anonymous 5 years ago 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.

1. anonymous

Which parts are confusing to you?

2. anonymous

Did you attempt anything so far?

3. anonymous

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

4. anonymous

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

5. anonymous

a is not quite right.

6. anonymous

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

7. anonymous

yes

8. anonymous

Oh, then yes that's right

9. anonymous

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.

10. anonymous

That should give you another nice system you can solve.

11. anonymous

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

12. anonymous

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?

13. anonymous

No, b has nothing to do with a.

14. anonymous

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

15. anonymous

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

16. anonymous

no.

17. anonymous

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

18. anonymous

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

19. anonymous

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

20. anonymous

What class is this for?

21. anonymous
22. anonymous

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

23. anonymous

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

24. anonymous

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

25. anonymous

so y should be 11.3 recurring?

26. anonymous

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.

27. anonymous

If 14−20+3y=0 then 3y = 6 so y = 2

28. anonymous

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

29. anonymous

Well your original post said x = 7.

30. anonymous

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

31. anonymous

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

32. anonymous

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

33. anonymous

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

34. anonymous

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

35. anonymous

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

36. anonymous

the angle between them is 90 actually

37. anonymous

damn

38. anonymous

0 would mean they were parallel.

39. anonymous

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

40. anonymous

=)

41. anonymous

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