zainriaz said:
but the confuzing thing is that in A=the units r i^and k^ but in B=the units r i^and j^
how can we solve it when the units r different?
They're not really "units," but "unit vectors." But the name doesn't matter as much as knowing what they are.
So here's your problem: you have the following two vectors:
\vec A=2\hat i+6\hat k
\vec B=6\hat i+5\hat j
and you need to take the dot product. To do this, you need to use the distributive property. Think about this: if you were multiplying
(m + n)(p + q)
you would know how to expand that out, right? You'd get
mp + mq + np + nq
Stop me if you're not 100% comfortable with that.
To take the dot product of two vectors, you do the same thing, except that you replace regular multiplication with the dot product. So in the example above, if m,n,p,q were vectors instead of regular variables, what you get would be
(\vec m + \vec n)\cdot(\vec p + \vec q) = \vec m\cdot\vec p + \vec m\cdot\vec q + \vec n\cdot\vec p + \vec n\cdot\vec q
Does that make sense? Please ask if it doesn't.
If that does make sense, try doing the same thing with your two vectors \vec A and \vec B. You should get four terms, and each one should contain a product of two unit vectors, like \hat i\cdot\hat i[/itex] or \hat j\cdot\hat k, etc.<br />
<br />
Now you need to evaluate those products of unit vectors, i.e. replace them with their numeric values. The relevant identities have been given already in this thread, and you should also be able to look them up in your textbook or other reference. But you should also think about why they're true. They come from the definition<br />
\vec{A}\cdot\vec{B} = \vert A\vert \vert B\vert \cos\theta<br />
Think about which direction the various unit vectors point in space and what the angles between them are.
