# Problem with vectors ?

Ok
Heres the text :Find the angle of vectors a=3p+2q and b=p+5q and if p and q are orthogonal unit ? ... ?
ok the formula is pretty simple cos=a*b/!a! * !b! =
But once i progress and I get 3p square + 15pq + 2qp +10q square ,now I see here that 15pq and 2qp are eliminated is just BEEEYOOOND MEEE I MEAN hows that possible ,thanks in advance ?!?!?

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mathman
p and q are orthogonal, which means (definition) that their dot product is 0.

p and q are orthogonal, which means (definition) that their dot product is 0.
More specific please Im trying to understand for future references ,thnx

Have you tried finding the vector sum of your two vectors? As the vector sum can be plugged into your formula, along with the two original vectors to find your angle, in which case you will need to do the inverse cos function. It.

theta=cos^-1((|A|*|B|)/AB))

Where * is dotted on.

Good luck.

mathman
More specific please Im trying to understand for future references ,thnx
Let a and b be arbitrary vectors (any dimension). The definition of orthogonal is the dot product is 0. Question for you - do you know what a dot product is?

Let a and b be arbitrary vectors (any dimension). The definition of orthogonal is the dot product is 0. Question for you - do you know what a dot product is?
dot product !a! * !b! *cos theta

But how in general can orthogonal plane have a dot product of 0 ?

rock.freak667
Homework Helper
dot product !a! * !b! *cos theta

But how in general can orthogonal plane have a dot product of 0 ?
If they are orthogonal, what is the angle between the vectors?

If they are orthogonal, what is the angle between the vectors?
like Zero ,lol ?

Mark44
Mentor
like Zero ,lol ?
Like, no.
Do you know what orthogonal means?

lorik said:
But how in general can orthogonal plane have a dot product of 0 ?
What you're asking here makes no sense.