# Is there any relation between Gram-Schmidt process in $\mathbb R^3$ and vector cross product?

Using Gram-Schmidt orthogonalization process we can find an orthogonal set of vectors from a given set of vectors,also we were taught previously that crossing between two non-collinear vectors gives a vector perpendicular to the two vectors.Is there any correlation between the two processes of find orthogonal system of vectors,are the two essentially the same?

• +1 Interesting question! Never thought of the connection before – gt6989b Jan 19 at 15:05

Note that the cross-product of two vectors is defined only on $$\mathbb R^3$$. So, I will assume that we are working on $$\mathbb R^3$$.

If you have $$3$$ linearly independent vectors $$v_1$$, $$v_2$$ and $$v_3$$, if you apply the Gram-Schmidt orthogonalization process to them and you obtain $$w_1$$, $$w_2$$, $$w_3$$, then$$w_3=\frac{v_1\times v_2}{\lVert v_1\times v_2\rVert}(=w_1\times w_2).\tag1$$So, if you are aware of the cross-product, it is enough to compute $$w_1$$ and $$w_2$$ and then to simply use $$(1)$$ to get $$w_3$$.

The answer is more no than yes.

In Gram-Schmidt 3D, the first step is to construct a vector $$v_2$$ orthogonal to $$u_1$$, using a linear combination of $$u_1$$ and $$u_2$$. This vector is coplanar with them. It has little to do with $$u_1\times u_2$$.

The second step is to construct a vector $$v_3$$ orthogonal to both $$u_1$$ and $$u_2$$, using a linear combination of $$u_1,u_2,u_3$$. This vector is indeed orthogonal to $$v_1$$ and $$v_2$$ but this is unavoidable. And it doesn't recover the length of $$v_1\times v_2$$.