Linear algebra problem!

Stuck on a particular question and can't seem to find the solution.
Help would be greatly appreciated.

just two lines...
apply the definitions of cross product and dot product
(note that the modulus of a unit vector is 1)

Factorize

well know trig identity

answer ...

I can get as far as using cauchy-schwarz and I end up with just the last two terms. I'm not sure what to do with the cross product one?

just two lines...
apply the definitions of cross product and dot product
(note that the modulus of a unit vector is 1)

Factorize

well know trig identity

answer ...

Are you sure you can do this? Don't you deduce the |axb| formula from this identity
(unless there is another way i don't know about), meaning this reasoning is circular.

Expanding coordinates seems the best way imo

~~

~~

~~

I think the definition of the cross product as $||\vec{a}||\cdot ||\vec{b}||\sin\theta \cdot\vec{n}$ with n a unit vector orthogonal to both a and b is.. simply terrible. It gives no method of computation, and the norm of the vector looks quite forced. There is a natural way to tie everything you (should) know about cross products together.

The task at hand is to take two vectors in R^3 and produce one which is orthogonal to both. Let $\vec{u}=(u_1,u_2,u_3)$ and $\vec{v}=(v_1,v_2,v_3)$ be a pair of linearly independent vectors, and consider the "determinant equation", $\begin{vmatrix} x & y & z \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} = 0.$

Expanding along the top row, we get:

$x\cdot \begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix} - y\cdot \begin{vmatrix} u_1 & u_3 \\ v_1 & v_3 \end{vmatrix} + z\cdot \begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix} = 0$

Which can be written

$(x,y,z)\cdot ( \begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix}, -\begin{vmatrix} u_1 & u_3 \\ v_1 & v_3 \end{vmatrix}, \begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix} )=0.$

Now in the original determinant, it is clear that if (x,y,z) is either u or v (or indeed any linear combination of these) then the equation is satisfied - this is just a property of deteminants. Having written it as a dot product, we've just shown that the vector $( \begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix}, -\begin{vmatrix} u_1 & u_3 \\ v_1 & v_3 \end{vmatrix}, \begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix} )=0$ is orthogonal to both u and v (and again, any linear combination of these - this is of course the plane defined by u and v). This vector is the cross product of u and v. This approach explains why we calculate cross products via a determinant.

Note we required that u and v be linearly independent. This is because otherwise the determinant is trivially zero. This is just saying u x au = 0 for any vector u and any scalar a.

Finally, if you put this cross product back into the first matrix, the determinant will no longer be zero (by construction, it couldn't be zero), it will instead be the volume of the parallelepiped spanned by u, v and their cross product. Numerically it comes out as $\begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix}^2 +\begin{vmatrix} u_1 & u_3 \\ v_1 & v_3 \end{vmatrix}^2+ \begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix}^2 =||u\times v||^2$

But since $u\times v$ is orthogonal to u and v, then the volume of the parallelepiped is $||u\times v||\cdot ||a||\cdot ||b||\cdot \sin\theta$, because ||u x v|| is just the height of the parallelepiped.

All together then, $||u\times v||^2 = ||u\times v||\cdot ||a||\cdot ||b||\cdot \sin\theta \implies ||u\times v|| = ||a||\cdot ||b||\cdot \sin\theta$


So hopefully that shows a few links that may have been unclear.