# Find an orthogonal basis of $R^3$ which contains a vector

Find an orthogonal basis of $${\rm I\!R}^3$$ which contains the vector $$v=\begin{bmatrix}1\\1\\1\end{bmatrix}$$.

I want to solve this without the use of the cross-product or G-S process. Please look at my solution and let me know if I did it right.

Let $$u$$ be an arbitrary vector $$u=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$$ that is orthogonal to $$v$$.

Thus,

$$v\ \bullet\ u = x_1 + x_2 + x_3 = 0$$
$$0= x_1 + x_2 + x_3$$
$$x_1= -x_2 -x_3$$

Any vector of the form $$\begin{bmatrix}-x_2 -x_3\\x_2\\x_3\end{bmatrix}$$ will be orthogonal to $$v$$.

Let $$x_2 = x_3 = 1$$
So, $$u=\begin{bmatrix}-2\\1\\1\end{bmatrix}$$ is orthogonal to $$v$$.

We now have two orthogonal vectors $$u$$ and $$v$$.

Put $$u$$ and $$v$$ as rows of a matrix, called $$A$$. Find a basis for $$A^\bot = null(A)^T$$:

Digression: I have memorized that when looking for a basis of $$A^\bot$$, we put the orthogonal vectors as the rows of a matrix, but I do not know why we put them as the rows and not the columns. Anyone care to explain the intuition?

$$A=\begin{bmatrix}1&1&1\\-2&1&1\end{bmatrix} \sim \begin{bmatrix}1&0&0\\0&1&1\end{bmatrix}$$
$$x_3 = x_3$$
$$x_2 = -x_3$$
$$x_1 = 0$$

A basis for $$null(A)$$ or $$A^\bot$$ with $$x_3$$ = 1 is: $$(0,-1,1)$$. Call this $$w$$.

Therefore, $$w$$ is orthogonal to both $$u$$ and $$v$$ and is a basis which spans $${\rm I\!R}^3$$.
By definition of orthogonal vectors, the set $$[u,v,w]$$ are all linearly independent.

Is this correct? If so, what is a more efficient way to do this? If not, how do you do this keeping in mind I can't use the cross product G-S process?

• 1st: I think you mean (Col A)$^\perp$ instead of A$^\perp$. Anyway, to answer your digression, when you multiply Ax = b, note that the i-th coordinate of b is the dot product of the i-th row of A with x. Now suppose x$\in$ Nul(A). Then b = 0, and so every row is orthogonal to x. – Joel Pereira Nov 12 '18 at 4:13

If you use the same reasoning to get $$w=(x_1,x_2,x_3)$$ (that you did to get $$v$$), then $$0=v\cdot w=-2x_1+x_2+x_3$$. So in general, $$(\frac{x_2+x_3}2,x_2,x_3)$$ will be orthogonal to $$v$$.
Now we get $$-x_2-x_3=\frac{x_2+x_3}2$$ (since $$w$$ needs to be orthogonal to both $$u$$ and $$v$$).
So, $$-2x_2-2x_3=x_2+x_3$$. Then $$x_2=-x_3$$.
So, say $$x_2=1,x_3=-1$$. Then we get $$w=(0,1,-1)$$.
• Can you clarfiy why $−x2−x3=\frac{x2+x3}{2}$ tells us that $w$ is orthogonal to both $u$ and $v$? I can't immediately see why. – udpcon Nov 12 '18 at 4:13
• We need a vector which simultaneously fits the patterns gotten by setting the dot products equal to zero. I've set $(-x_2-x_3,x_2,x_3)=(\frac{x_2+x_3}2,x_2,x_3)$. – Chris Custer Nov 12 '18 at 4:29