Angles between sides of triangle ABC and unit vectors

[cr7einstein]

I was going through this link - https://books.google.co.in/books?id=bGnGtHkyWSAC&pg=SL1-PA46&lpg=SL1-PA46&dq=a+course+in+iitjee+mathematics+3+forces+along+the+sides+of+equilateral+triangle+find+resultant&source=bl&ots=T2jf9Vk4sR&sig=BmFQLhIQCm-6z8MeE-7Fgw1An1g&hl=en&sa=X&ei=0cucVcbCLM7muQTn-IvADA&ved=0CB4Q6AEwAA#v=onepage&q=a course in iitjee mathematics 3 forces along the sides of equilateral triangle find resultant&f=false

Chapter 2, page A35. Below the diagram on the right hand side, they give the formulae to compute unit vectors along the sides of EQUILATERAL triangle. I am unable to understand how do they get the formulas. Can anyone help? Moreover, is there a how do they get the angle as 240 degrees
in the second equation?

Please explain how are the equations derived, and how do you get the angles for unit vectors.

Thanks in advance!

vectors

 

[RUber]

If you look at the orientation of the vectors like in the diagram on p.A34 in the text you posted, you will see that AB+BC - AC = 0.
Then if you imagine each of those vectors, AB, BC, and -AC starting at the origin, you might find it easier to understand the angle values they are plugging into the equation. Of course, this is based on BC being at angle zero and the vectors being evenly spaced around the unit circle.

 

[jedishrfu]

I couldn't view the page you posted. I think Google limits the number of viewers and that limit has been exceeded.

Perhaps you could type what you want us to discuss or provide a screen capture of the page.

 

[HallsofIvy]

I cannot see the picture but if you are talking about an equilateral triangle then, of course, each of its interior angles is 60 degrees. If a general planar vector of unit length makes angle $\theta$ with the x-axis then its components are $cos(\theta)$ and $sin(\theta)$ so taking the base If of the triangle along the positive x-axis, one leg would be represented by the vector $cos(\theta)\vec{i}+ sin(\theta)\vec{j}= (\sqrt{3}/2)\vec{i}+ \frac{1}{2}\vec{j}$. The next leg, a vector starting from the tip of that one will make a 60 degree angle from the tip. And it will make an angle of 60 degrees with the third vector which lies along the x-axis but is directed back toward the negative. It makes angle 180- 60= 120 degrees with the positive x-axis so a vector in the opposite direction to the second leg is $cos(120)\vec{i}+ sin(120)\vec{j}= -\frac{1}{2}\vec{i}+ (\sqrt{3}/2)\vec{j}$. Since that is in the opposite direction, the second leg is given by $\frac{1}{2}\vec{i}- (-\sqrt{3}/2)\vec{j}$. Of course, the last leg, lying on the x-axis but directed back toward the first vertex is $-\vec{i}$

 

[SammyS]

Here is what OP is referring to:

That seems to be in reference to a figure above problem 2,4.

Will OP return ?