DLA Examples

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Section 14.4 Examples

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In this section.

Subsection 14.4.1 Orthogonal vectors
Subsection 14.4.2 Orthogonal projection
Subsection 14.4.3 Cross product

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Subsection 14.4.1 Orthogonal vectors

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Example 14.4.1. Testing for orthogonality.

As in Discovery 14.2, and as discussed in Subsection 14.3.2, it's fairly easy to find orthogonal vectors in $\R^2\text{.}$ And it's not that much more difficult in $\R^3\text{.}$

The vectors $\uvec{u} = (3,7)$ and $\uvec{v} = (-7,3)$ are orthogonal in $\R^2\text{,}$ because
$\begin{equation*} \udotprod{u}{v} = 3\cdot (-7) + 7\cdot 3 = -21 + 21 = 0 \text{.} \end{equation*}$
The vectors $\uvec{u} = (3,7,1)$ and $\uvec{v} = (-7,2,7)$ are orthogonal in $\R^3\text{,}$ because
$\begin{equation*} \udotprod{u}{v} = 3\cdot (-7) + 7\cdot 2 + 1\cdot 7 = -21 + 14 + 7 = 0 \text{.} \end{equation*}$

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Example 14.4.2. Orthogonality of the standard basis vectors.

In $\R^n\text{,}$ the standard basis vectors are always orthogonal to each other. When we compute $\dotprod{\uvec{e}_i}{\uvec{e}_j}$ with $i\neq j\text{,}$ the $1$ in the $\nth[i]$ component of $\uvec{e}_i$ won't line up with the $1$ in the $\nth[j]$ component of $\uvec{e}_j\text{,}$ and we'll get a computation something like

$\begin{align*} \dotprod{\uvec{e}_i}{\uvec{e}_j} \amp = 0\cdot 0 + \dotsb + 0\cdot 0 + \overbrace{1\cdot 0}^{\nth[i]\text{ times }\nth[i]} + 0 \cdot 0 + \dotsb + 0\cdot 0 + \overbrace{0\cdot 1}^{\nth[j]\text{ times }\nth[j]} + 0 \cdot 0 + \dotsb + 0\cdot 0\\ \amp = 0\text{.} \end{align*}$

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Subsection 14.4.2 Orthogonal projection

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Let's complete the computations from Discovery 14.3.

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Example 14.4.3. Using orthogonal projection to compute distance from a point to a line in $\R^2$ .

The line through the origin and parallel to $\uvec{a} = (3,1)$ consists of all scalar multiples of $\uvec{a}\text{.}$ We would like to know the following.

What is the point on this line closest to the point $Q(4,4)\text{?}$
What is the distance from $Q$ to the line?

We know that the point we are looking for is at the terminal point of $\proj_{\uvec{a}} \uvec{u}\text{,}$ where $\uvec{u} = \abray{OQ} = (4,4)\text{.}$ So compute

$\begin{equation*} \proj_{\uvec{a}} \uvec{u} = \frac{4\cdot 3 + 4\cdot 1}{3^2 + 1^2}\; (3,1) = \frac{8}{5} (3,1) = \left(\frac{24}{5}, \frac{8}{5}\right)\text{,} \end{equation*}$

which tells us that the point on the line closest to $Q$ is $P(24/5,8/5)\text{.}$ Now, the vector

$\begin{equation*} \uvec{n} = \abray{PQ} = (-4/5,12/5) \end{equation*}$

will be a normal vector for the line, extending from $P$ to $Q\text{,}$ and so the norm of this vector represents the (perpendicular) distance between $Q$ and the line:

$\begin{equation*} d = \unorm{n} = \sqrt{\left(-\frac{4}{5}\right)^2 + \left(\frac{12}{5}\right)^2} = \sqrt{\frac{160}{25}} = \frac{4\sqrt{10}}{5}\text{.} \end{equation*}$

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Subsection 14.4.3 Cross product

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Here is an example of using the cross product to answer a geometry question in

$\R^3\text{.}$

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Example 14.4.4. Using cross product to determine the equation of a plane in $\R^3$ .

Suppose we would like to determine the equation of the plane in $\R^3$ that passes through the point $(3,3,3)$ and is parallel to the vectors $\uvec{u} = (1,2,-3)$ and $\uvec{v} = (0,2,5)\text{.}$

The equation we are looking for is of the form $a x + b y + c z = d\text{.}$ We know that $a,b,c$ can be taken to be the components of any normal vector for the plane. A normal vector for the plane must be orthogonal to the plane, and hence must be orthogonal to each of $\uvec{u}$ and $\uvec{v}\text{.}$ We can use the cross product to compute such a vector:

$\begin{align*} \uvec{n} = \ucrossprod{u}{v} \amp = \left\lvert\begin{array}{rrr} \ivec{} \amp \jvec{} \amp \kvec{} \\ 1 \amp 2 \amp -3 \\ 0 \amp 2 \amp 5 \end{array}\right\rvert\\ \amp = \ivec{}\bbrac{2\cdot 5 - (-3)\cdot 2} - \jvec{}\bbrac{1\cdot 5 - (-3)\cdot 0} + \kvec{}(1\cdot 2 - 2\cdot 0)\\ \amp = \injkvec{16}{5}{2} \text{.} \end{align*}$

So we can use $16 x - 5 y + 2 z = d$ as the equation of the plane, for some as-yet-to-be-determined value of $d\text{.}$ But we also know that the plane passes through the point $(3,3,3)\text{,}$ so we must have

$\begin{equation*} 16 \cdot 3 - 5 \cdot 3 + 2 \cdot 3 = d \qquad\implies\qquad d = 39 \text{.} \end{equation*}$

Thus, the plane can be described algebraically by the equation $16 x - 5 y + 2 z = 39\text{,}$ or in point-normal form by the equation $\dotprod{\uvec{n}}{(\uvec{x}-\uvec{x}_0)} = 0\text{,}$ where $\uvec{n}$ is as computed above, $\uvec{x}_0$ is the “base” point $(3,3,3)\text{,}$ and $\uvec{x} = (x,y,z)$ is a variable point.