Question

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.

Answer 1

Which parts are confusing to you?

Answer 2

Did you attempt anything so far?

Answer 3

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

Answer 4

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

Answer 5

a is not quite right.

Answer 6

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

Answer 7

yes

Answer 8

Oh, then yes that's right

Answer 9

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.

Answer 10

That should give you another nice system you can solve.

Answer 11

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

Answer 12

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?

Answer 13

No, b has nothing to do with a.

Answer 14

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

Answer 15

oh no, so i'm meant to do this?:


-7i+5j +yk . q... cos 90 = 0

Answer 16

no.

Answer 17

The dot product of two vectors:
\[<a,b,c> \cdot <e,f,g>\ = ae + bf + cg\]

Answer 18

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

Answer 19

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

Answer 20

What class is this for?

Answer 21

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

Answer 22

This is something you probably would have covered in a linear algebra class or some other introduction to vectors

Answer 23

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

Answer 24

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

Answer 25

so y should be 11.3 recurring?

Answer 26

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.

Answer 27

If 14−20+3y=0 then 3y = 6 so y = 2

Answer 28

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

Answer 29

Well your original post said x = 7.

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

Yes. That will give you a vector that is orthogonal to each. You recall how to take the cross product?

Answer 35

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

Answer 36

the angle between them is 90 actually

Answer 37

damn

Answer 38

0 would mean they were parallel.

Answer 39

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

Answer 40

=)

Answer 41

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