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

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    Trig Vectors
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Discussion Overview

The discussion revolves around finding the counterclockwise angle that the vector difference B - A makes with the +x axis. Participants explore the mathematical approach to calculating this angle using trigonometric functions and clarify the components of the vectors involved.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant presents the components of the vector difference B - A as 2.77 and -5.95, and describes their initial approach using the tangent function to find the angle.
  • Another participant questions the clarity of the original question and seeks confirmation on whether the provided components correspond to the x and y components of the vector.
  • A later reply clarifies the original question and suggests that the tangent should be calculated as tan(θ) = 5.95 / 2.77, indicating a potential misunderstanding in the initial calculations.
  • One participant identifies a mistake in the calculation process, suggesting that the signs of the components can be adjusted based on the quadrant in which the angle lies.
  • There is a mention of needing to consider the direction of vector A when determining the final angle, implying that the angle should be in the first quadrant.

Areas of Agreement / Disagreement

Participants express differing views on the calculations and the interpretation of the vector components. There is no consensus on the correct angle calculation, as participants are still clarifying their understanding and addressing mistakes.

Contextual Notes

There are unresolved aspects regarding the initial calculations and the interpretation of the vector components, particularly concerning the signs and the quadrant in which the angle lies.

ashcash
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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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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.
 
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?
 
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.
 
I understand my mistake now!
Thank you so much!
 

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