DLA Examples

Section 13.4 Examples

In this section.

Subsection 13.4.1 Orthogonal vectors
Subsection 13.4.2 Orthogonal projection
Subsection 13.4.3 Cross product

Subsection 13.4.1 Orthogonal vectors

Example 13.4.1. Testing for orthogonality.

As in Discovery 13.2, and as discussed in Subsection 13.3.2, it’s fairly easy to form orthogonal vectors in

R^{2} .

And it’s not that much more difficult in

R^{3} .

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

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Example 13.4.2. Orthogonality of the standard basis vectors.

R^{n},

the standard basis vectors are always orthogonal to each other. When we compute

e_{i} \cdot e_{j}

with

i \neq j,

the

1

in the

i^{th}

component of

e_{i}

won’t line up with the

1

in the

j^{th}

component of

e_{j},

and we’ll get a computation something like

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\begin{aligned} e_{i} \cdot e_{j} & = 0 \cdot 0 + \dots + 0 \cdot 0 + \overset{i^{th} times i^{th}}{\overset{⏞}{1 \cdot 0}} + 0 \cdot 0 + \dots + 0 \cdot 0 + \overset{j^{th} times j^{th}}{\overset{⏞}{0 \cdot 1}} + 0 \cdot 0 + \dots + 0 \cdot 0 \\ = 0 . \end{aligned}

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Subsection 13.4.2 Orthogonal projection

Let’s complete the computations from Discovery 13.3.

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Example 13.4.3. Using orthogonal projection to compute distance from a point to a line in $R^{2}$ .

The line through the origin and parallel to

a = (3, 1)

consists of all scalar multiples of

a .

We would like to know the following.

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

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We know that the point we are looking for is at the terminal point of

{proj}_{a} u,

where

u = \vec{O Q} = (4, 4) .

So compute

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{proj}_{a} u = \frac{4 \cdot 3 + 4 \cdot 1}{3^{2} + 1^{2}} (3, 1) = \frac{8}{5} (3, 1) = (\frac{24}{5}, \frac{8}{5}),

which tells us that the point on the line closest to

Q

P (24 / 5, 8 / 5) .

Now, the vector

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n = \vec{P Q} = (- 4 / 5, 12 / 5)

will be a normal vector for the line, extending from

P

Q,

and so the norm of this vector represents the (perpendicular) distance between

Q

and the line:

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d = ‖ n ‖ = \sqrt{{(- \frac{4}{5})}^{2} + {(\frac{12}{5})}^{2}} = \sqrt{\frac{160}{25}} = \frac{4 \sqrt{10}}{5} .

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Subsection 13.4.3 Cross product

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

R^{3} .

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Example 13.4.4. 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

u = (1, 2, - 3)

and

v = (0, 2, 5) .

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The equation we are looking for is of the form

a x + b y + c z = d .

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

u

and

v .

We can use the cross product to compute such a vector:

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\begin{aligned} n = u \times v & = | \begin{array}{rrr} i & j & k \\ 1 & 2 & - 3 \\ 0 & 2 & 5 \end{array} | \\ = i (2 \cdot 5 - (- 3) \cdot 2) - j (1 \cdot 5 - (- 3) \cdot 0) + k (1 \cdot 2 - 2 \cdot 0) \\ = 16 i - 5 j + 2 k . \end{aligned}

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 .

But we also know that the plane passes through the point

(3, 3, 3),

so we must have

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16 \cdot 3 - 5 \cdot 3 + 2 \cdot 3 = d ⟹ d = 39 .

Thus, the plane can be described algebraically by the equation

16 x - 5 y + 2 z = 39,

or in point-normal form by the equation

n \cdot (x - x_{0}) = 0,

where

n

is as computed above,

x_{0}

is the “base” point

(3, 3, 3),

and

x = (x, y, z)

is a variable point.

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