Math 511: Linear Algebra

5.1 Orthogonality

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5.1.1 Geometry

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We discussed the dot product and cross product in vector space $\mathbb{R}^n$ from the point of view of a linear transformation. In this chapter we want to add geometry to the algebra of our vector spaces. The physicist perspective of a vector in $\mathbb{R}^n$ is an arrow in this space. This implies both a direction that the arrow is pointing and length of the arrow.

Recall that the standard position of a vector is when the tail of the vector sits at the origin and the tip sits at the coordinates of the vector with respect to the canonical or standard basis $\left\{\,\mathbf{\hat{\imath}},\ \mathbf{\hat{\jmath}}\,\right\}$. The direction of the vector is the angle the vector makes with the positive horizontal axis, span$\left\{\,\mathbf{\hat{\imath}}\,\right\}$.

Figure 1

The length or magnitude of the vector is the linear length from the tail to the tip of the vector. In standard position vector with standard coordinates

$$ \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} $$

is associated with the triangle with vertices $(0,0)$, $(x_1,0)$, and $(x_1,x_2)$. The length or magnitude of the vector is the length of the hypotenuse. This length is also called the modulus of the vector. Using the Pythagorean theorem

$$ |\mathbf{x}| := \sqrt{x_1^2 + y_1^2} $$

Figure 2

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5.1.2 Determining the Angle of a Vector

The angle or direction can be computed in standard coordinates $\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ using the right triangle with vertices $(0,0)$, $(x_1,0)$, and $(x_1,x_2)$.

Figure 3

Notice that the lengths of the legs of the right triangle associated with vector $\mathbf{x}$ are $|x_1|$ and $|x_2|$. Since vector $\mathbf{x}$ is in the second quadrant, the acute angle we obtain from the associated right triangle is the reference angle $\pi-\theta$. The reference angle is given by

$$ \begin{align*} \pi-\theta &= \tan^{-1}\left(\dfrac{|x_2|}{|x_1|}\right) = \cos^{-1}\left(\dfrac{|x_1|}{x_1^2 + x_2^2}\right) = \sin^{-1}\left(\dfrac{|x_2|}{x_1^2 + x_2^2}\right) \\ \\ &= \tan^{-1}\left(\dfrac{|x_2|}{|x_1|}\right) = \cos^{-1}\left(\dfrac{|x_1|}{|\mathbf{x}|}\right) = \sin^{-1}\left(\dfrac{|x_2|}{|\mathbf{x}|}\right) \end{align*} $$

Vector $\mathbf{y}$ is in the first quadrant so we do not need find the absolute values of the coordinates for the lengths of the legs. The acute angle of the right triangle is our direction or angle of vector $\mathbf{y}$,

$$ \begin{align*} \phi &= \tan^{-1}\left(\dfrac{y_2}{y_1}\right) = \cos^{-1}\left(\dfrac{y_1}{y_1^2 + y_2^2}\right) = \sin^{-1}\left(\dfrac{y_2}{y_1^2 + y_2^2}\right) \\ \\ &= \tan^{-1}\left(\dfrac{y_2}{y_1}\right) = \cos^{-1}\left(\dfrac{y_1}{|\mathbf{y}|}\right) = \sin^{-1}\left(\dfrac{y_2}{|\mathbf{y}|}\right) \\ \end{align*} $$

One can compute the trigonometric functions of the angle of a vector from its standard coordinates,

$$ \begin{align*} \cos(\theta) &= \dfrac{x_1}{|\mathbf{x}|}\qquad\text{and} &\sin(\theta) &= \dfrac{x_2}{|\mathbf{x}|} \\ \\ \cos(\phi) &= \dfrac{y_1}{|\mathbf{y}|}\qquad\text{and} &\sin(\phi) &= \dfrac{y_2}{|\mathbf{y}|} \end{align*} $$

However computing inverse trigonometric functions requires consideration. Computing the acute reference angle using the standard inverse trigonometric functions with the absolute values of the coordinates, determining the quadrant of the vector, and determining the direction from the reference angle is simpler and more accurate.

There are computing environments with inverse trigonometric functions which require two inputs, the coordinates $x_1$ and $x_2$, so that the correct direction angle can be returned.

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5.1.3 Dot Product

In his video Dot Products and Duality, Grant Sanderson explains the definition of intuition for the dot product in vector spaces $\mathbb{R}^n$, and their relationship with linear algebra.

If we have two vectors $\mathbf{x}$ and $\mathbf{y}$ in standard position, we draw a line segment from the tip of vector $\mathbf{y}$ perpendicular to vector $\mathbf{x}$,

Figure 4

The vector with the same direction as vector $\mathbf{x}$ and length equal to the leg of the resulting right triangle is call the projection of vector $\mathbf{y}$ onto vector $\mathbf{x}$ is defined by the vector

As $0\le\theta\le\pi$ is the angle between two vectors, the length of vector $\mathbf{p}$ is given by

$$ |\mathbf{p}| = |\mathbf{y}||\cos(\theta)|, $$
where an acute angle implies that the projection of $\mathbf{y}$ onto $\mathbf{x}$ has the same direction as $\mathbf{x}$, and an obtuse angle signifies that the projection of $\mathbf{y}$ onto $\mathbf{x}$ has the opposite direction as $\mathbf{x}$.

Definition Dot Product

The scalar product or dot product of two vectors $\mathbf{x}$ and $\mathbf{y}$, is the product of the lengths of vector $\mathbf{x}$ and the projection $\ \mathbf{p}$ of vector $\ \mathbf{y}$ onto vector $\ \mathbf{x}$.

$$ \left<\mathbf{x},\mathbf{y}\right> := |\mathbf{x}|\,|\mathbf{p}| = |\mathbf{x}|\,|\mathbf{y}|\cos(\theta), $$

where $\theta$ is the included angle between the two vectors.

We learn in the video above that, for a fixed vector $\mathbf{y}\in\mathbb{R}^n$, this is a linear transformation

$$ L[\mathbf{x}] = \left<\mathbf{x},\mathbf{y}\right> $$

from the vector space $\mathbb{R}^n$ to the real numbers $\mathbb{R}$.

$$ L\,:\,\mathbb{R}^n\rightarrow\mathbb{R} $$

Any linear transformation from one finite dimensional vector space to another can be represented by a matrix with respect to the canonical bases, and this matrix is $\mathbf{y}^T$,

$$ \left<\mathbf{x},\mathbf{y}\right> = L[\mathbf{x}] = \mathbf{y}^T\mathbf{x} = x_1y_1 + x_2y_2 + \cdots + x_ny_n = \mathbf{x}\cdot\mathbf{y} $$

From this identity, one obtains equations for the vector projection and the length of the vector projection, the scalar component of vector $\mathbf{y}$ onto $\mathbf{x}$.

Definition Component

The scalar component of one vector $\mathbf{y}$ onto vector $\mathbf{x}$ in $\mathbb{R}^n$ is give by

$$ \text{Comp}_{\mathbf{x}}\mathbf{y} := |\mathbf{p}| = |\mathbf{y}|\cos(\theta) = \dfrac{|\mathbf{x}||\mathbf{y}|\cos(\theta)}{|\mathbf{x}|} = \dfrac{\mathbf{x}\cdot\mathbf{y}}{|\mathbf{x}|} = \dfrac{\mathbf{y}^T\mathbf{x}}{|\mathbf{x}|} $$

This definitions for the vector projection and component rely on the fact that the vector $\mathbf{u} = \dfrac{\mathbf{x}}{|\mathbf{x}|}$ is a vector that points in the same direction as vector $\mathbf{x}$, but has a length of one. Here length is the Euclidean length $\left|\mathbf{x}\right|=\sqrt{x_1^2 + \dots + x_n^2}$. If one scales a unit vector by a $\alpha\in\mathbb{R}$, the product $\alpha\mathbf{y}$ is a vector parallel to $\mathbf{u}$ with length $\alpha$.

If one scales the vector $\frac{\mathbf{x}}{\left|\mathbf{x}\right|}$ by the scalar $d = \text{Comp}_{\mathbf{x}}\mathbf{y}$, then one obtains a vector parallel $\mathbf{x}$ with length $d$.

This product vector is the vector $\mathbf{p}$ or vector projection of $\mathbf{y}$ onto $\mathbf{x}$.

Definition Vector Projection

The vector projection of vector $\mathbf{y}$ onto vector $\mathbf{x}$ in $\mathbb{R}^n$ is give by

$$ \text{Proj}_{\mathbf{x}}\mathbf{y} := \mathbf{p} = \left(\dfrac{\mathbf{x}\cdot\mathbf{y}}{|\mathbf{x}|}\right)\left(\dfrac{\mathbf{x}}{|\mathbf{x}|}\right) = \dfrac{\mathbf{x}\cdot\mathbf{y}}{|\mathbf{x}|^2}\mathbf{x} = \dfrac{\mathbf{x}\cdot\mathbf{y}}{\mathbf{x}\cdot\mathbf{x}}\mathbf{x} = \dfrac{\mathbf{y}^T\mathbf{x}}{\mathbf{x}^T\mathbf{x}}\mathbf{x} $$

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5.1.4 Euclidean Space

We define a new mathematical structure, a Euclidean Space, $\left(\mathbb{R}^n,\cdot\right)$

Definition Euclidean Vector Space $\mathbb{E}^n$

Euclidean Space $\mathbb{E}^n$ is the vector space $\mathbb{R}^n$ endowed with the Euclidean dot product

$$ \mathbf{x}\cdot\mathbf{y} := \mathbf{y}^T\mathbf{x} = \displaystyle\sum_{j=1}^n x_jy_j = \mathbf{x}^T\mathbf{y} $$

The very important difference is that Euclidean space has a definition of length and direction, while the vector space $\mathbb{R}^n$ does not. However we immediately ignore this rule and think of them as the same thing. There are even references to the geometry of vectors in $\mathbb{R}^n$ in previous chapters. Remember the physicist's perspective requires we think of "an arrow in space" which has magnitude and direction. Keep in mind that everything we discussed in the first four chapters did not require us to talk about angles or length, we just used them as examples.

Now we must formally add the notion of angle and direction to our vector spaces. When we add an inner product to a vector space, we are adding a notion of geometry to our algebraic vector space that did not necessarily already exist. Clearly in our first examples of vector spaces $\mathbb{R}^2$, $\mathbb{R}^3$, and $\mathbb{R}^n$ we should already be familiar with Euclidean geometry.

Definition

The modulus, length or norm of a vector in a Euclidean Space $\mathbb{E}^n$ is given by

$$ \lVert \mathbf{x} \rVert := \lvert \mathbf{x} \rvert = \sqrt{\displaystyle\sum_{j=1}^n x_i^2} $$

Example 1

if $\mathbf{x} = \begin{bmatrix}\ 3\ \\ \ 4\ \end{bmatrix}$, then $\left\| \mathbf{x} \right\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$.

In a normed vector space such as Euclidean space $\mathbb{E}^n$, the distance between two vectors is defined to be the distance between the tips of the vectors in standard position; that is the length of vector $\mathbf{y}-\mathbf{x}$.

Figure 5

The vector from the tip of $\mathbf{x}$ to the tip of $\mathbf{y}$ is called $\mathbf{y}-\mathbf{x}$ because it is the vector one adds to vector $\mathbf{x}$ to obtain vector $\mathbf{y}$.

$$ \mathbf{y} = \mathbf{x} + \left(\mathbf{y} - \mathbf{x}\right) $$

Thus the distance between vector $\mathbf{x}$ to vector $\mathbf{y}$

$$ d\left(\mathbf{x},\mathbf{y}\right) := \lVert \mathbf{y} - \mathbf{x} \rVert $$

Example 2

The distance between vector $\begin{bmatrix} 3 \\ 4 \end{bmatrix}$ and vector $\begin{bmatrix} 1 \\ 3 \end{bmatrix}$

$$ d\left(\begin{bmatrix} 3 \\ 4 \end{bmatrix},\ \begin{bmatrix} 1 \\ 3 \end{bmatrix}\right) = \left\| \begin{array}{c} 3-1 \\ 4-3 \end{array} \right\| = \left\| \begin{array}{c} 2 \\ 1 \end{array} \right\| = \sqrt{2^2 + 1^2} = \sqrt{5} $$

Example 3

If $\mathbf{x} = \begin{bmatrix} -2\ \\ \ \ 5\ \end{bmatrix}$ and $\mathbf{y} = \begin{bmatrix} \ \ 6\ \\ \ \ 4\ \end{bmatrix}$, then

(a) The unit vector with the same direction as vector $\mathbf{x}$ is

$$ \mathbf{u} = \dfrac{\mathbf{x}}{\|\mathbf{x}\|} = \dfrac{1}{\sqrt{(-2)^2 + 5^2}}\begin{bmatrix} -2\ \\ \ \ 5\ \end{bmatrix} = \dfrac{1}{\sqrt{29}}\begin{bmatrix} -2\ \\ \ \ 5\ \end{bmatrix} = \begin{bmatrix} -\frac{2}{\sqrt{29}}\ \\ \ \ \frac{5}{\sqrt{29}}\ \end{bmatrix} $$

(b) The unit vector in the direction of $\mathbf{y}$ is

$$ \mathbf{v} = \dfrac{\mathbf{y}}{\|\mathbf{y}\|} = \dfrac{1}{\sqrt{6^2 + 4^2}} = \dfrac{1}{\sqrt{52}}\begin{bmatrix} \ \ 6\ \\ \ \ 4\ \end{bmatrix} = \begin{bmatrix} \ \ \frac{3}{\sqrt{13}}\ \\ \ \ \frac{2}{\sqrt{13}}\ \end{bmatrix} $$

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5.1.5 Properties of Dot Product

Definition

The Angle Between Two Vectors in Euclidean Space

If $\mathbf{x}$ and $\mathbf{y}$ are two vector in Euclidean Space $\mathbb{R}^2$ or $\mathbb{R}^3$, then the angle between the two vectors is given by

$$ \cos(\theta) = \dfrac{\mathbf{x}\cdot\mathbf{y}}{\lVert \mathbf{x} \rVert\,\lVert \mathbf{y} \rVert} $$

Since the cosine of an angle is a number in the interval $[-1,1]$, we have that $0 \le \left|\cos(\theta)\right|\le 1$ for all angles $\theta\in\mathbb{R}$.

Theorem 5.1.1 The Cauchy-Schwarz Inequality

If $\mathbf{x}$ and $\mathbf{y}$ are two vector in Euclidean Space $\mathbb{R}^2$ or $\mathbb{R}^3$, then

$$ \begin{align*} \lVert\,\mathbf{x}\cdot\mathbf{y}\,\rVert &= \left| \, \lVert \mathbf{x} \rVert \, \lVert \mathbf{y} \rVert\, \lvert \cos(\theta)\, \right| \\ \\ &= \lVert \mathbf{x} \rVert \, \lVert \mathbf{y} \rVert \, \left|\cos(\theta)\right| \\ \\ &\le \lVert \mathbf{x} \rVert \, \lVert \mathbf{y} \rVert \end{align*} $$ with equality holding if and only if one of the vectors is zero or one of the vectors is a multiple of the other.

Proof:

If one or both of the vectors is zero then either $\lVert \mathbf{x} \rVert = 0$ and/or $\lVert \mathbf{y} \rVert = 0$. In any case, $\mathbf{x}\cdot\mathbf{y}= 0$ and the product $\lVert \mathbf{x} \rVert \, \lVert \mathbf{y} \rVert = 0$. So

$$ \lVert \,\mathbf{x}\cdot\mathbf{y}\, \rVert = 0 = \lVert \mathbf{x} \rVert \, \lVert \mathbf{y} \rVert $$
If the one vector is a multiple of the other, then they both belong to the same line or span. Either the angle between them is zero or $\pi$. In this case $\left|\cos(\theta)\right|=1$, so

$$ \left|\,\mathbf{x}\cdot\mathbf{y}\,\right| = \lVert \mathbf{x} \rVert \, \lVert \mathbf{y} \rVert \, \left|\cos(\theta)\right| \le \lVert \mathbf{x} \rVert \, \lVert \mathbf{y} \rVert $$ $\tombstone$

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5.1.6 The Triangle Inequality

Theorem 5.1.3 The Triangle Inequality

If $\mathbf{u}$ and $\mathbf{v}$ are vectors in $\mathbb{R}^n$, then

$$ \lVert \mathbf{u} + \mathbf{v} \rVert \le \lVert \mathbf{u} \rVert + \lVert \mathbf{v} \rVert $$

Proof:

We will use the dot product definition of length

$$ \begin{align*} \lVert \mathbf{u} + \mathbf{v} \rVert^2 &= \left(\mathbf{u} + \mathbf{v}\right)\cdot\left(\mathbf{u} + \mathbf{v}\right) \\ \\ &= \left(\mathbf{u} + \mathbf{v}\right)\cdot\mathbf{u} + \left(\mathbf{u} + \mathbf{v}\right)\cdot\mathbf{v} \\ \\ &= \mathbf{u}\cdot\mathbf{u} + \mathbf{v}\cdot\mathbf{u} + \mathbf{u}\cdot\mathbf{v} + \mathbf{v}\cdot\mathbf{v} \\ \\ &= \lVert \mathbf{u} \rVert^2 + 2\mathbf{u}\cdot\mathbf{v} + \lVert \mathbf{v} \rVert^2 \\ \\ &= \lVert \mathbf{u} \rVert^2 + 2 \lVert \mathbf{u} \rVert \, \lVert \mathbf{v} \rVert \,\cos(\theta) + \lVert \mathbf{v} \rVert^2 \\ \\ &\le \lVert \mathbf{u} \rVert^2 + 2 \lVert \mathbf{u} \rVert \, \lVert \mathbf{v} \rVert \, |\cos(\theta)| + \lVert \mathbf{v} \rVert^2 \\ \\ &\le \lVert \mathbf{u} \rVert^2 + 2 \lVert \mathbf{u} \rVert \, \lVert \mathbf{v} \rVert + \lVert \mathbf{v} \rVert^2 \\ \\ &= \left( \lVert \mathbf{u} \rVert + \lVert \mathbf{v} \rVert \right)^2 \\ \end{align*} $$ Thus

$$ \begin{align*} \lVert \mathbf{u} + \mathbf{v} \rVert^2 &\le \left( \lVert \mathbf{u} \rVert + \lVert \mathbf{v} \rVert \right)^2 \\ \\ \lVert \mathbf{u} + \mathbf{v} \rVert &\le \lVert \mathbf{u} \rVert + \lVert \mathbf{v} \rVert \\ \end{align*} $$
The last equation is true because both sides of this equation are the squares of non-negative values. This establishes the triangle inequality. ∎

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5.1.7 Orthogonality

If the scalar product of two vectors is zero, and neither of the vectors is the zero vector, then

$$ 0 = \mathbf{x}\cdot\mathbf{y} = \lVert \mathbf{x} \rVert \, \lVert \mathbf{y} \rVert \,\cos(\theta) $$

If neither $\lVert \mathbf{x} \rVert = 0$ nor $\lVert \mathbf{y} \rVert = 0$, then $\cos(\theta)=0$. Thus the angle between the two vectors is $\frac{\pi}{2}$.

Definition

Orthogonal Vectors

If the scalar product of two vectors in $\mathbb{R}^n$ is zero, then the vectors are perpendicular. Perpendicular vectors are said to be orthogonal.

This means that the zero vector in $\mathbb{R}^n$ is perpendicular to all nonzero vectors in $\mathbb{R}^n$.

Theorem 5.1.4 The Pythagorean Theorem

If $\mathbf{u}$ and $\mathbf{v}$ are vectors in $\mathbb{R}^n$, then $\mathbf{u}$ and $\mathbf{v}$ are orthogonal if and only if

$$ \lVert \mathbf{u} + \mathbf{v} \rVert^2 = \lVert \mathbf{u} \rVert^2 + \lVert \mathbf{v} \rVert^2 $$

Proof:

The proof is very similar to the proof for the triangle inequality.

$$ \begin{align*} \left\|\mathbf{u} + \mathbf{v}\right\|^2 &= \left(\mathbf{u} + \mathbf{v}\right)\cdot\left(\mathbf{u} + \mathbf{v}\right) \\ \\ &= \left(\mathbf{u} + \mathbf{v}\right)\cdot\mathbf{u} + \left(\mathbf{u} + \mathbf{v}\right)\cdot\mathbf{v} \\ \\ &= \mathbf{u}\cdot\mathbf{u} + \mathbf{v}\cdot\mathbf{u} + \mathbf{u}\cdot\mathbf{v} + \mathbf{v}\cdot\mathbf{v} \\ \\ &= \lVert \mathbf{u} \rVert^2 + 2\mathbf{u}\cdot\mathbf{v} + \lVert \mathbf{v} \rVert^2 \\ \\ &= \lVert \mathbf{u} \rVert^2 + 2\cdot0 + \lVert \mathbf{v} \rVert^2 \\ \\ &= \lVert \mathbf{u} \rVert^2 + \lVert \mathbf{v} \rVert^2 \end{align*} $$ ∎

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5.1.8 Determine the Angle Between Two Vectors

Example 4

Determine the angle between vectors $\mathbf{x} = \begin{bmatrix} -3\ \\ -1\ \end{bmatrix}$ and $\mathbf{y} = \begin{bmatrix}\ \ 3\ \\ \ \ 8\ \end{bmatrix}$.

$$ \begin{align*} \cos(\theta) &= \dfrac{\mathbf{x}\cdot\mathbf{y}}{ \lVert \mathbf{x} \rVert \, \lVert \mathbf{y} \rVert} \\ \\ &= \dfrac{ (-3)(3) + (-1)(8) }{\sqrt{ (-3)^2 + (-1)^2}\,\sqrt{3^2 + 8^2}} \\ \\ &= \dfrac{ -9 - 8 }{\sqrt{10}\,\sqrt{73}} = -\dfrac{17}{\sqrt{730}} \\ \\ \theta &\approx 129^{\circ} \end{align*} $$

Figure 6

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5.1.9 The Distance From a Point to a Line

Example 5

Find the distance from the point $P = (3,1)$ to the line $y=-\frac{3}{4}x$.

Figure 7

• A line has the form $\mathbf{y} = \mathbf{m}t + \mathbf{b}$ in vector form, where direction vector $\mathbf{m} = \langle 4, -3 \rangle$, and $\mathbf{b} = \mathbf{0}$.

• The line $\mathbf{y} = \mathbf{m^{\perp}}t$ is perpendicular to the direction vector $\mathbf{m^{\perp}} = \langle 3, 4 \rangle$.

• The scalar component of vector $\overline{OP}$ onto $\mathbf{m^{\perp}}$ is the distance from the line to point $P$. If $\mathbf{b}\neq\mathbf{0}$, then we would need the scalar component of vector $\mathbf{p}-\mathbf{b}$.

• The distance $d = \text{Comp}_{\mathbf{m^{\perp}}} \overline{OP}$

$$ d = \frac{\mathbf{m^{\perp}}\cdot\overline{OP}}{\left\|\mathbf{m^{\perp}}\right\|} = \frac{3\cdot 3 + 4\cdot 1 }{\sqrt{9 + 16}} = \frac{13}{5} $$

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5.1.10 The Equation of a Plane

Example 6

Find the equation of the plane that passes through the points

$$ P_0 = (1,0,0),\qquad P_1 = (0,1,0),\qquad P_2 = (0,0,1) $$

• The vector equation of a plane in three dimensional space is

$$ \mathbf{n}\cdot\left(\mathbf{p}-\mathbf{p}_0\right) = 0 $$

• We need two vectors on the plane. We can use these vectors to determine a normal vector to the plane.

• Select one point to be the base of both vectors

$$ \overrightarrow{P_0P_1} = \mathbf{P_1} - \mathbf{P_0} = \begin{bmatrix} 0-1 \\ 1-0 \\ 0-0 \end{bmatrix} = \begin{bmatrix} -1\ \\ \ \ 1\ \\ \ \ 0\ \end{bmatrix} $$

$$ \overrightarrow{P_0P_2} = \mathbf{P_2} - \mathbf{P_0} = \begin{bmatrix} 0-1 \\ 0-0 \\ 1-0 \end{bmatrix} = \begin{bmatrix} -1\ \\ \ \ 0\ \\ \ \ 1\ \end{bmatrix} $$

Figure 8

• A normal vector is a vector that is orthogonal to all of the vectors on the plane. The cross product of two vectors in $\mathbb{R}^3$ is orthogonal to both vectors so,

$$ \mathbf{n} = \overrightarrow{P_0P_1} \times \overrightarrow{P_0P_2} = \begin{vmatrix} \mathbf{\hat{\imath}}\ & -1\ & -1\ \\ \mathbf{\hat{\jmath}}\ &\ \ 1\ &\ \ 0\ \\ \mathbf{\hat{k}}\ &\ \ 0\ &\ \ 1\ \end{vmatrix} = 1\mathbf{\hat{\imath}} + 1\mathbf{\hat{\jmath}} + 1\mathbf{\hat{k}} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} $$

Figure 9

• We denote any point on the plane by $P = (x,y,z)$ and the vector equation of the plane is

$$ \mathbf{n}\cdot\overrightarrow{PP_0} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\cdot\begin{bmatrix} x-1 \\ y-0 \\ z-0 \end{bmatrix} = (x-1) + y + z = 0 $$

Figure 10

• The scalar equation of the plane is then $$ x + y + z = 1 $$

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5.1.11 The Distance From a Point to a Plane

Example 7

Find the distance from point $P = (1,1,2)$ to the plane $2x + 2y + z = 0$.

Figure 11

Point $P = (1,1,2)$ is not on the plane as $2(1) + 2(1) + 1(2) = 6 \neq 0$.

• The distance from the point to the plane is the length of the perpendicular vector from the closest point on the plane $Q$ to point $P$.

• The normal vector $\mathbf{n}$ and the vector $\mathbf{v} = \overrightarrow{QP}$ are both orthogonal to the plane.

• The distance from the point $P$ to the plane is the length of the projection vector of $\mathbf{p}$ onto the normal vector $\mathbf{n}$. That is, the distance from the point $P$ to the plane is the scalar component of vector $\mathbf{p}$ onto the normal vector $\mathbf{n}$.

$$ \text{Comp}_{\mathbf{n}}\mathbf{v} = \dfrac{\mathbf{v}\cdot\mathbf{n}}{\left\|\mathbf{n}\right\|} = \dfrac{ (2)(1) + (2)(1) + (1)(2)}{\sqrt{1^2 + 2^2 + 2^2}} = \dfrac{6}{\sqrt{9}} = 2 $$

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5.1.12 Dot Products as a Linear Transformation

It is often useful to think of dot product in terms of matrix multiplication. For vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^n$,

$$ \mathbf{u}\cdot\mathbf{v} = \mathbf{v}^T\mathbf{u} = \displaystyle\sum_{k=1}^n u_kv_k $$

This was explained in the video at the beginning of this section. We obtain a linear transformation $L_{\mathbf{v}}\,:\,\mathbb{R}^n\rightarrow\mathbb{R}$ that maps every vector in $\mathbb{R}^n$ to the real line. The result of this map is the scalar product of $\mathbf{u}$ and $\mathbf{u}$.

$$ L_{\mathbf{v}}(\mathbf{u}) := \mathbf{v}^T\mathbf{u} = \mathbf{u}\cdot\mathbf{v} $$

Since it is a linear transformation from $n$-dimensional vector space $\mathbb{R}^n$ to 1-dimensional vector space $\mathbb{R}$, it can be represented by a $1\times n$ matrix. That matrix is $\mathbf{v}^T$. The linear transformation represented by the $1\times n$ matrix $\mathbf{v}^T$ is called the dual of vector $\mathbf{v}$. Although the author indicates that the properties of dot product are a direct consequence of the properties of matrix multiplication; it is rather

The properties of the scalar product yield a linear transformation from a finite dimensional vector space to the real line. Thus it can be represented by a $1\times n$ matrix and the properties of dot product and matrix multiplication correspond with each other.

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