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Which parts are confusing to you?

Did you attempt anything so far?

does that mean i have done a) and then im stuck on b?

a is not quite right.

wait, is the q vector 2i -4j + 3k ?

yes

Oh, then yes that's right

That should give you another nice system you can solve.

haha ok, i might be back to confirm in a few minutes (maybe 10) :)

No, b has nothing to do with a.

just plug in the given value for x
and solve
\[p \cdot q = 0\]

oh no, so i'm meant to do this?:
-7i+5j +yk . q... cos 90 = 0

no.

The dot product of two vectors:
\[ \cdot \ = ae + bf + cg\]

In your case you have x=7 so:
p = <7,5,y>
q = <2,-4,3>
\(\implies p \cdot q = 14 - 20 + 3y = 0\)

oh, turns out i didn't know how to use dot product nor do i understand it.

What class is this for?

http://www.maths.uq.edu.au/courses/MATH1050/

The dot product of two vectors is the sum of the product of their components.

gah
sorry, i have no memory of doing vectors in high school, and this is 8 years later now

so y should be 11.3 recurring?

If 14−20+3y=0 then 3y = 6 so y = 2

oh right, but you made x=-7 into x=7 there :)

Well your original post said x = 7.

I can't help it if I solve the problem you give me ;p

so it does, but thats not what the actual question says on the pdf i copied :) damn

Ok well then it'd be -14 - 20 + 3y = 0 so 3y = 34 so y = 34/3

so to finish the question i have ot take the cross product of q and latest version of p?

the angle between them is 90 actually

damn

0 would mean they were parallel.

thats cool then,
now ill only have 4 more questions like that to get through today

=)

so just to make sure im not making more silly mistakes, is the new vector for b) -14i - 20j + 34k ?