this question is crushing my brain space!!! . Consider the vectors p = xi + 5j + yk and q = 2i 4j + 3k. a) For what values of x and y are the vectors p and q parallel? b) If x = 7 for what value of y are the vectors p and q perpendicular? Given that x and y takes these values nd a third vector that is perpendicular to both pand q.

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this question is crushing my brain space!!! . Consider the vectors p = xi + 5j + yk and q = 2i 4j + 3k. a) For what values of x and y are the vectors p and q parallel? b) If x = 7 for what value of y are the vectors p and q perpendicular? Given that x and y takes these values nd a third vector that is perpendicular to both pand q.

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Which parts are confusing to you?
Did you attempt anything so far?
i did, and got something with a) if parallel then they are scalar multiples of each other... so ap=q then axi = 2j a5=-4 ay= 3 q=(2 -4 3) p= (x 5 y) a=-4/5 x = -0.4 y=-4/15

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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
So for b) you know that the dot product of two vectors is 0 if and only if they are perpendicular. So if they are perpendicular you know that the dot product of the two must be 0.
That should give you another nice system you can solve.
haha ok, i might be back to confirm in a few minutes (maybe 10) :)
does that mean i should be able to find j and k directions' magnitude when x=-7 by comparing the other values i got previously?
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/
This is something you probably would have covered in a linear algebra class or some other introduction to vectors
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?
np. Basically you just multiply each component pair, and add all those products. So if p = i + 2j + 3k and q = 4i + 3j - 2k Then \(p \cdot q = (1\cdot4) + (2\cdot 3) + (3 \cdot -2) = 4 + 6 -6 = 4\) The notable thing about the dot product is that if the two vectors are orthogonal (perpendicular) then the dot product will equal 0.
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?
Yes. That will give you a vector that is orthogonal to each. You recall how to take the cross product?
add the sum of the products of the components of the two vectors multiplied by the sine of the angle between them ... but that doesnt seem like it'll work cause sine 0 is 0 so i will end up with 0 which doesn't help
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 ?

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