The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

This discussion is now closed.

Check out other Related discussions

vector question

Could someone give me some guidance on how to do this question

2 unit vectors a and b have acute angle gamma between them
show that b-(a.b)a is perperdicular to a and of length sin(gamma)

Thanks
Reply 1
Avatar for ttoby
ttoby
15
For the first part, recognise that b-(a.b)a = b-(cos gamma)a

You want to show that b-(cos gamma)a . a = 0

Remember that (b-(cos gamma)a) . a = b.a - (cos gamma)a.a and work it out from there.

For the second part, draw a diagram of the vectors a, b, (a.b)a and b-(a.b)a
Reply 2
Avatar for abcd9876abcd
abcd9876abcd
OP
ttoby
For the first part, recognise that b-(a.b)a = b-(cos gamma)a

You want to show that b-(cos gamma)a . a = 0

Remember that b-(cos gamma)a . a = b.a - (cos gamma)a.a and work it out from there.

For the second part, draw a diagram of the vectors a, b, (a.b)a and b-(a.b)a


I understand how u did the first part to get b-(cos gamma)a but don't get how u did the next part??
Reply 3
Avatar for ttoby
ttoby
15
For the next part, you multiply out the brackets to give (b−(cos⁡γ)a)⋅a=(b⋅a)−((cos⁡γ)a⋅a)(b-(\cos \gamma)a) \cdot a = (b \cdot a) - ((\cos \gamma)a\cdot a) which you can then work out, since you know the lengths of the vectors and the angle between them. I'll edit my post above to make it a bit clearer.
Reply 4
Avatar for sonic23
sonic23
11
ttoby
For the next part, you multiply out the brackets to give (b−(cos⁡γ)a)⋅a=(b⋅a)−((cos⁡γ)a⋅a)(b-(\cos \gamma)a) \cdot a = (b \cdot a) - ((\cos \gamma)a\cdot a) which you can then work out, since you know the lengths of the vectors and the angle between them. I'll edit my post above to make it a bit clearer.


sorry im stil not sure how you do the second part.

thanks
Reply 5
Avatar for mathz
mathz
8
if you dot a vector with itself you have the length squared, so
look at
[b-(a.b)a].[b-(a.b)a]

can you exapnd this ?

Related discussions

Latest

Trending

Trending

Articles for you