Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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

4 Answers 4

up vote 4 down vote accepted

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, $\mathbf d = \mathbf 0 \iff \mathbf b = \mathbf c$.

share|improve this answer

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.

share|improve this answer
    
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 $b - c = 0$.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.