Recent content by FilipaE

Sorry didnt think the first message had sent!

So the answer is 108? Thankyou! For [(2a+b) x(a+2b)]^2 is this the same as (2a+b) x(a+2b) dot (2a+b) x(a+2b) = |(2a+b) x(a+2b)|^2 ?? I know i have to use some sort of distributive law here but i am not sure on how to do it would it be 2a x a + 2a x 2b + b x a + b x 2b ? where 2a x a = 0...

Thankyou! I have got the answer 108? So for [(2a+b) x(a+2b)]^2 I do [(2a+b)x(a+2b)] dot [(2a+b)x(a+2b)] = |(2a+b)x(a+2b)|^2 I know to do this i need to use the distributive law. Is it 2a x a + 2a x2b + b xa + b x2b? Where 2a x a = 0 and bx2b=0?

|v|^2?

Is it the dot product of the two vectors, creating a scalar?

yes i do now, but given this information how would i write the vector a x b?

Oh i see, so how does the formula differ without the modulus sign? and what happens where there is addition involved as in Q1 part 2

This is the definition of cross product online |a x b|=|a||b|sinx I have no idea what you mean vectors is a very confusing topic for me

and (a x b)^2 means the cross product of a and b all sqaured

I am using x as the cross product, meaning the cross product of a and b is |a||b|sinx, is this not correct?

Homework Statement #1 Given that the angle between the vectors a and b is 2Pi/3 and |a|=3 and |b|=4 calculate: (axb)^2 [(2a+b)x(a+2b)]^2 #2 Given three unit vectors, a, b, c such that a+b+c=0 find (a dot b) + (b dot c) + (c dot a) #3 Given AB=a+2b BC=-4a-b CD= -5a-3b...