Finding the Counter Clockwise Angle of Vector Difference B-A with the +x Axis

In summary, the conversation discusses finding the counterclockwise angle that the vector difference B-A makes with the +x axis. The given components for B-A are 2.77 and -5.95, and the correct answer is found to be 65 degrees using the inverse tangent function. The mistake was made when taking the tangent of the components separately instead of dividing the magnitude of the two.
  • #1
ashcash
3
0
I have a question on finding the counter clockwise angle the vector difference B-A makes with the +x axis
I already have the components of vector difference for B-A and I have checked that they are correct they are 2.77,-5.95
I started with doing the tan function (opp/adj) which was -5.95/2.77
I got -2.148 which I found the tan(-2.148) to be -1.536
I then found the arctan to be 56.934
I then subtracted 360-56.93 and got an answer to be 303.07
This proved to be wrong and I am wondering if someone could help find where my mistake is.
 
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  • #2
I find that your question is a bit unclear, do you have the original question?

You got $2.77$ and $-5.95$; are those the lengths of the $x$ and $y$ components of the vector respectively?

$$\tan\left({\theta}\right)=\frac{5.95}{2.77}$$

I do not get $\theta$ the be $57$ degrees.
 
  • #3
Sorry for the confusing question, the question word for word is
"Find the counterclockwise angle the vector difference B⃗ −A⃗ makes with the +x axis."
2.77 is for B and -5.95 is for A
So my math must be off, I am assuming I went wrong getting
tan(θ)=-5.95/2.77
First I took what I got from dividing the two which was -2.148, and plugged it in my calculator and like this (tan -2.148) and got 1.536
then I do arctan(1.536) am I right?
 
  • #4
I see your mistake now.

For all intents and purposes, the signs don't don't matter since we're dealing with magnitudes here. We can just adjust accordingly depending on what quadrant we are on.

$$\tan\left({\theta}\right)=\frac{5.95}{2.77}$$

Theta is computed by the inverse tangent:

$$\tan^{-1}\left({\frac{5.95}{2.77}}\right)\approx65 \text{ degrees}$$

If your calculator is on radians, use the conversion factor $\frac{180}{\pi}$

- - - Updated - - -

Notice, however, that your question wants $\vec{B-A}$, so we need to flip the vector $\vec{A}$ in the opposite directions. Therefore, the vector lies on the first quadrant.
 
  • #5
I understand my mistake now!
Thank you so much!
 

Related to Finding the Counter Clockwise Angle of Vector Difference B-A with the +x Axis

What is a vector in trigonometry?

A vector in trigonometry is a mathematical quantity that has both magnitude and direction. It is typically represented by an arrow pointing in the direction of the vector's magnitude.

How do I solve a trigonometry problem involving vectors?

To solve a trigonometry problem with vectors, you will need to break down the vectors into their components, typically in the x and y directions. Then, you can use trigonometric functions such as sine, cosine, and tangent to find the magnitudes and directions of the vectors.

Can vectors be used in three-dimensional space?

Yes, vectors can be used in three-dimensional space. In addition to having x and y components, vectors in three-dimensional space also have a z component to represent movement in the vertical direction.

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A scalar in trigonometry is a quantity that has only magnitude, while a vector has both magnitude and direction. Examples of scalars include temperature, distance, and time, while examples of vectors include velocity, force, and displacement.

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