Find an Orthonormal Basis of $\R^3$ Containing a Given Vector

Vector Space Problems and Solutions

Problem 600

Let $\mathbf{v}_1=\begin{bmatrix}
2/3 \\ 2/3 \\ 1/3
\end{bmatrix}$ be a vector in $\R^3$.

Find an orthonormal basis for $\R^3$ containing the vector $\mathbf{v}_1$.

 
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The first solution uses the Gram-Schumidt orthogonalization process.
On the other hand, the second solution uses the cross product.

Solution 1 (The Gram-Schumidt Orthogonalization)

First of all, note that the length of the vector $\mathbf{v}_1$ is $1$ as
\[\|\mathbf{v}_1\|=\sqrt{\left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\right)^2+\left(\frac{1}{3}\right)^2}=1.\]


We want to find two vectors $\mathbf{v}_2, \mathbf{v}_3$ such that $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is an orthonormal basis for $\R^3$.
The vectors $\mathbf{v}_2, \mathbf{v}_3$ must lie on the plane that is perpendicular to the vector $\mathbf{v}_1$.
So consider the subspace
\[W=\left \{\,\begin{bmatrix} x\\y\\z \end{bmatrix} \in \R^3 \quad \middle | \quad \begin{bmatrix} x\\y\\z \end{bmatrix} \cdot \begin{bmatrix}
2/3 \\ 2/3 \\ 1/3
\end{bmatrix}=0 \,\right\}\]

Note that $W$ consists of all vectors that are perpendicular to $\mathbf{v}_1$, hence $W$ is a plane that is perpendicular to $\mathbf{v}_1$.


The relation
\[ \begin{bmatrix} x\\y\\z \end{bmatrix} \cdot \begin{bmatrix}
2/3 \\ 2/3 \\ 1/3
\end{bmatrix}=0\] can be written as
\[\frac{2}{3}x+\frac{2}{3}y+\frac{1}{3}z=0,\] or equivalently
\[z=-2x-2y.\] Hence the vectors in $W$ can be written as
\[\begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} x\\y\\-2x-2y \end{bmatrix}=x\begin{bmatrix}1\\0\\-2\end{bmatrix}+y\begin{bmatrix}0\\1\\-2\end{bmatrix}.\]

It follows that
\[\left\{\,\begin{bmatrix}1\\0\\-2\end{bmatrix}, \begin{bmatrix}0\\1\\-2\end{bmatrix} \,\right\}\] is a basis for the subspace $W$. Let us call these vectors $\mathbf{u}_1, \mathbf{u}_2$, respectively.


We apply the Gram-Schmidt orthogonalization to this basis $\{\mathbf{u}_1, \mathbf{u}_2\}$ and obtain an orthogonal basis as follows.
We do not change the first vector: let $\mathbf{w}_1=\mathbf{u}_1$.

Next, we set
\[\mathbf{w}_2=\mathbf{u}_2+a\mathbf{u}_1\] for some scalar $a$.
To determine $a$, we compute
\begin{align*}
0&=\mathbf{w}_1\cdot \mathbf{w}_2=\mathbf{u}_1\cdot(\mathbf{u}_2+a\mathbf{u}_1) \\
&=\mathbf{u}_1\cdot\mathbf{u}_2+a\mathbf{u}_1\cdot\mathbf{u}_1\\
&=(1\cdot 0+0\cdot 1+(-2)\cdot(-2))+a(1\cdot 1+ 0\cdot 0 +(-2)\cdot (-2)\\
&=4+5a.
\end{align*}
Hence, $a=-4/5$ and we obtain
\begin{align*}
\mathbf{w}_2&=\mathbf{u}_2-\frac{4}{5}\mathbf{u}_1\\[6pt] &=\begin{bmatrix}0\\1\\-2\end{bmatrix} -\frac{4}{5}\begin{bmatrix}1\\0\\-2\end{bmatrix}.
\end{align*}
(Note that you may use the Gram-Schumidt orthogonalization formula, instead of the above method.)


As scaling does not change the orthogonality, consider $5\mathbf{w}_2$, instead of $\mathbf{w}_2$ (to avoids fractions).
We have
\[5\mathbf{w}_2=\begin{bmatrix}0\\5\\-10\end{bmatrix} -4\begin{bmatrix}1\\0\\-2\end{bmatrix}=\begin{bmatrix}
-4\\ 5\\-2
\end{bmatrix}.\]

Therefore, $\{\mathbf{w}_1, 5\mathbf{w}_2\}$ is an orthogonal basis of $W$.


We obtain an orthonormal basis of $W$ by normalizing the length of these basis vectors.
As
\[\|\mathbf{w}_1\|=\sqrt{1^2+0^2+(-2)^2}=\sqrt{5}\] and
\[\|5\mathbf{w}_2\|=\sqrt{(-4)^2+5^2+(-2)^2}=\sqrt{45}=3\sqrt{5},\] the vectors
\[\mathbf{v}_2:=\frac{\mathbf{w}_1}{\|\mathbf{w}_1\|}=\frac{1}{\sqrt{5}}\begin{bmatrix}1\\0\\-2\end{bmatrix}\] and
\[\mathbf{v}_3:=\frac{5\mathbf{w}_2}{\|5\mathbf{w}_2\|}=\frac{1}{3\sqrt{5}}\begin{bmatrix}-4\\5\\-2\end{bmatrix}\] form an orthonormal basis of $W$.
Note that as the vectors $\mathbf{v}_2, \mathbf{v}_3$ lie in $W$, they are still perpendicular to the vector $\mathbf{v}_1$.


It follows that $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is an orthonomal set in $\R^3$, thus it is an orthonormal basis for $\R^3$.

Solution 2 (Cross Product)

Next, we solve the problem using the cross product.

Let $\mathbf{v}_1=\frac{1}{3}\mathbf{u}_1$, where $\mathbf{u}_1=\begin{bmatrix}
2 \\
2 \\
1
\end{bmatrix}$.
Our first goal is to find the vectors $\mathbf{u}_2$ and $\mathbf{u}_3$ such that $\{\mathbf{u}_1,\mathbf{u}_2, \mathbf{u}_3\}$ is an orthogonal basis for $\R^3$.


Let $\mathbf{x}=\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}$ be a vector that is perpendicular to $\mathbf{u}_1$.
Then we have $\mathbf{x}\cdot \mathbf{u}_1=0$, and hence we have the relation
\[2x+2y+z=0.\] For example, the vector $\mathbf{u}_2:=\begin{bmatrix}
1 \\
0 \\
-2
\end{bmatrix}$ satisfies the relation, and hence $\mathbf{u}_2\cdot \mathbf{u}_1=0$.
(So far, it is not so different from Solution 1.)


Now, let us define the third vector $\mathbf{u}_3$ to be the cross product of $\mathbf{u}_1$ and $\mathbf{u}_2$:
\[\mathbf{u}_3:=\mathbf{u}_1\times \mathbf{u}_2=\begin{bmatrix}
2 \\
2 \\
1
\end{bmatrix}\times \begin{bmatrix}
1 \\
0 \\
-2
\end{bmatrix}=\begin{bmatrix}
-4 \\
5 \\
-2
\end{bmatrix}.\] By the property of the cross product, the vector $\mathbf{u}_3$ is perpendicular to both $\mathbf{u}_1, \mathbf{u}_2$.

Therefore, the set
\[\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\}=\left\{\,\begin{bmatrix}
2 \\
2 \\
1
\end{bmatrix}, \begin{bmatrix}
1 \\
0 \\
-2
\end{bmatrix}, \begin{bmatrix}
-4 \\
5 \\
-2
\end{bmatrix} \,\right\}\] is an orthogonal basis for $\R^3$ as it consists of three nonzero orthogonal vectors.


Finally, to obtain an orthonormal basis for $\R^2$, we just need to normalize the lengths of these vectors.
The lengths are
\begin{align*}
\|\mathbf{u}_1\|&=\sqrt{2^2+2^2+1^2}=\sqrt{9}=3 \\
\|\mathbf{u}_2\|&=\sqrt{1^2+0^2+(-2)^2}=\sqrt{5}\\
\|\mathbf{u}_3\|&=\sqrt{(-4)^2+5^2+(-2)^2}=\sqrt{45}=3\sqrt{5}.
\end{align*}
Then we have $\mathbf{v}_1=\frac{\mathbf{u}_1}{\|\mathbf{u}_1\|}$.
Let $\mathbf{v}_2:=\frac{\mathbf{u}_2}{\|\mathbf{u}_2\|}$ and $\mathbf{v}_3:=\frac{\mathbf{u}_3}{\|\mathbf{u}_3\|}$.

It follows that the set
\[\{\mathbf{v}_1\mathbf{v}_2\mathbf{v}_3\}=\left\{\, \frac{1}{3}\begin{bmatrix}
2 \\
2 \\
1
\end{bmatrix}, \, \frac{1}{\sqrt{5}}\begin{bmatrix}1\\0\\-2\end{bmatrix}, \, \frac{1}{3\sqrt{5}}\begin{bmatrix}-4\\5\\-2\end{bmatrix} \,\right\}\] is an orthonormal basis for $\R^3$.


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