# For vectors in three-dimensional space, if $a \cdot b$ and $a \cdot c$ are equal, and $a \times b$ and $a \times c$ are equal, are b and c equal?

For vectors in three-dimensional space, if $a \cdot b$ and $a \cdot c$ are equal, and $a \times b$ and $a \times c$ are equal, are b and c equal? I tried looking for counter-examples or using coordinate-by-coordinate proofs, but that didn't get me anywhere.

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Obviously you want $a \ne 0$. In that case, $a \times b = a \times c$ implies $a \times (b-c) = 0$. Which vectors have 0 cross product with $a$ ? – Ted Sep 26 '11 at 4:45
why dont uu try expanding the given products and check for yourself? – Bhargav Sep 26 '11 at 4:45
@168355: I did, but couldn't achieve anything. Ted you are right. – multivar Sep 26 '11 at 4:46

We can conclude that $\mathbf b= \mathbf c$ provided $\mathbf a \neq \mathbf 0$.

Write $\mathbf d = \mathbf b - \mathbf c$. The given equations can be written as $$\mathbf a \cdot \mathbf d = 0, \ \ \ \mathbf a \times \mathbf d = \mathbf 0.$$ Now, suppose $\theta$ is the angle between $\mathbf a$ and $\mathbf d$. $$|\mathbf a \cdot \mathbf d| = |\mathbf a| |\mathbf d| |\cos \theta|, \ \ \ |\mathbf a \times \mathbf d| = |\mathbf a| |\mathbf d| |\sin \theta|, \ \ \$$ Then, squaring and adding, $$0 = 0 + 0 = |\mathbf a \cdot \mathbf d|^2 + |\mathbf a \times \mathbf d|^2 = |\mathbf a|^2 |\mathbf d|^2 (\cos^2 \theta + \sin^2 \theta) = |\mathbf a|^2 |\mathbf d|^2,$$ which implies that either $\mathbf a = \mathbf 0$ or $\mathbf d = \mathbf 0$. Finally, $\mathbf d = \mathbf 0 \iff \mathbf b = \mathbf c$.

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Yes. The equality of the dot products says the projection of $b$ and $c$ on $a$ are the same. The equality of the cross products says the perpendicular components are equal. When trying to check, you can rotate and scale so that $a=(1,0,0), b=(b_x,b_y,0)$ to make it easier.

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That makes sense because I couldn't find a counter-example, but how do I prove it? I got a big system of equations and not sure if I can show much – multivar Sep 26 '11 at 4:48

Yes: $a \cdot (b - c) = 0$, so $a$ is orthogonal to $b-c$. Also, $a \times (b - c) = 0$, so since $a$ and $b-c$ are perpendicular, $||a|| ||b - c|| = 0$. So either $a = 0$ or $b - c = 0$.

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Yes (provided that $a$ is not the zero vector). A somewhat different reasoning than the ones given above goes as follows. We can view the vectors as (Hamilton's) quaternions. In the ring of quaternions we have $$ab=-(a\cdot b)+a\times b=-(a\cdot c)+a\times c=ac.$$ The quaternions form a skewfield, so left cancellation of a non-zero element is legal.

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