Vector operations(year 1 uni material)

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?

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 $\mathbf{a} = \mathbf{b}$

can you expand on why it doesnt necessarily need to be zero

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.

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.