If a and b are vectors and a = (a.b)b, find the magnitude of vector b.?

Question

The equation are all in vectors. I know a.b is a scalar, question is can I just take the first b x the b beside it to form |b|^2 ? If so then what about the dot? I'm confused about the part how to expand (a.b)b. Can somebody pls help clarify? Thanks!

Answer 1

It's not a trick question, but it's tricky one.

a and b point in the same direction; they're just multiples of each other.

Based on that, you know that a.b has magnitude magnitude(a) * magnitude (b).

So it turns out that magnitude(a) * unit vector = mag (a) * mag (b) * mag (b) * unit vector.

And that means mag (b) has to equal 1.

Answer 2

No, u cant separate b from (a.b) like that!! The definition of dot product does not allow that & remember - NEVER question the definitions in Sciences !!!
Solution:
Because (a.b) is a scalar the given equation implies:
vector a is parallel to vector b ------- (1)
Now, taking modulus on both sides of given equation:
mod (a) = mod (a.b) * mod (b)
i.e, mod (a) = mod (a) * mod (b) cos (0 deg) * mod (b)
[ For cos (0 deg) - refer equation (1) ]
whence we get,
mod (b) ^2 = 1
implies: mod (b) = 1 [ as modulus of a vector can't be negative!! ]
i.e. MAGNITUDE of vector b = 1.

Answer 3

You don't expand it, and you certainly don't take (a·b)b to be equal to a(b·b). One of these quantities is a scalar multiple of b, and the other is a scalar multiple of a, which is impossible, unless b is a scalar multiple of a, which certainly isn't true in general (although it happens to be true in this case).

The key here is to take the dot product of both sides with b. Then you have:

a·b=((a·b)b)·b

Now, inner products, including dot products, are linear, so ((a·b)b)·b=(a·b)(b·b)=a·b.

Now, there are two possibilities. One of them is that (a·b)=0, which, since a is a sclar multiple of b, implies that a=b=0. If this is not the case, however, we may divide both sides by it, thereby obtaining:

b·b=1

Which implies:

|b|=1

So |b|=1 or 0