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 $\textbf{E}\cdot d\textbf{l}=-(\nabla V)\cdot d\textbf{l}$ .

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    $\begingroup$ Nope, A does not have to be the same vector as B. A and B only need to have the same parallel component along C for A. C=B. C. $\endgroup$ – Luo Zeyuan Oct 31 '20 at 7:00
  • $\begingroup$ But texts (Griffith's Introduction to Electrodynamics for instance) assume that the vectors are equal. $\endgroup$ – Toba Oct 31 '20 at 7:28
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    $\begingroup$ The argument in electromagnetism is a little more nuanced than just comparing dot products: the argument there involves the integral. Since the two integrals are always equal, irrespective of the path taken, no matter how small or how it's oriented, it is equivalent to saying the quantities that are being integrated are equal. It's not the result of a dot product identity. $\endgroup$ – Philip Oct 31 '20 at 7:39
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    $\begingroup$ On this site Mathjax is the expected way to enter and display mathematical expressions. Images of text and equations are very strongly discouraged. Please edit your question to use Mathjax. $\endgroup$ – StephenG Oct 31 '20 at 8:12
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    $\begingroup$ @LuoZeyuan: even if $A$ and $B$ have the same length, and even in two dimensions, we can't deduce that $A=B$. Take for instance $A=(1,1), B=(-1,1), C=(0,1)$. $\endgroup$ – TonyK Nov 2 '20 at 14:02

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.

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    $\begingroup$ Another proof goes as follows (though yours is as good!). Since $\vec{A}\cdot\vec{C}=\vec{B}\cdot\vec{C}$ for all $\vec{C}$, we have $(\vec{A}-\vec{B})\cdot\vec{C}=0$ for all $\vec{C}$. Hence taking $\vec{C}=\vec{A}-\vec{B}$, we have $(\vec{A}-\vec{B})\cdot(\vec{A}-\vec{B})=|\vec{A}-\vec{B}|^2=0$. Hence $\vec{A}-\vec{B}=\vec{0}$, since it has norm equal to zero. $\endgroup$ – Jilal Jahangir Oct 31 '20 at 20:08

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 $\vec{C}$.

  • $\begingroup$ I agree, the equality of the dot products is not enough to guarantee that A=B. To conclude that A=B, we need the additional condition that A.C=B.C for all values of C. Photon gave an intuitive proof for the statement. $\endgroup$ – Toba Oct 31 '20 at 12:32

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 $\mathbf{A} = \mathbf{B}$.

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    $\begingroup$ An equally great answer, an alternative (dare I say more rigorous) proof to that given by Photon. I would've marked it too if this site allowed marking more than one answer. $\endgroup$ – Toba Oct 31 '20 at 19:24

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.

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.

  • $\begingroup$ There is still an implicit assumption that E.dl=-(grad V).dl . This is because in the proof of the statement you gave, a very short path was considered. If we take a very short path in the equation given in Griffiths, the integral signs disappear (since there's nothing to sum along such a path) and E.dl=-(grad V).dl results. Pardon my clumsy notation, I'm unable to use Mathjax on my smartphone. $\endgroup$ – Toba Oct 31 '20 at 19:18
  • $\begingroup$ The difference is that the "short" path used in this argument isn't infinitesimal. It has finite length. I didn't want to spell it out here, but this is basically the same argument as the one presented here: math.stackexchange.com/questions/274702/… $\endgroup$ – Charles Hudgins Oct 31 '20 at 19:24
  • $\begingroup$ Because of the smoothness of $\vec{F}$ (which we assume), there must be some path of finite length along which $\vec{F} \cdot d\vec{l} > 0$. So we really are talking about an integral and not merely $\vec{F} \cdot d\vec{l}$. $\endgroup$ – Charles Hudgins Oct 31 '20 at 19:28
  • $\begingroup$ The short path is not infinitesimal in the argument I gave. It only has to be small enough that E and grad V can be considered constant along its length. $\endgroup$ – Toba Oct 31 '20 at 19:33
  • $\begingroup$ The path would have to be infinitesimal for $\vec{E}$ and $\nabla V$ to be constant along its length. Rather, we choose the path short enough that the deviations from being constant are not large enough to make the integrand nonpositive. Thus the integrand is strictly positive on a set of strictly positive measure, which means its integral is strictly positive, contradicting our initial assumption. If I'm not making sense, I can make the proof fully rigorous if you'd like. $\endgroup$ – Charles Hudgins Oct 31 '20 at 19:37

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.

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 $$\vec{B}\ne\vec{C}$$


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