Thread: Vector Projection Derivation problem, why is this wrong

I've attached an image file. On it are two separate derivations for vector projection.
On the left side, I understand why the derivation is right, but I can't see why the derivation on the right is wrong.
Any ideas?

(I use v and b as vectors, and ||v|| and ||b|| for their scalar values)

Originally Posted by Tclack

I've attached an image file. On it are two separate derivations for vector projection.
On the left side, I understand why the derivation is right, but I can't see why the derivation on the right is wrong.
Any ideas? (I use v and b as vectors, and ||v|| and ||b|| for their scalar values)

BTW: note that is meaningless.

as Plato points out, you have to be sure when squaring, that you are squaring a scalar, as vectors can't be "squared" (vector multiplication is not, in general, defined).

the reason the derivation on the right is wrong, is because you wind up with a vector pointing in the direction of v, and you want one pointing in the direction of b.

the derivation on the right should start with:

proj_bv = |v|cosθ (b/|b|)

b/|b| is the UNIT vector pointing in the direction of b, and the "b-component" of v has MAGNITUDE |v|cosθ.

then, substituting in cosθ = (b.v)/(|b||v|) gives us:

[(b.v)/(|b|²)]b, as expected.

Thanks, I'm new to using the formal language of vectors. I've always tried to get around it, like just using scalar projection for things like blocks sliding down planes or law of sines and cosines to find out the angle between two vectors in 3-space, now that I'm starting vector valued functions, these shortcomings have come back to bite me in the ass. :]

i used to struggle with: what IS a vector?

they are "sort of" like numbers, we can add them and subtract them. where they differ, is how multiplication works. basically we have two "kinds" of multiplication:

A) "scalar multiplication": scalar x vector = vector. geometrically, this is "stretching/shrinking" a vector (stretching if our scalar is a > 1, shrinking if our scalar is 0 ≤ a < 1, and "stretching/shrinking in the opposite direction" if our scalar is negative).

B) "inner product" (the first kind you meet is usually called the "dot product"): vector x vector = scalar. this is most often used to give meaning to the term "angle between two vectors".

so vectors are a kind of "hybrid" structure, we have these things (vectors) which we can add, but multiplication requires interaction with scalars (which have their own numerical structure, that of a field, which we are more used to). often, to keep the two different TYPES of things (scalars and vectors) distinct, textbooks will use boldface letters for vectors, so you don't get mixed-up.

what makes vectors useful, is that they allow us to do a kind of arithmetic in "more dimensions than one" (so we are not limited to a "number line" for computation). since, in our world, things often happen "in space", vectors give a way of modelling interactions of more complicated things than just "amounts" (like sailing north, but the wind is blowing west, so how do you compensate?).

the underlying logic of vector spaces "makes more sense" if you know more about the different kinds of mathematical structure there can be, but it does require a sizable "detour". and for most practical applications, seeing a vector as an n-tuple of real (or complex) numbers (usually n = 2 or 3, but higher "dimensions" aren't any harder than these, there's just "more coordinates") is usually sufficient, it's not until one gets quite far along in mathematics that other, "weirder" vector spaces need to be tackled.

That's some deep stuff... allow me to bring the subject to a shallower level

do you have to take care to keep track of the direction of a vector? for example if you normalize a vector pointed in the negative direction, v= -i-j-k you get it's value, which will be a positive amount. That makes sense, but I was doing a problem and I almost got an answer that was the correct value but wrong sign, here's the problem and my work:

A force F= 4i - 6j + k Newtons applied to a point that moves a distance 15 meters in the direction b= i + j + k How much work is done?

apparently the answer is -5(Sqrt3)

Taking a step back and imagining the problem as a whole I can see that the vector b is in the positive direction +i,+j,+k direction and that the projection of F onto b is in the -1/3i,-1/3j,-1/3k direction, so I can see why is would be negative, but if b was in the direction +i,-j,+k then how would you know if work done was positive or negative?

the two vectors v and -v have the same magnitude (size, how "long the arrow is") but point in opposite directions. the thing is, there's not just "two directions" in space, there's 6 (actually 3, where each can be positive or negative):

+/- i
+/- j
+/- k

for a total of 8 different "octants" of 3-dimensional space. so just characterizing direction as positive or negative is too simplistic.

in your working-out above,

what you want is W = F.d = F.15(b/|b|) (b = (1,1,1) is not a unit vector, so we have to divide it by its magnitude, to get something we multiply by 15).

= (4,-6,1).(15/√3,15/√3,15/√3) = (15/√3)[(4,6,-1).(1,1,1)] = (5√3)(4 - 6 + 1) = -5√3

if you want to just consider what happens along the vector b = (1,1,1), we can resolve F into it's "b-component" by finding proj_bF:

proj_bF = [(F.b)/|b|²]b = [(4,-6,1).(1,1,1)/3](1,1,1) = (-1/3)(1,1,1) = (-1/3,-1/3,-1/3) = (-1/3)b = (-1/3)|b|(b/|b|)

= (-1/√3)b/|b| <---negative force in the direction of b.

so in the line determined by b, we have a force of -1/√3, which goes a distance 15, so

W = Fd = (-1/√3)(15) = (-3/√3)(5) = -5√3

what you did is calculate |W|, which conveys "how big the amount of work was", but not which WAY.

a thousand thanks for you help on both problems today Deveno... only 36 more to go in this chapter