Question

Please show that the vector a is orthogonal to the hyperplane H = H(a,€); that is, if u and v are in H, then a is orthogonal to u - v.

Answer 1

Please show that the vector $\mathbf{a}$ is orthogonal to the hyperplane $\mathsf{H} = \mathsf{H}(\mathsf{a},\mathsf{E})$; that is, if $\mathbf{u}$ and $\mathbf{v}$ are in $\mathsf{H}$, then $\mathbf{a}$ is orthogonal to $\mathbf{u} - \mathbf{v}$.

**Solution.**

We present an example to illustrate this statement.

If the vector $\vec{a}$ is orthogonal to the $H$, then $\vec{a}$ is a normal vector for $H$. Let $\vec{a} = (2,3)$ and $H$ is a straight line with equation

$$2x + 3y = 1$$

Find points on $H$. Suppose point $(x; y)$ lies on $H$. We note that when $x = 2$, $y = -1$, so $\vec{u} = (2, -1)$ lies on $H$. Thus,

$$(2,3) \cdot ((x, y) - (2, -1)) = 0,$$

or, equivalently,

$$(2,3) \cdot (x - 2, y + 1) = 0,$$

is a normal equation for $H$. Since $\vec{v} = (-1,1)$ also lies on $H$, one of directions of the straight line $H$ is $\vec{v} - \vec{u} = (-3,2)$.

Note that

$$\vec{a} \cdot (\vec{v} - \vec{u}) = (2,3) \cdot (-3,2) = 0,$$

so $\vec{a}$ is orthogonal to $\vec{v} - \vec{u}$.