Vector operations(year 1 uni material)

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localmemelord
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Let c be a non-zero vector and suppose that a and b satisfy a·c=b·c(∗) Does it follow that a=b? Give a geometrical meaning to equation (∗).

i said yes it does follow by using the scalar product on both sides of the equation but this is wrong.
the answer was to multiply both sides by c/|c|^2 and that proceeds to say that vector a and b have the same projections on c, i.e., Proj c a=Proj c b.

can someone explain how they arrived at the final answer and why they used c/|c|^2?
Last edited by localmemelord; 11 months ago
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RDKGames
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(Original post by localmemelord)
Let c be a non-zero vector and suppose that a and b satisfy a·c=b·c(∗) Does it follow that a=b? Give a geometrical meaning to equation (∗).

i said yes it does follow by using the scalar product on both sides of the equation but this is wrong.
the answer was to multiply both sides by c/|c|^2 and that proceeds to say that vector a and b have the same projections on c, i.e., Proj c a=Proj c b.

can someone explain how they arrived at the final answer and why they used c/|c|^2?
The statement is false and can be shown by choosing a,b,c to be the three unit vectors in \mathbb{R}^3.

Anyway, I find their reasoning not too clear.

I would just note that

 \mathbf{a}\cdot \mathbf{c} = \mathbf{b} \cdot \mathbf{c} \implies(\mathbf{a}-\mathbf{b}) \cdot\mathbf{c} = 0

which means that \mathbf{a} - \mathbf{b} is perpendicular to \mathbf{c}. It does not necessarily need to be zero, so we do not necessarily have \mathbf{a} = \mathbf{b}
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DFranklin
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(Original post by localmemelord)
Let c be a non-zero vector and suppose that a and b satisfy a·c=b·c(∗) Does it follow that a=b? Give a geometrical meaning to equation (∗).

i said yes it does follow by using the scalar product on both sides of the equation but this is wrong.
the answer was to multiply both sides by c/|c|^2 and that proceeds to say that vector a and b have the same projections on c, i.e., Proj c a=Proj c b.

can someone explain how they arrived at the final answer and why they used c/|c|^2?

If c is a unit vector, then (a.c) is the length of the projection of a on c, and then c(a.c) is the actual projected vector.

The "/|c^2|" is what you get when you adjust for c not being a unit vector (i.e. scaling both occurrences of c in c(a.c) to be unit vectors).
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localmemelord
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(Original post by RDKGames)
The statement is false and can be shown by choosing a,b,c to be the three unit vectors in \mathbb{R}^3.

Anyway, I find their reasoning not too clear.

I would just note that

 \mathbf{a}\cdot \mathbf{c} = \mathbf{b} \cdot \mathbf{c} \implies(\mathbf{a}-\mathbf{b}) \cdot\mathbf{c} = 0

which means that \mathbf{a} - \mathbf{b} is perpendicular to \mathbf{c}. It does not necessarily need to be zero, so we do not necessarily have \mathbf{a} = \mathbf{b}
can you expand on why it doesnt necessarily need to be zero
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DFranklin
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(Original post by localmemelord)
can you expand on why it doesnt necessarily need to be zero
To take a really simple case: suppose c = i and we want a.c = 0. If x, y, z are the components of a (i.e. a = xi + yj + zk), then a.c = x. So requiring that a.c = 0 only tells us with x = 0; we are left with a completely free choice for y and z. Explicitly, we could have c = j or c = k, or any linear combination of these solutions.

On a more generic level, in an n-dimensional space, the condition a.c = 0 is equivalent to a single condition of the form a_1c_1 + a_2c_2 + ... a_nc_n = 0, and requiring this to be true can only reduce the dimension of possible values for c by 1. I.e. a.c = 0 defines an n-1 dimensional subspace of \mathbb{R}^n.
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RDKGames
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(Original post by localmemelord)
can you expand on why it doesnt necessarily need to be zero
Let \theta be the angle between (\mathbf{a} - \mathbf{b}) and \mathbf{c}.

(\mathbf{a}-\mathbf{b}) \cdot \mathbf{c} \equiv |\mathbf{a} - \mathbf{b}| |\mathbf{c}| \cos \theta

Recall this??

So if this is zero, then either |\mathbf{a} - \mathbf{b}| = 0 (in which case \mathbf{a} = \mathbf{b}), OR we simply need \cos \theta = 0.

Hence it is not necessary that we must have the first condition.
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localmemelord
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(Original post by RDKGames)
Let \theta be the angle between (\mathbf{a} - \mathbf{b}) and \mathbf{c}.

(\mathbf{a}-\mathbf{b}) \cdot \mathbf{c} \equiv |\mathbf{a} - \mathbf{b}| |\mathbf{c}| \cos \theta

Recall this??

So if this is zero, then either |\mathbf{a} - \mathbf{b}| = 0 (in which case \mathbf{a} = \mathbf{b}), OR we simply need \cos \theta = 0.

Hence it is not necessary that we must have the first condition.
that makes sense because for (a-b).c to be equal to zero, (a-b) has to be perpendicular to c (cos90 = 0)or the length of (a-b) has to be zero. also since c is a non zero vector, c cannot have a zero-length.

one more thing, in the above comment you said their reasoning is unclear, why do you think they used c/|c|^2?
and finally what does it mean for something to have a projection on something else, for e.g. projection of a on c?
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DFranklin
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(Original post by localmemelord)
that makes sense because for (a-b).c to be equal to zero, (a-b) has to be perpendicular to c (cos90 = 0)or the length of (a-b) has to be zero. also since c is a non zero vector, c cannot have a zero-length.

one more thing, in the above comment you said their reasoning is unclear, why do you think they used c/|c|^2?
and finally what does it mean for something to have a projection on something else, for e.g. projection of a on c?
I have explained this; if you don't know what projection means (for vectors), this is the relevant wiki page:

https://en.m.wikipedia.org/wiki/Vector_projection

Some of that page is a bit abstract but the diagram showing projection and rejection should be helpful.
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localmemelord
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(Original post by DFranklin)
I have explained this; if you don't know what projection means (for vectors), this is the relevant wiki page:

https://en.m.wikipedia.org/wiki/Vector_projection

Some of that page is a bit abstract but the diagram showing projection and rejection should be helpful.
thanks!
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