Vectors(i think they belong here)

[kathryn]

Hi, I have a question saying 'using knowledge of the dot products calculate the angle between these 2 vectors'.

a=1i+1j
b=-2i+2j

welll i think you work it out using the dot product rule which gives 0. (I THINK)
Then Im not sure what to do, do you use

a.b=ab cos(theta)
thanks

 

[steve_b]

a=1i+1j
b=-2i+2j

welll i think you work it out using the dot product rule which gives 0. (I THINK)
Then Im not sure what to do, do you use

a.b=ab cos(theta)

--------------

The dot produce does give 0, so substitute it into the equation you gave:

0 = ab cos(theta)

The "a" and "b" are the magnitudes of the vectors, and neither of them is 0, so the cosine(theta) must be 0. The first angle at which cos(x) = 0 is 90 degrees.

Hope that helps...

Steve

 

[soroban]

Hello, kathryn!

Using knowledge of the dot products calculate the angle between these two vectors:
\(\displaystyle \;\;\;\vec{a}\:=\:i\,+\,j,\;\;\vec{b}\:=\:-2i\,+\,2j\)

The formula is: \(\displaystyle \L\,\cos\theta\:=\:\frac{a\cdot b}{|a||b|}\)

We have: \(\displaystyle \L\,\cos\theta\;=\;\frac{\langle1,\,1\rangle\cdot\langle-2,\,2\rangle}
{\sqrt{1^2+1^2}\cdot\sqrt{(-2)^2+2^2}} \;=\;\frac{(1)(-2)\,+\,(1)(2)}{\sqrt{2}\cdot\sqrt{8}}\;=\;
\frac{-2\,+\,2}{\sqrt{16}}\;=\;\frac{0}{4}\;=\;0\)

Since \(\displaystyle \cos\theta\,=\,0\), then \(\displaystyle \,\theta\,=\,90^o\,\) (or \(\displaystyle 270^o)\)

 

[kathryn]

Thats great thanks, just one quick question can you use the same formula if you want to find the angle using the axb method instead? Or does sin replace cos?
Thanks

 

[pka]

Both of these are true:
\(\displaystyle \theta = \arccos \left( {\frac{{\left( {\vec A \cdot \vec B} \right)}}{{\left\| {\vec A} \right\|\,\left\| {\vec B} \right\|}}} \right)
\quad \& \quad
\theta = \arcsin \left( {\frac{{\left\| {\vec A \times \vec B} \right\| \cdot {{\rm sgn}} \left( {\vec A \cdot \vec B} \right)}}{{\left\| {\vec A} \right\|\,\left\| {\vec B} \right\|}}} \right)\).

However, I always suggest the use of the arccosine because it gives values \(\displaystyle 0 \le \theta \le \pi\) whereas arcsine returns values \(\displaystyle \frac{{ - \pi }}{2} \le \theta \le \frac{\pi }{2}\) which must be adjusted.