Completing an orthonormal basis of a plane to a basis for $\mathbb{R}^3$

Question

I was asked to find an orthonormal basis for the plane $x + 2y +3z =0$. I found a regular basis, $(-2,1,0),(-3,0,1)$, and then performed the Gram-Schmidt process to find 2 orthonormal vectors that span the plane. I got the vectors $\frac{1}{\sqrt{5}}(-2,1,0), \frac{1}{\sqrt{70}}(-3,-6,5)$.

Then, I was asked to complete the basis to a basis of $\mathbb{R}^3$. So, I am looking for a vector $(x,y,z)$ such that $(x,y,z)\cdot\frac{1}{\sqrt{5}}(-2,1,0) = 0,\ (x,y,z)\cdot\frac{1}{\sqrt{70}}(-3,-6,5) = 0,\ x^2 + y^2 + z^2 = 1$.

From the first 2 equations, I made a system, and after reducing I got the matrix $$ \begin{pmatrix} -2 & 1 & 0\\ -3 & 0 & 1 \end{pmatrix} $$ After looking, I realized that the rows of this matrix are the original basis I found for the plane.

Does this happen for a reason or is it just a coincidence?

Answer 1

the vector $\pmatrix{1,&0,&0}^\top$ is not on the plane $x + 2y + 3z = 0.$ i can project this vector onto the normal to the plane and get the third ortonormal vector to complete the basis you have.

so $$(1,0,0)^\top = k(1,2,3) + b$$ choose $k$ so that $b$ is orthogonal to $(1,2,3)^\top.$ taking the dot product give you $k = \frac1{14}.$ so $$b = \frac{1}{14}\pmatrix{13,&-2, &-3}.$$ make the vector $b$ of length $1.$

Answer 2

What you notice is not a coincidence. It is just a consequence of the way the dot product is expressed using the vector coordinates.

And your last vector is the cross product of the previous two. Which avoids the computation you mention.

Answer 3

Why don't you perform Gram-Schmidt again?

If $u$ and $v$ are the vectors of the orthonormal basis of the plane, choose any $a$ that is not in the plane, and set $$b=a-\langle a,u\rangle u-\langle a,v\rangle v,$$ then normalise: $$w=\frac b{\lVert b\rVert}.$$