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.

currently I am working on Linear Algebra in University. I have a keen interest in mathematics and I also know that it can become frustrating at times. Right know I am frustrated and I seek for help in understanding the following:

  1. I know how to evaluate the dot/cross product and projection, but my biggest problem is when it comes to the exercises. I can't figure out what the results on the dot/cross product represent. My textbook is full with formal proofs, but for the sake of usability is does not a great job. I have my exam by the end of December and I need some advice so I can continue my studies and actually understand the more advanced topics such as Euclidean Vector Spaces. I apologize if this is not the usually question, but I am desperate for help. Thank you very much for your time and effort.


share|improve this question
Here is a related question: math.stackexchange.com/questions/77/…, –  Jonas Meyer Dec 8 '12 at 22:02
I appreciate this helpful link. Thanks –  Southpaw Dec 9 '12 at 5:02
add comment

1 Answer

Geometrically Speaking:

The dot product is vital for measuring lengths and angles. First ${\bf u} \cdot {\bf u} = ||{\bf u}||^2$ and second, assuming ${\bf u}$ and ${\bf v}$ are non-zero, we have ${\bf u} \cdot {\bf v} = 0$ if and only if ${\bf u}$ and ${\bf v}$ are perpendicular.

The cross product of two non-zero, non-parallel vectors gives a third vector that is perpendicular to the first two. Thus: ${\bf u} \perp ({\bf u} \times {\bf v}) \perp {\bf v}.$

Moreover, the dot product and the cross product are related by the triple scalar product:

$$[{\bf u},{\bf v},{\bf w}] = ({\bf u} \times {\bf v})\cdot {\bf w}$$

The triple scalar product is the volume of the parallelepiped spanned by ${\bf u}$, ${\bf v}$ and ${\bf w}$.

Abstractly Speaking:

The dot product can be used to show that a vector space $V$ is isomorphic to its dual vector space $V^*$. Recall that $V^*$ consists of all linear maps $f : V \to \mathbb{R}.$ The isomorphism $\phi : V \to V^*$ is given by $\phi : {\bf u} \mapsto f_{{\bf u}}$ where $f_{{\bf u}}({\bf v}) = {\bf u} \cdot {\bf v}$ for all ${\bf v}$ in $V$.

share|improve this answer
Thank you very much for your answer, it helped me a lot. –  Southpaw Dec 9 '12 at 5:03
add comment

Your Answer


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.