Calculus III Problem: find smallest angle

[lizbeth]

Homework Statement

Find the smallest angle between 3x-2y-4z = 5 and {x = 1 + 2t, y = 3 + 4t, z = 5 + 6t}.

Homework Equations

3x-2y-4z=5 is a plane, the other 3 equations constitute a line.

The Attempt at a Solution

My understanding is that the angle between a plane and a line can be found by using the direction vector of the line and the normal vector of the plane, and applying the dot product formula,
n . v = |n||v|cos Θ.
Doing this, I get that theta is the inverse cosine of -29 / (square root of 29 * square root of 56), which comes out to 130.18 degrees. Since the smallest angle would be the acute angle, I subtracted this from 180 and got 49.82 degrees.

Professor says this is the wrong answer. Can anyone help steer me in the right direction?

 

[Simon Bridge]

You should look to the details of your working - could be as simple as an arithmetic mistake.
Note: if n is the normal vector to the plane, and v is the vector pointing along the line, then arccos(n.v/|n||v|) gives the angle between n and v, not between v and the plane.

 

[lizbeth]

Thanks, Simon. I have checked the arithmetic of n . v and also |n| and |v|, and those are correct. The inverse cosine of course I have to rely on my calculator for. :)

I think what you're saying is that since the angle that results here is the angle between n and v, that the angle I found by subtracting from 180 is still the angle between n and v, just on the other "side" of the plane, and that this further needs to be subtracted from 90 to find the actual angle -- is that right??

 

[Simon Bridge]

Well done - you should be able to verify this by a sketch.

 

[lizbeth]

Great! Thank you!