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Higher Maths Propertities of scalar vectors help?

I came across this question in the leckie and leckie higher student book and I am unsure on what to do, can anyone help? I tried to find the sides but then realised it wouldn't work as I am only told the angle.

a=2u+3v where u and v are vectors of unit length inclined at an angle of 60 degrees to each other. By first calculating a.a find the length of the vector a.
Original post by sabrinaclark23
I came across this question in the leckie and leckie higher student book and I am unsure on what to do, can anyone help? I tried to find the sides but then realised it wouldn't work as I am only told the angle.

a=2u+3v where u and v are vectors of unit length inclined at an angle of 60 degrees to each other. By first calculating a.a find the length of the vector a.

u and v are unit length, so you do know two of the triangle's sides and the angle, so can solve for the third side.
Original post by sabrinaclark23
By first calculating a.a find the length of the vector a.

Have you tried this? (2u+3v).(2u+3v)
Original post by RogerOxon
Have you tried this? (2u+3v).(2u+3v)

I wouldnt be able to use the unit lengths to find a side though, nope but i will go and try that! Thank you
Original post by RogerOxon
Have you tried this? (2u+3v).(2u+3v)


I still don't know what to do, could you try and work it out please? Thanks
Original post by sabrinaclark23
I still don't know what to do, could you try and work it out please? Thanks

I can give a little more info, but you really need to think about this and understand it.

Expand a.a:

a.a = (2u + 3v).(2u + 3v) = 4u.u + 9v.v + 6u.v + 6v.u = 4u.u + 9v.v + 12u.v

u and v are unit vectors, so you know u.u and v.v
You have the angle between u and v, so can calculate u.v
You can calculate the length of a from a.a
Original post by RogerOxon
I can give a little more info, but you really need to think about this and understand it.

Expand a.a:

a.a = (2u + 3v).(2u + 3v) = 4u.u + 9v.v + 6u.v + 6v.u = 4u.u + 9v.v + 12u.v

u and v are unit vectors, so you know u.u and v.v
You have the angle between u and v, so can calculate u.v
You can calculate the length of a from a.a


Thank you, I managed to do it by subsituting u and v in and then multiplying it by cos 60 and i got the square root of 169 which is 13. Do you think that is right as the answers say square root of 19.
Original post by sabrinaclark23
Thank you, I managed to do it by subsituting u and v in and then multiplying it by cos 60 and i got the square root of 169 which is 13. Do you think that is right as the answers say square root of 19.

I got sqrt(19). Could you post your workings?
Original post by RogerOxon
I got sqrt(19). Could you post your workings?


I seemed to get a different answer this time haha :frown: I've attacted it.
Earlier:
a.a = (2u + 3v).(2u + 3v) = 4u.u + 9v.v + 6u.v + 6v.u = 4u.u + 9v.v + 12u.v

u.u = 1
v.v = 1
u.v = cos(60) = 1/2

Therefore, a.a = 4 + 9 + 12/2 = 19

a.a = |a|^2, so the length of a is sqrt(a.a) = sqrt(19)

You went wrong with the dot products, substituting 2 and 3 for u and v. If there is still an issue, could you explain why you did that?

Edit: When writing vector symbols, I underline them, so that they do not get confused with scalars. I should have done that when I typed them too.
(edited 7 years ago)
Original post by RogerOxon
Earlier:
a.a = (2u + 3v).(2u + 3v) = 4u.u + 9v.v + 6u.v + 6v.u = 4u.u + 9v.v + 12u.v

u.u = 1
v.v = 1
u.v = cos(60) = 1/2

Therefore, a.a = 4 + 9 + 12/2 = 19

a.a = |a|^2, so the length of a is sqrt(a.a) = sqrt(19)

You went wrong with the dot products, substituting 2 and 3 for u and v. If there is still an issue, could you explain why you did that?

Edit: When writing vector symbols, I underline them, so that they do not get confused with scalars. I should have done that when I typed them too.


Thank you, I know where I went wrong now, could I just ask though, how did you work out that u and v were 1?
Original post by sabrinaclark23
Thank you, I know where I went wrong now, could I just ask though, how did you work out that u and v were 1?

u and v are not 1, but are of magnitude / length 1, from the question:

Original post by sabrinaclark23
where u and v are vectors of unit length inclined at an angle of 60 degrees to each other. By first calculating a.a find the length of the vector a.

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