algebra, 5th Ed., Howard Anton

10. Dec 2, 2014

### Staff: Mentor

Yes, I misunderstood what was being asked here.

11. Dec 2, 2014

### mafagafo

Thanks, and no problem about that. But it was somewhat settled and I was only waiting for LCKurtz to comment my on my answer. Also thanks for the books, I guess the first one is a bit out of what I can grasp but will look into the others.

12. Dec 2, 2014

### ehild

It looks correct.

13. Dec 2, 2014

### mafagafo

Make up your mind, I am writing a "proof" here. I was distracted but I still think I got it right.

14. Dec 2, 2014

### Staff: Mentor

Using the original two vectors, you had $cos(\theta) = 1/6$, from which we get $\theta \approx 80.4°$. If $cos(\theta) = -1/6$, you get $\theta \approx 99.6°$. Does it make sense to subtract 90° here if you want the angle between the two vectors (not the angle between the line direction and the plane)?

15. Dec 2, 2014

### mafagafo

Imagine the "normalized" normal being [0, 1, 0]. Let the vector v be [a, b, c] where are b is negative. The dot product is b, therefore negative. The cos^-1 gives me the angle (obtuse) between the normal and v. Let $\alpha$ be this angle. Let $\beta + \gamma = \alpha$ and Let $\beta$ be the angle between the normal and the plane (90 degrees) and $\gamma$ the angle between the direction vector and the plane. I get rid of beta and all that remains is the angle between the plane and the vector.

But I want the angle between the line and the plane (the angle of intersection), right?

16. Dec 2, 2014

### Staff: Mentor

Yes, in that case subtracting 90° gives you the angle between the plane and the line.

In your example with a normal of <0, 1, 0>, the plane in question is parallel to the x-z plane. A simpler example in two-d might be helpful. Instead of <0, 1, 0> let's look at the vector <1, 0>, a vector that is normal to the y-axis. For the line direction, let's use <-1, 1>.

Using the formula in your first post, which still applies here, we get $cos(\theta) = -1/2$ so $\theta$ = 120°. The acute angle that <-1, 1> makes with the normal is 180° - 120° = 60°. The angle that <-1, 1> makes with the y-axis (standing in for the plane in your example) is 120° - 90° = 30°.

I hope everything is clear now.

17. Dec 2, 2014

### mafagafo

Yes it is (and somewhat was) clear. I just wanted to know that my answer to Kurtz was correct, and it was. I have read so many examples with xyz that I forgot to use just xy in my "demonstration".

18. Dec 2, 2014

### LCKurtz

Sorry I didn't get back to you mafagafo. I had to go to Chicago last week so was away for a while. But, yes, your answer to my question was correct.

19. Dec 2, 2014

### mafagafo

Thanks and no problem, it is nice that you commented another possible case.