# Thread: Cross product and dot product relationships

1. ## Cross product and dot product relationships

The question is the form of true or false
If $\displaystyle \underline a \bullet \underline b = \underline a \bullet \underline c$ then $\displaystyle \underline b = \underline c$

The answer is given as false, I cannot understand why though. If the dot product of lhs and rhs equal each other and they share vector a in the calculation of the dot product how can vector b and c not equal each other ?

and similarly for if $\displaystyle \underline a \times \underline b = \underline a \times \underline c$ then $\displaystyle \underline b = \underline c$

why does b not equal c in this case aswell?

2. Originally Posted by Craka
The question is the form of true or false
If $\displaystyle \underline a \bullet \underline b = \underline a \bullet \underline c$ then $\displaystyle \underline b = \underline c$

The answer is given as false, I cannot understand why though. If the dot product of lhs and rhs equal each other and they share vector a in the calculation of the dot product how can vector b and c not equal each other ?
what if a is the zero vector?

and similarly for if $\displaystyle \underline a \times \underline b = \underline a \times \underline c$ then $\displaystyle \underline b = \underline c$

why does b not equal c in this case aswell?
if $\displaystyle a \times b = a \times c$ then $\displaystyle |a \times b | = |a \times c|$. that means $\displaystyle |a||b| \sin \theta_1 = |a||c| \sin \theta_2$

clearly if $\displaystyle a$ is the zero vector, we have the same problem as we did with the dot product.

if $\displaystyle a$ is not the zero vector, then we can divide both sides by $\displaystyle |a|$ to get

$\displaystyle |b| \sin \theta_1 = |c| \sin \theta_2$

does that mean $\displaystyle |b| = |c|$? if not, then clearly $\displaystyle b \ne c$

3. Sorry I missed putting that part in that question, a is not a zero vector.

Think I follow that with the cross product, and similarly I would do the same with the dot product though it would be in terms of cosine?

4. Originally Posted by Craka
Sorry I missed putting that part in that question, a is not a zero vector.
ok, then pick b such that it is perpendicular to a. so the dot product is zero. it means therefore, i can pick c to be any multiple of +/- b and i would still get the dot product to be zero (since multiples of b are parallel to be and would therefore also be perpendicular to a)

thus, we found b and c that make the claim false, so it is false

5. see here. look under "algebraic properties"