# Angle at Ox

• Sep 20th 2009, 04:08 AM
Viral
Angle at Ox
I'm really confused on which angle here is "the angle at Ox", and how you find out.

That's a quick image I made of it. I have that as the vector $|-3i + 2j|$ but I'm not sure whether the angle is $a$ or $a + b$ or $b + c$ or $a + b + c$.
• Sep 20th 2009, 04:40 AM
Plato
Quote:

Originally Posted by Viral
That's a quick image I made of it. I have that as the vector $|-3i + 2j|$ but I'm not sure whether the angle is $a$ or $a + b$ or $b + c$ or $a + b + c$.

From your graphic, the usual understanding of $O_x=b+c$.
The that angle is $\pi + \arctan \left( {\frac{2}{{ - 3}}} \right)$.

However, you need to check your text material for the wanted definition.
• Sep 20th 2009, 04:49 AM
Viral
That's the problem =\ , the only information given is:

Quote:

Given that $a = i + j$ and $b = 2i - j$ find the magnitude of each of the following vectors and the angle each makes with Ox.
So, for the above example, it's the same as $180 - arctan(\frac{2}{3})$ ?

What about with this example: ?

I've been answering those ones as just $\theta = arctan(\frac{j}{i})$
I can give a standard approach. Consider the vector $p\vec{i}+q\vec{j}$.
The angle associated with that vector is: $\left\{ {\begin{array}{rl} {\arctan \left( {\frac{q}{p}} \right),} & {p > 0} \\ {\pi + \arctan \left( {\frac{q}{p}} \right),} & {p < 0\;\& \;q > 0} \\ { - \pi + \arctan \left( {\frac{q}{p}} \right),} & {p < 0\;\& \;q < 0} \\ \end{array} } \right.$