Question

Let $\mathbf{u}=\left(\begin{array}{r}5\\ -6\\ 7\end{array}\right)$, and let $W$ be the set of all $\mathbf{x}$ in ${\mathbb{R}}^3$ such that $\mathbf{u}\cdot \mathbf{x}=0$. What theorem in Chapter 4 can be used to show that $W$ is a subspace of ${\mathbb{R}}^3$? Describe $W$ in geometric language.

Short answer

The theorem that can be used in chapter 4 is theorem 2. And geometrically, $W$ is a plane through the origin.

Step-by-step solution

Definition of Orthogonal sets

The two vectors $\mathbf{u}\mathrm{a}\mathrm{n}\mathrm{d}\mathbf{v}$ are Orthogonal if:

$\begin{array}{r}l{\Vert \mathbf{u}+\mathbf{v}\Vert }^2={\Vert \mathbf{u}\Vert }^2+{\Vert \mathbf{v}\Vert }^2\\ \mathrm{a}\mathrm{n}\mathrm{d}\\ \mathbf{u}\cdot \mathbf{v}=0\end{array}$.

Check whether $W$ is a subspace of ${\mathbb{R}}^3$ or not

The given vector is, $\mathbf{u}=\left(\begin{array}{r}\ast 20c5\\ -6\\ 7\end{array}\right)$ and $W=\left\{x\in {\mathbb{R}}^3|\mathbf{u}\cdot \mathbf{x}=0\right\}$.

Since is a null space of the $1\times 3$ matrix ${\mathbf{u}}^T$.

Therefore, Theorem 2 can be used to verify that $W$ is a subspace of ${\mathbb{R}}^3$, which is possible only, if $\mathbf{u}\cdot \mathbf{x}=0$ or ${\mathbf{u}}^T\cdot \mathbf{x}=0$, this shows that $W$ is a null-pace of ${\mathbf{u}}^T$. Hence $W$ is a subspace of ${\mathbb{R}}^3$.

Define geometrically

As $W$ has all the vectors which are perpendicular to $\mathbf{u}$. So find $\mathbf{u}\cdot \mathbf{x}=0$ by letting $\mathbf{x}=\left(\begin{array}{r}\ast 20c{x}_1\\ x_2\\ x_3\end{array}\right)$.

$\begin{array}{r}c\left(\begin{array}{c}\ast 20c5\\ -6\\ 7\end{array}\right)\cdot \left(\begin{array}{c}\ast 20c{x}_1\\ x_2\\ x_3\end{array}\right)=0\\ 5x_1-6x_2+7x_3=0\end{array}$

So geometrically, the subspace $W$ is a plane passing through the origin.