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I'm doing maths homework sheet and nearly finished it...
It is on vectors and the first question is easy but I don't know what to do for some reason...
Calculate length of vector a:
a = 5i + 2/2j - 4/3k
/ = square root
So what do I do?
It is on vectors and the first question is easy but I don't know what to do for some reason...
Calculate length of vector a:
a = 5i + 2/2j - 4/3k
/ = square root
So what do I do?
Page:
I'm doing maths homework sheet and nearly finished it...
It is on vectors and the first question is easy but I don't know what to do for some reason...
Calculate length of vector a:
a = 5i + 2/2j - 4/3k
/ = square root
So what do I do?
It is on vectors and the first question is easy but I don't know what to do for some reason...
Calculate length of vector a:
a = 5i + 2/2j - 4/3k
/ = square root
So what do I do?
Hmmm, I got a B in Higher Maths but I can't remember how to do that one. Vectors were pretty dificult. Usually I used to copy someone else's at school before class.
Just leave it out.
The tacher will probably never realise if it's right at the end.
But if they do just start whining that you didn't understand and they should have helped you.
That usually works
The tacher will probably never realise if it's right at the end.
But if they do just start whining that you didn't understand and they should have helped you.
That usually works
Erm...
Do you know the coordinates of i,j and k?
if so, sum the x and y values, i.e;
ax = 5*ix + 2/2*jx - 4/3*kx
ay = 5*iy + 2/2*jy - 4/3*ky
Then find the length of a by pythagoras
a = /[ax^2 + ay^2]
?
And to think i'm doing this for you on a sunday!
Do you know the coordinates of i,j and k?
if so, sum the x and y values, i.e;
ax = 5*ix + 2/2*jx - 4/3*kx
ay = 5*iy + 2/2*jy - 4/3*ky
Then find the length of a by pythagoras
a = /[ax^2 + ay^2]
?
And to think i'm doing this for you on a sunday!
SHEEPY wrote:
> So what do I do?
Ask Sonicrav; he's doing more A-level Maths modules than is healthy and if he hasn't covered vectors then I'd be very surprised.
> So what do I do?
Ask Sonicrav; he's doing more A-level Maths modules than is healthy and if he hasn't covered vectors then I'd be very surprised.
bah, hate damn mechanics...
Tried this stuff 2 years ago and failed it completly. Think I only got somthin like 5% on the final exam...Statistics are my strong point.
Tried this stuff 2 years ago and failed it completly. Think I only got somthin like 5% on the final exam...Statistics are my strong point.
Sheepy, just do it by Pythagorus
Length of Vector = sqr(5^2 + (2sqr2)^2 + (4sqr3)^2)
= sqr(25 + 8 + 48)
= sqr(81)
= 9
Length of Vector = sqr(5^2 + (2sqr2)^2 + (4sqr3)^2)
= sqr(25 + 8 + 48)
= sqr(81)
= 9
I take it i, j & k are the unit vectors on the x, y& z axes.
They should be. Then its just Pythagorus in three dimensions:
For eg
a = xi + yj + zk
gives length of a to be:
/(x^2 + y^2 + z^2)
/ is sqrt again and x^2 means x squared. Just stick in the values for x, y and z (5, 2/2 and 4/3 in this case)
I wont give you the answer, but it should be a whole number I think.
Any other qu's, I'm always happy to help.
They should be. Then its just Pythagorus in three dimensions:
For eg
a = xi + yj + zk
gives length of a to be:
/(x^2 + y^2 + z^2)
/ is sqrt again and x^2 means x squared. Just stick in the values for x, y and z (5, 2/2 and 4/3 in this case)
I wont give you the answer, but it should be a whole number I think.
Any other qu's, I'm always happy to help.
This is why I dropped Maths in school...
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