# Cross product and dot product relationships

• October 15th 2008, 05:21 PM
Craka
Cross product and dot product relationships
The question is the form of true or false
If $\underline a \bullet \underline b = \underline a \bullet \underline c$ then $\underline b = \underline c$

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 $\underline a \times \underline b = \underline a \times \underline c$ then $\underline b = \underline c$

why does b not equal c in this case aswell?
• October 15th 2008, 05:31 PM
Jhevon
Quote:

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

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?

Quote:

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

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

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

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

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

does that mean $|b| = |c|$? if not, then clearly $b \ne c$
• October 15th 2008, 05:35 PM
Craka
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?
• October 15th 2008, 05:38 PM
Jhevon
Quote:

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
• October 15th 2008, 05:45 PM
Jhevon
see here. look under "algebraic properties"