Properties of the dot product

Question

Suppose we have three vectors $\textbf{A}$, $\textbf{B}$ and $\textbf{C}$. If $\textbf{A}\cdot\textbf{C}=\textbf{B}\cdot\textbf{C}$, does that mean that $\textbf{A}$ must be equal to $\textbf{B}$? If so, can this property be proven?

Though the question is mainly mathematical, it has occurred to me a number of times when studying physics and I'll like a good explanation.

Now, the fundamental theorem for gradients states that $$ V (\textbf{b}) - V (\textbf{a}) = \int_\textbf{a}^\textbf{b}(\nabla V)\cdot d\textbf{l}, $$ so $$ \int_\textbf{a}^\textbf{b}(\nabla V)\cdot d\textbf{l} = -\int_\textbf{a}^\textbf{b}\textbf{E}\cdot d\textbf{l}. $$ Since, finally, this is true for any points $\textbf{a}$ and $\textbf{b}$, the integrands must be equal: $$ \bbox[10px,border:1px solid black]{\textbf{E} = -\nabla V.}\tag{2.23} $$

As an example of such a case, I have added an excerpt from Griffiths' Introduction to Electrodynamics. In the calculations, it was assumed that ${\textbf{E}}$ is equal to $-\nabla V$ based on the fact that $\textbf{E}\cdot d\textbf{l}=-(\nabla V)\cdot d\textbf{l}$ .

Answer 1

From $$\vec{A}\cdot\vec{C}=\vec{B}\cdot\vec{C}$$ you can conclude $$\vec{A}\cdot\vec{C}-\vec{B}\cdot\vec{C}=0$$ or $$(\vec{A}-\vec{B})\cdot\vec{C}=0.$$ However, this does not necessarily mean $\vec{A}-\vec{B}=\vec{0}$.

You can only conclude (from the definition of the dot product) that $\vec{A}-\vec{B}$ is perpendicular to $\vec{C}$.

Answer 2

Here is a proof. If $\mathbf{A}\cdot\mathbf{C} =\mathbf{B}\cdot\mathbf{C}$ for all $\mathbf{C}$, then $(\mathbf{A}-\mathbf{B})\cdot\mathbf{C} = 0$ for all $\mathbf{C}$. In particular we can choose $\mathbf{C} = \mathbf{A}-\mathbf{B}$ so that $(\mathbf{A}-\mathbf{B})\cdot(\mathbf{A}-\mathbf{B})=0$. Since the dot product is positive definite, $\mathbf{v} \cdot \mathbf{v} = 0$ only if $\mathbf{v} = 0$. We conclude $\mathbf{A} - \mathbf{B} = 0$, so $\mathbf{A} = \mathbf{B}$.

Answer 3

If for given $\vec A$ and $\vec B$ the equality $\vec A\cdot\vec C = \vec B\cdot\vec C$ holds for all vectors $\vec C$, or at least for a set of generators (say, a basis), then we can conclude that the two vectors are equal, otherwise we can't.

I will try to make it plausible: If we take the standard basis $\{\vec e_x, \vec e_y, \vec e_z\}$ for vector $\vec C$, then we get

$$\vec A\cdot\vec e_x = \vec B\cdot\vec e_x$$ $$\vec A\cdot\vec e_y = \vec B\cdot\vec e_y$$ $$\vec A\cdot\vec e_z = \vec B\cdot\vec e_z$$

But $\vec A \cdot \vec e_i=A_i$, the i-th component of the vector. So we have just shown that $A_x = B_x$, $A_y=B_y$ and $A_z=B_z$ and thus $\vec A = \vec B$.

On the other hand, let us assume that the equality holds for the first two basis vectors but doesn't for the last one, $\vec e_z$, then we know that the first two components of $\vec A$ and $\vec B$ coincide but the z components don't and thus the two vectors aren't equal.

The approach works for any basis, not necessarily the standard basis. Then you will get components with respect to the given basis -- if they all coincide, then the vectors also coincide.

It's not a rigorous proof but maybe helps to make the statement intuitive.

Answer 4

The argument Griffiths is making (I think. My copy of the text isn't with me right now), is not $$ -\vec{E} \cdot d\vec{l} = \nabla V \cdot d\vec{l} \implies-\vec{E} = \nabla V $$ Rather, he is arguing that if $$ -\int_{\vec{a}}^{\vec{b}} \vec{E} \cdot d\vec{l} = \int_{\vec{a}}^{\vec{b}} \nabla V \cdot d\vec{l} $$ for all $\vec{a},\vec{b} \in \mathbb{R}^3$ (really all paths between all such $\vec{a}$ and $\vec{b}$), then $$ -\vec{E} = \nabla V $$

In other words, Griffiths' argument doesn't really hinge on any property of the dot product. It has a lot more to do with the properties of integrals and continuous functions.

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Proof of Griffiths' Claim

Observe that if we take $\vec{F} = \nabla V - \vec{E}$, all we have to prove is that if $$ \int_\gamma \vec{F} \cdot d\vec{l} = 0 $$ for all paths $\gamma$ in $\mathbb{R}^3$, then $\vec{F} = 0$.

We will need to assume that $\vec{F}$ is continuous for the argument to work. We will prove the contrapositive. Suppose $\vec{F} \neq 0$. This means there exists $\vec{x}_0 \in \mathbb{R}^3$ such that $\vec{F}(\vec{x}_0) \neq 0$. Set $\vec{a} = \vec{F}(\vec{x}_0)/|\vec{F}(\vec{x}_0)|$. Define $h(\vec{x}) = \vec{F}(\vec{x}) \cdot \vec{a}$. Observe that $h(\vec{x}_0) = |\vec{F}(\vec{x}_0)| > 0$. Set $\epsilon = h(\vec{x}_0)/2 > 0$. Since $h$ is continuous, $h^{-1}(\epsilon , \infty)$ is open and contains $\vec{x}_0$. Thus, there exists a ball $B_r(\vec{x}_0)$ of radius $r > 0$ centered at $\vec{x}_0$ contained in $h^{-1}(\epsilon, \infty)$. Now $$ \vec{x}_0 + t \vec{a} \in B_r(\vec{x}_0) $$ for all $t \in (-r , r)$, from which it follows $$ h\left(\vec{x}_0 + t\vec{a}\right) > \epsilon $$ for all $t \in (-r, r)$.

Let $\gamma(t) = \vec{x}_0 + t\vec{a}$ for $t \in [-r/2, r/2]$. Then, by definition $$ \int_\gamma \vec{F} \cdot d\vec{l} = \int_{-r/2}^{r/2} \vec{F}(\gamma(t)) \cdot \gamma ' (t) \, dt = \int_{-r/2}^{r/2} \vec{F}(\gamma(t)) \cdot \vec{a} \, dt = \int_{-r/2}^{r/2} h(\gamma(t)) dt \geq \epsilon \cdot r > 0 $$ This completes the proof of the contrapositive, which is equivalent to our desired claim.

Answer 5

Consider this: $$\vec{A}.\vec{B}=AB\cos{\theta_1}$$ $$\vec{A}.\vec{C}=AC\cos{\theta_2}$$ If both the results are equal: $$AB\cos{\theta_1}=AC\cos{\theta_2}$$ or$$B\cos{\theta_1}=C\cos{\theta_2}$$ As you can see that we have two factors on which the result depends. So we cn easily manipulative them to be different vectors.

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For eg. let $$\vec{A}=i + j$$ $$ \vec{B}=2i + 3j$$ $$\vec{C}=3i + 2j$$ Clearly $$\vec{A}.\vec{B}=\vec{A}.\vec{C}= 5$$ But $$\vec{B}\ne\vec{C}$$