# Calculus III Problem: find smallest angle

## 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
Homework Helper
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.

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
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