Dot product and cross product

**Pinkk** · August 30th 2010, 06:55 PM

Given vectors $a,b\in \mathbb{R}^{3}$ and that $a\cdot c = b\cdot c$ and $a \times c = b\times c$ for some nonzero vector $c$ , prove that $a=b$ .

So I showed that if $a = 0$ then $b=0$ , so now I assume $a,b$ are nonzero vectors. So $|a||c|\cos\theta = |b||c|\cos\phi$ and $|a||c|\sin\theta = |b||c|\sin\phi$ (where $\theta$ is the angle between $a$ and $c$ and $\phi$ is the angle between $b$ and $c$ ) and since all three vectors are nonzero, I can solve for one of the magnitudes and show that $|a| = |b|$ and then from there I get $cos\theta = \cos\phi$ . Now, can I assume that the angles must be in the range of $0\le \theta , \phi \le \pi$ and therefore that $\theta = \phi$ . If that conclusion can be made, is it sufficient to show that if two vectors have the same magnitude and that the angles between them and the vector $c$ are the same, then they are the same vector? If not, how would I go about proving this? Thanks.

**Ackbeet** · August 31st 2010, 03:02 AM

If that conclusion can be made, is it sufficient to show that if two vectors have the same magnitude and that the angles between them and the vector are the same, then they are the same vector?

I don't think so. You have to use more information about the cross product than just the magnitudes. Counterexample to this claim:

$a=\hat{x}, b=\hat{y}, c=\hat{z}.$

The dot products are all zero because of orthogonality. The magnitudes are all the same, and the angle between a and c is the same as between b and c: 90 degrees. But a is not equal to b.

You have to use all the hypotheses: the dot products are equal, the vector products are all equal (and not just the magnitudes, but the directions of the vector products are equal!), and you have to use the fact that c is nonzero.

I would probably go long here, and starting writing out equations component-wise, and see where I get.

**Plato** · August 31st 2010, 06:25 AM

Originally Posted by Pinkk

Given vectors $a,b\in \mathbb{R}^{3}$ and that $a\cdot c = b\cdot c$ and $a \times c = b\times c$ for some nonzero vector $c$ , prove that $a=b$ .

Using a well known identity, $c \times \left( {a \times c} \right) = \left( {c \cdot c} \right)a - \left( {c \cdot a} \right)c\;\;\& \;\; c \times \left( {b \times c} \right) = \left( {c \cdot c} \right)b - \left( {c \cdot b} \right)c$

From which we get $\left( {c \cdot c} \right)a - \left( {c \cdot c} \right)b = 0\; \Rightarrow \;a = b$

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