# Angle Between Two Vectors: Solving for θ

• nabelekt

## Homework Statement

Vectors and have scalar product -7.00 and their vector product has magnitude 3.00.
What is the angle between these two vectors?

## Homework Equations

|A| |B| cosθ = -7
|A| |B| sinθ = 3

## The Attempt at a Solution

tanθ = (-3/7)
tan-1(-3/7)=θ

When I enter tan-1(-3/7) into my calculator, I get -23.199, but when I enter 23.199 into MasteringPhysics, it tells me that I am incorrect. What am I doing wrong?

Thanks!

Is the angle expected in degrees or radians?

The answer is required to be in degrees.

What is the range of arctan? From the given values, can you determine whether the angle is really in that range?

I believe that the range of arctan is 270°<θ<90°. Correct? 23.199° seems to be in that range…

Thanks for the help. It's been a while since I've done this stuff.

The range given does not make sense. The range should be (-π/2,π/2).

I guess it would actually lie in quadrant II whereas the range of arctan is quadrants I and IV. So what should I do?

You should express you angle via some auxiliary angle that is in the range of arctan, obtain the value of the auxiliary angle, and then get your angle.

## Homework Statement

Vectors and have scalar product -7.00 and their vector product has magnitude 3.00.
What is the angle between these two vectors?

## Homework Equations

|A| |B| cosθ = -7
|A| |B| sinθ = 3

## The Attempt at a Solution

tanθ = (-3/7)
tan-1(-3/7)=θ

When I enter tan-1(-3/7) into my calculator, I get -23.199, but when I enter 23.199 into MasteringPhysics, it tells me that I am incorrect. What am I doing wrong?

Thanks!

You should take care to notice two things here:

(1) If |A| |B| cosθ = -7, then cosθ must be negative. In which quadrants is cosθ negative?

(2) tanθ has a period of ∏ radians, so tanθ = tan(θ+n∏) for any integer n, and tan-1(tanθ) = θ +n∏. Can you find a value of n that gives you an angle in the correct quadrant for θ +n∏=tan-1(-3/7)?