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's 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$\left\{\,\mathbf{\hat{\imath}}\,\right\}$.
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
$$|\mathbf{x}| := \sqrt{x_1^2 + x_2^2}$$
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 $(x_1,x_2)$.
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
$$ \begin{align*} \pi-\theta &= \tan^{-1}\left(\dfrac{|x_2|}{|x_1|}\right) = \cos^{-1}\left(\dfrac{|x_1|}{\sqrt{x_1^2 + x_2^2}}\right) = \sin^{-1}\left(\dfrac{|x_2|}{\sqrt{x_1^2 + x_2^2}}\right) \\ \\ &= \tan^{-1}\left(\dfrac{|x_2|}{|x_1|}\right) = \cos^{-1}\left(\dfrac{|x_1|}{|\mathbf{x}|}\right) = \sin^{-1}\left(\dfrac{|x_2|}{|\mathbf{x}|}\right) \end{align*} $$
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 $\mathbf{y}$,
$$ \begin{align*} \phi &= \tan^{-1}\left(\dfrac{y_2}{y_1}\right) = \cos^{-1}\left(\dfrac{y_1}{\sqrt{y_1^2 + y_2^2}}\right) = \sin^{-1}\left(\dfrac{y_2}{\sqrt{y_1^2 + y_2^2}}\right) \\ \\ &= \tan^{-1}\left(\dfrac{y_2}{y_1}\right) = \cos^{-1}\left(\dfrac{y_1}{|\mathbf{y}|}\right) = \sin^{-1}\left(\dfrac{y_2}{|\mathbf{y}|}\right) \\ \end{align*} $$
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.
Video Lecture 1: Dot Products and Duality.¶
Dot Products and DualityIf 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 $\mathbf{x}$,
The vector with the same direction as vector $\mathbf{x}$ and length equal to the leg of the resulting right triangle is called the projection of vector $\mathbf{y}$ onto vector $\mathbf{x}$, and is defined by the vector
$$\mathbf{p} = |\mathbf{y}|\cos(\theta)\,\frac{\mathbf{x}}{|\mathbf{x}|}.$$
Figure 4 does not depict the projection if angle $\theta$ is obtuse. In this case the projection vector points in the opposite direction to $\mathbf{x}$.
As $0\le\theta\le\pi$ is the angle between two vectors and $\dfrac{\mathbf{x}}{|\mathbf{x}|}$ is a unit vector, 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 in $\mathbb{R}^2$
The scalar product or dot product of two vectors $\mathbf{x}$ and $\mathbf{y}$, is the signed product of the lengths of both vector $\mathbf{x}$ and the projection $\mathbf{p}$ of vector $\ \mathbf{y}$ onto vector $\ \mathbf{x}$. The dot product is positive if the angle between $\mathbf{x}$ and $\mathbf{y}$ is acute. The dot product is negative if the angle between $\mathbf{x}$ and $\mathbf{y}$ is obtuse.
$$\mathbf{x}\cdot\mathbf{y} :=|\mathbf{x}|\,|\mathbf{y}|\cos(\theta),$$
where $\theta$ is the included angle between the two vectors.
We learn in the video above that, for a fixed vector $\mathbf{y}\in\mathbb{R}^2$, this is a linear transformation
$$L[\mathbf{x}] = \left<\mathbf{x},\mathbf{y}\right>$$
from the vector space $\mathbb{R}^2$ to the real numbers $\mathbb{R}$.
$$L\,:\,\mathbb{R}^2\rightarrow\mathbb{R}$$
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 $\mathbf{y}^\mathsf{T}$,
$$\left<\mathbf{x},\mathbf{y}\right> = L[\mathbf{x}] = \mathbf{y}^\mathsf{T}\mathbf{x} = x_1y_1 + x_2y_2 = \mathbf{x}\cdot\mathbf{y}$$
From this identity, one obtains equations for the vector projection, the length of the vector projection, and the scalar component of vector $\mathbf{y}$ onto $\mathbf{x}$.
Definition¶
Component in $\mathbb{R}^2$
The scalar component of one vector $\mathbf{y}$ onto vector $\mathbf{x}$ in $\mathbb{R}^2$ is given by
$$ \text{Comp}_{\mathbf{x}}\mathbf{y} := |\mathbf{y}|\cos(\theta) = \dfrac{|\mathbf{x}||\mathbf{y}|\cos(\theta)}{|\mathbf{x}|} = \dfrac{\mathbf{x}\cdot\mathbf{y}}{|\mathbf{x}|} = \dfrac{\mathbf{y}^\mathsf{T}\mathbf{x}}{|\mathbf{x}|} $$
Notice that scalar component has the same sign convention as dot product due to the cosine of the included angle.
These 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 $\left|\mathbf{x}\right|=\sqrt{x_1^2 + x_2^2}$. If one scales a unit vector by scalar $\alpha$, the product $\alpha\mathbf{u}$ 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 to $\mathbf{x}$ with length $|d|$.
This product vector is the vector $\mathbf{p}$ or vector projection of $\mathbf{y}$ onto $\mathbf{x}$.
Definition¶
Vector Projection in $\mathbb{R}^2$
The vector projection of vector $\mathbf{y}$ onto vector $\mathbf{x}$ in $\mathbb{R}^2$ is given by
$$ \text{Proj}_{\mathbf{x}}\mathbf{y} := \mathbf{p} = \left(\dfrac{\mathbf{x}\cdot\mathbf{y}}{|\mathbf{x}|}\right)\left(\dfrac{\mathbf{x}}{|\mathbf{x}|}\right) = \dfrac{\mathbf{x}\cdot\mathbf{y}}{|\mathbf{x}|^2}\mathbf{x} = \dfrac{\mathbf{x}\cdot\mathbf{y}}{\mathbf{x}\cdot\mathbf{x}}\mathbf{x} = \dfrac{\mathbf{y}^\mathsf{T}\mathbf{x}}{\mathbf{x}^\mathsf{T}\mathbf{x}}\mathbf{x} $$
Exercise 1¶
Compute $\langle 5,\,2\rangle\cdot\langle 3,\,-4\rangle$.
View Solution
$$\langle 5,\,2\rangle\cdot\langle 3,\,-4\rangle = (5)(3) + (2)(-4) = 15 - 8 = 7$$Exercise 2¶
For $\mathbf{x}=\langle 3,\,4\rangle$ and $\mathbf{y}=\langle 7,\,1\rangle$, find the scalar component of $\mathbf{y}$ onto $\mathbf{x}$ and the projection of $\mathbf{y}$ onto $\mathbf{x}$.
View Solution
$\mathbf{x}\cdot\mathbf{y} = 21 + 4 = 25$ and $|\mathbf{x}| = \sqrt{9+16} = 5$, so $$\text{comp}_{\mathbf{x}}\mathbf{y} = \frac{\mathbf{x}\cdot\mathbf{y}}{|\mathbf{x}|} = \frac{25}{5} = 5, \qquad \text{proj}_{\mathbf{x}}\mathbf{y} = \frac{\mathbf{x}\cdot\mathbf{y}}{|\mathbf{x}|^2}\,\mathbf{x} = \frac{25}{25}\langle 3,\,4\rangle = \langle 3,\,4\rangle$$Our geometric discussion is somewhat different in $\mathbb{R}^n$ when $n > 2$. We can arrange two linearly independent vectors in standard position in $\mathbb{R}^n$. This yields three points,
- The base point of both vectors at the origin.
- The tip of one vector in standard coordinates.
- The tip of the other vector in standard coordinates.
These three points define a plane in $\mathbb{R}^n$. In this plane there is an angle between the two vectors. The same 2-dimensional geometry in this embedded plane motivates the geometric definitions in $\mathbb{R}^n$. This plane is the span of the two vectors.
If the span of the two vectors is a line because they are linearly dependent, then the angle between the two vectors is zero or $\pi$. The figure above does not describe their geometry because there are infinitely many planes that contain colinear vectors in standard position.
Definition¶
Dot Product in $\mathbb{R}^n$
The scalar product or dot product of two vectors $\mathbf{x}$ and $\mathbf{y}$, is the signed product of the lengths of vector $\mathbf{x}$ and the projection $\ \mathbf{p}$ of vector $\ \mathbf{y}$ onto vector $\ \mathbf{x}$.
$$\mathbf{x}\cdot\mathbf{y} := |\mathbf{x}|\,|\mathbf{y}|\cos(\theta),$$
where $\theta$ is the included angle between the two vectors. If the two vectors are colinear, then the angle between the vectors is zero or $\pi$ radians. If the two vectors are linearly independent, the angle is defined in $\mathrm{Span}\left\{ \mathbf{x},\ \mathbf{y} \right\}$.
This means that the geometry of dot product has not changed much. Dot product is important for its geometric applications in STEM, and its more abstract applications in machine learning and robotics.
I want to make proving Theorem 2 easier by first proving a theorem that illustrates how we can use Chapter 4 and rotations.
Theorem 1¶
Rotation of a Unit Vector to $\mathbf{e}_1$
Any unit vector can be mapped to $\mathbf{e}_1$ in $\mathbb{R}^n$ using a rotation matrix.
Proof of Theorem 1¶
For rotation in $\mathbb{R}^3$¶
1. Construct rotation matrix $R_z$ that maps unit $\mathbf{w}$ to the $xz$-plane.
Suppose that $\mathbf{w}$ has standard coordinates $\langle\, w_1,\ w_2,\ w_3 \,\rangle$. Define $$r = \sqrt{w_1^2 + w_2^2}$$ and define $$R_z(\alpha) = \begin{bmatrix}\ \cos(\alpha)\ & -\sin(\alpha)\ &\ \ 0\ \\ \ \sin(\alpha)\ &\ \cos(\alpha)\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ \end{bmatrix}$$
Choose $\alpha$ so that $$\cos(\alpha)=\frac{w_1}{r},\ \sin(\alpha)=-\frac{w_2}{r}\qquad\left( \alpha = \mathrm{atan2}(-w_2,w_1) \right)$$
This results in $$R_z = \begin{bmatrix}\ \ w_1/r\ &\ \ w_2/r\ &\ \ 0\ \\ -w_2/r\ &\ \ w_1/r\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ \end{bmatrix},\qquad R_z\mathbf{w} = \begin{bmatrix}\ \ r\ \\ \ \ 0\ \\ \ \ w_3\ \end{bmatrix}$$
2. Construct rotation matrix $R_y$ that maps to $\mathbf{e}_1$. Using $$R_y(\beta) = \begin{bmatrix}\ \cos(\beta)\ &\ \ 0\ &\ \sin(\beta)\ \\ \ \ 0\ &\ \ 1\ &\ \ 0\ \\ -\sin(\beta)\ &\ \ 0\ &\ \cos(\beta)\ \end{bmatrix}$$
Choose $\beta$ so that $$\cos(\beta) = r,\ \sin(\beta)=w_3\qquad\left( \beta=\mathrm{atan2}(w_3,r) \right)$$
Then $$R_y = \begin{bmatrix}\ \ r \ &\ \ 0\ &\ \ w_3\ \\ \ \ 0\ &\ \ 1\ &\ \ 0\ \\ -w_3\ &\ \ 0\ &\ \ r\ \end{bmatrix},\qquad R_y\begin{bmatrix}\ \ r\ \\ \ \ 0\ \\ \ \ w_3\ \end{bmatrix} = \begin{bmatrix}\ \ 1\ \\ \ \ 0\ \\ \ \ 0\ \end{bmatrix}$$
3. The composite $$R_w = R_yR_z = \begin{bmatrix}\ \ w_1\ &\ \ w_2\ &\ \ w_3\ \\ -w_2/r\ &\ w_1/r\ &\ \ 0\ \\ -w_1w_3/r\ & -w_2w_3/r &\ \ r\ \end{bmatrix},\quad R_w\mathbf{w} = \mathbf{e}_1$$
4. If $\mathbf{w} =\langle 0,\ 0,\ \pm1 \rangle = \pm\mathbf{e}_3$, the $z$-rotation angle is indeterminate so $R_z=I$. When $\mathbf{w}=-\mathbf{e}_1$, then $$R_w = \begin{bmatrix} -1\ &\ \ 0\ &\ \ 0\ \\ \ \ 0\ & -1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ \end{bmatrix}$$.
For rotation in $\mathbb{R}^n$¶
In this case we must construct $n-1$ $n\times n$ rotation matrices for each of the $n-1$ coordinate planes containing the $\mathbf{e}_1$-axis. The product (composition) of the $n-1$ rotations fulfills the same function as in the 3-dimensional case. $$R_w = R_{\mathbf{e}_2}\cdots R_{\mathbf{e}_n}$$ If $\mathbf{w} = \mathbf{e}_k$ for $2\le k\le n$, then the $\mathbf{e}_k$-rotation angle is indeterminate so $R_{\mathbf{e}_k}=I$. When $\mathbf{w}=-\mathbf{e}_1$, then $$R_w = \begin{bmatrix} -\mathbf{e}_1 & -\mathbf{e}_2 & \mathbf{e}_3 & \cdots & \mathbf{e}_n \end{bmatrix}\quad \blacksquare$$
Theorem 2¶
Dot Product is Matrix Multiplication
The dot product of two vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^n$ is given by
$$\mathbf{u}\cdot\mathbf{v} = \mathbf{v}^\mathsf{T}\mathbf{u} = \mathbf{u}^\mathsf{T}\mathbf{v}$$
Proof of Theorem 2¶
Suppose that $\mathbf{u}$, $\mathbf{v}$ and $\mathbf{w}$ are three vectors in $\mathbb{R}^n$ and $c$ is a positive scalar. Let us first prove the definition of dot product is commutative and linear.
$$\mathbf{0}\cdot\mathbf{u} = \mathbf{u}\cdot\mathbf{0} = (0\mathbf{v})\cdot\mathbf{u} = 0$$
Let us consider nonzero vectors $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$, and a positive scalar $c$.
1. Commutativity
$$\mathbf{u}\cdot\mathbf{v} = |\mathbf{u}|\,|\mathbf{v}|\cos(\theta) = |\mathbf{v}|\,|\mathbf{u}|\cos(\theta) = \mathbf{v}\cdot\mathbf{u}$$
2. Homogeneity
$$\begin{align*} (c\mathbf{u})\cdot\mathbf{v} &= |c\mathbf{u}|\,|\mathbf{v}|\cos(\theta) = |c|\,|\mathbf{u}|\,|\mathbf{v}|\cos(\theta) = c\,|\mathbf{u}|\,|\mathbf{v}|\cos(\theta) = |\mathbf{u}|\,|c\mathbf{v}|\cos(\theta) = \mathbf{u}\cdot(c\mathbf{v}) \\ \\ \end{align*}$$
If we multiply $\mathbf{u}$ by $-c < 0$, then the angle changes to $\pi - \theta$ and since
$$-c\cos(\pi-\theta) = c\cos(\theta),$$
only the sign of the result changes. Hence for any scalar $c$,
$$(c\mathbf{u})\cdot\mathbf{v} = \mathbf{u}\cdot(c\mathbf{v}) = c(\mathbf{u}\cdot\mathbf{v})$$
3. Additivity
We use Theorem 1 and rotation $R_{\mathbf{w}/|\mathbf{w}|}$ that maps $\mathbf{w}$ to the $x$-axis. Rotations do not change length of vectors so dot product will be unaffected. In the new coordinate system where $\dfrac{\mathbf{w}}{|\mathbf{w}|}\mapsto\mathbf{e}_1$, $$\mathbf{u}\cdot\mathbf{w} = u_1|\mathbf{w}|,\quad\mathbf{v}\cdot\mathbf{w} = v_1|\mathbf{w}|$$
Now it is clear that $$\left(\mathbf{u}+\mathbf{v}\right)\cdot\mathbf{w} = (u_1 + v_1)|\mathbf{w}| = \mathbf{u}\cdot\mathbf{w} + \mathbf{v}\cdot\mathbf{w}$$
Thus if we fix $\mathbf{w}$, the map $B\,:\,\mathbb{R}^n\rightarrow\mathbb{R}$ defined by $B(\mathbf{u}) = \mathbf{u}\cdot\mathbf{w}$ is a linear map. Likewise the map $B'\,:\,\mathbb{R}^n\rightarrow\mathbb{R}$ defined by $B'(\mathbf{u}) = \mathbf{w}\cdot\mathbf{u}$ is linear. Therefore dot product is bilinear.
4. Matrix Multiplication
To complete the proof we need only construct the matrix that represents $B$. $B$ maps $\mathbb{R}^n$ to $\mathbb{R}$, so the matrix representation of $B$ is a $1\times n$ matrix. $$\begin{align*} B(\mathbf{e}_1) &= \dfrac{w_1}{|\mathbf{w}|}|\mathbf{w}| = w_1 \\ &\ddots \\ B(\mathbf{e}_k) &= \frac{w_k}{|\mathbf{w}|}|\mathbf{w}| = w_k \\ &\ddots \\ B(\mathbf{e}_n) &= \dfrac{w_n}{|\mathbf{w}|}|\mathbf{w}| = w_n \end{align*}$$ Thus $\left[ w_1\ \dots\ w_k\ \dots\ w_n \right] = \mathbf{w}^\mathsf{T}$. For fixed $\mathbf{w}\in\mathbb{R}^n$, $\mathbf{u}\cdot\mathbf{w} = \mathbf{w}^\mathsf{T}\mathbf{u}$ for all $\mathbf{u}\in\mathbb{R}^n$. Similarly, $\mathbf{u}\cdot\mathbf{w} = \mathbf{w}\cdot\mathbf{u} = \mathbf{u}^\mathsf{T}\mathbf{w}$. $\blacksquare$
Example 1¶
If $\mathbf{u}=\langle 2,\,-1,\,3\rangle$ and $\mathbf{v}=\langle 4,\,5,\,1\rangle$, then $\mathbf{u}\cdot\mathbf{v} = 8 - 5 + 3 = 6$, which is exactly the matrix product
$$\mathbf{v}^\mathsf{T}\mathbf{u} = \begin{bmatrix} 4 & 5 & 1 \end{bmatrix}\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} = 6.$$
Exercise 3¶
For $\mathbf{u}=\langle 2,\,-1,\,3\rangle$ and $\mathbf{v}=\langle 4,\,5,\,1\rangle$, compute $\mathbf{u}\cdot\mathbf{v}$, and confirm that $\mathbf{v}^\mathsf{T}\mathbf{u}$ gives the same result.
View Solution
$$\mathbf{u}\cdot\mathbf{v} = (2)(4) + (-1)(5) + (3)(1) = 8 - 5 + 3 = 6$$ $$\mathbf{v}^\mathsf{T}\mathbf{u} = \begin{bmatrix} 4 & 5 & 1 \end{bmatrix}\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} = 8 - 5 + 3 = 6$$Exercise 4¶
In $\mathbb{R}^4$, compute $\langle 1,\,2,\,2,\,1\rangle\cdot\langle 3,\,-1,\,0,\,4\rangle$.
View Solution
$$\langle 1,\,2,\,2,\,1\rangle\cdot\langle 3,\,-1,\,0,\,4\rangle = 3 - 2 + 0 + 4 = 5$$The combination of a vector space with a dot product defined on the vector space constitutes a new mathematical structure.
Definition¶
Euclidean Space
Euclidean Space $\mathbb{E}^n$ is the vector space $\mathbb{R}^n$ endowed with the Euclidean dot product
$$ \mathbf{x}\cdot\mathbf{y} := \lvert\mathbf{x}\rvert\,\lvert\mathbf{y}\rvert\cos(\theta) = \mathbf{y}^\mathsf{T}\mathbf{x} = \displaystyle\sum_{j=1}^n x_jy_j = \mathbf{x}^\mathsf{T}\mathbf{y} $$
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¶
Norm
The modulus, length or norm of a vector in a Euclidean Space $\mathbb{E}^n$ is given by
$$\lvert \mathbf{x} \rvert := \lvert \mathbf{x} \rvert = \left( \mathbf{x}\cdot\mathbf{x} \right)^{1/2} = \sqrt{\displaystyle\sum_{i=1}^n x_i^2}$$
Example 2¶
If $\mathbf{x} = \begin{bmatrix}\ 3\ \\ \ 4\ \end{bmatrix}$, then $\lvert \mathbf{x} \rvert = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$.
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 $\mathbf{y}-\mathbf{x}$.
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 $\mathbf{x}$ and vector $\mathbf{y}$
$$d\left(\mathbf{x},\mathbf{y}\right) := \left| \mathbf{y} - \mathbf{x} \right|$$
Definition¶
Distance Between Vectors
If $\mathbf{u}$ and $\mathbf{v}$ are vectors in $\mathbb{R}^n$, then the distance $d$ between the vectors is defined by
$$d(\mathbf{u},\mathbf{v}) = \left| \mathbf{v} - \mathbf{u} \right|$$
Example 3¶
The distance between vector $\begin{bmatrix} 3 \\ 4 \end{bmatrix}$ and vector $\begin{bmatrix} 1 \\ 3 \end{bmatrix}$
$$d\left(\begin{bmatrix} 3 \\ 4 \end{bmatrix},\ \begin{bmatrix} 1 \\ 3 \end{bmatrix}\right) = \left| \left[ \begin{array}{c} 3-1 \\ 4-3 \end{array} \right] \right| = \left| \left[ \begin{array}{c} 2 \\ 1 \end{array} \right] \right| = \sqrt{2^2 + 1^2} = \sqrt{5}$$
Example 4¶
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
$$ \mathbf{v} = \dfrac{\mathbf{y}}{|\mathbf{y}|} = \dfrac{1}{\sqrt{52}}\begin{bmatrix} \ \ 6\ \\ \ \ 4\ \end{bmatrix} = \begin{bmatrix} \ \ \frac{3}{\sqrt{13}}\ \\ \ \ \frac{2}{\sqrt{13}}\ \end{bmatrix} $$
Definition¶
Angle between Vectors
If $\mathbf{x}$ and $\mathbf{y}$ are two vectors in $\mathbb{R}^n$, then the angle between the vectors is given by
$$\cos(\theta) = \frac{\mathbf{x}\cdot\mathbf{y}}{\left|\mathbf{x}\right|\,\left|\mathbf{y}\right|}$$
Example 5¶
Determine the angle between vectors $\mathbf{x} = \begin{bmatrix} -3\ \\ -1\ \end{bmatrix}$ and $\mathbf{y} = \begin{bmatrix}\ \ 3\ \\ \ \ 8\ \end{bmatrix}$.
$$ \begin{align*} \cos(\theta) &= \dfrac{\mathbf{x}\cdot\mathbf{y}}{ \lvert \mathbf{x} \rvert \, \lvert \mathbf{y} \rvert} \\ \\ &= \dfrac{ (-3)(3) + (-1)(8) }{\sqrt{ (-3)^2 + (-1)^2}\,\sqrt{3^2 + 8^2}} \\ \\ &= \dfrac{ -9 - 8 }{\sqrt{10}\,\sqrt{73}} = -\dfrac{17}{\sqrt{730}} \\ \\ \theta &\approx 129^{\circ} \end{align*} $$
The use of $\lvert\cdot\rvert$ and $\lVert\cdot\rVert$
In the next section we will generalize our concept of dot product to abstract vector spaces.
When denoting length or magnitude in Euclidean space $\mathbb{E}^n$ or $\mathbb{R}^n$ we mostly use the single absolute value sign, $\lvert\cdot\rvert$.
When denoting magnitude or norm when using a non-Euclidean geometry in $\mathbb{R}^n$ we use the double absolute value sign, $\lVert\cdot\rVert$.
When denoting magnitude or norm when using a geometry defined on an abstract vector space $V$ we use the double absolute value sign, $\lVert\cdot\rVert$.
The double absolute value sign is always correct to use. The use of single absolute value sign in $\mathbb{E}^n$ is a historical convention.
Exercise 5¶
Find $|\langle 6,\,8\rangle|$ and $|\langle 2,\,3,\,6\rangle|$.
View Solution
$$|\langle 6,\,8\rangle| = \sqrt{36 + 64} = \sqrt{100} = 10, \qquad |\langle 2,\,3,\,6\rangle| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$$Exercise 6¶
Find the distance between $\langle 5,\,6\rangle$ and $\langle 2,\,2\rangle$, and the distance between $\langle 3,\,5,\,7\rangle$ and $\langle 1,\,2,\,1\rangle$.
View Solution
$\langle 5,\,6\rangle - \langle 2,\,2\rangle = \langle 3,\,4\rangle$, so the distance is $\sqrt{9+16} = 5$.$\langle 3,\,5,\,7\rangle - \langle 1,\,2,\,1\rangle = \langle 2,\,3,\,6\rangle$, so the distance is $\sqrt{4+9+36} = 7$.
Exercise 7¶
Give the unit vector in the direction of $\langle 6,\,8\rangle$.
View Solution
$|\langle 6,\,8\rangle| = 10$, so the unit vector is $$\frac{1}{10}\langle 6,\,8\rangle = \left\langle \tfrac{3}{5},\ \tfrac{4}{5} \right\rangle$$Exercise 8¶
Find the cosine of the angle between $\langle 2,\,1\rangle$ and $\langle 3,\,-1\rangle$.
View Solution
$\mathbf{x}\cdot\mathbf{y} = 6 - 1 = 5$, $|\mathbf{x}| = \sqrt{5}$, $|\mathbf{y}| = \sqrt{10}$, so $$\cos(\theta) = \frac{5}{\sqrt{5}\,\sqrt{10}} = \frac{5}{\sqrt{50}} = \frac{1}{\sqrt{2}} \qquad (\theta = 45^\circ)$$Theorem 3¶
The Cauchy-Schwarz Inequality
If $\mathbf{x}$ and $\mathbf{y}$ are two vectors in Euclidean Space $\mathbb{R}^n$, then
$$ \begin{align*} \lvert\,\mathbf{x}\cdot\mathbf{y}\,\rvert &= \left| \, \lvert \mathbf{x} \rvert \, \lvert \mathbf{y} \rvert\, \cos(\theta)\, \right| \\ \\ &= \lvert \mathbf{x} \rvert \, \lvert \mathbf{y} \rvert \, \left|\cos(\theta)\right| \\ \\ &\le \lvert \mathbf{x} \rvert \, \lvert \mathbf{y} \rvert \end{align*} $$ 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 of Theorem 3¶
If one or both of the vectors is zero then either $\lvert \mathbf{x} \rvert = 0$ or $\lvert \mathbf{y} \rvert = 0$. In any case, $\mathbf{x}\cdot\mathbf{y}= 0$ and the product $\lvert \mathbf{x} \rvert \, \lvert \mathbf{y} \rvert = 0$. Thus
$$ \lvert \,\mathbf{x}\cdot\mathbf{y}\, \rvert = 0 = \lvert \mathbf{x} \rvert \, \lvert \mathbf{y} \rvert $$
If 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 the result is equality. Hence
$$ \left|\,\mathbf{x}\cdot\mathbf{y}\,\right| = \lvert \mathbf{x} \rvert \, \lvert \mathbf{y} \rvert \, \left|\cos(\theta)\right| = \lvert \mathbf{x} \rvert \, \lvert \mathbf{y} \rvert $$
If neither vector is the zero vector and they are not colinear, then $0<\left|\cos(\theta)\right|<1$. Therefore
$$ \left|\,\mathbf{x}\cdot\mathbf{y}\,\right| = \lvert \mathbf{x} \rvert \, \lvert \mathbf{y} \rvert \, \left|\cos(\theta)\right| < \lvert \mathbf{x} \rvert \, \lvert \mathbf{y} \rvert \qquad\blacksquare $$
For example, with $\mathbf{x}=\langle 3,\,4\rangle$ and $\mathbf{y}=\langle 5,\,12\rangle$, we have $\left|\,\mathbf{x}\cdot\mathbf{y}\,\right| = |15 + 48| = 63$, while $|\mathbf{x}|\,|\mathbf{y}| = (5)(13) = 65$, so $63 \le 65$ — a strict inequality, since the vectors are not parallel.
Theorem 4¶
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 of Theorem 4¶
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. $\qquad\blacksquare$
Exercise 9¶
Verify the Cauchy-Schwarz inequality for $\mathbf{x}=\langle 3,\,4\rangle$ and $\mathbf{y}=\langle 5,\,12\rangle$.
View Solution
$|\mathbf{x}\cdot\mathbf{y}| = |15 + 48| = 63$ and $|\mathbf{x}|\,|\mathbf{y}| = (5)(13) = 65$, so $$|\mathbf{x}\cdot\mathbf{y}| = 63 \le 65 = |\mathbf{x}|\,|\mathbf{y}|$$Exercise 10¶
Verify the triangle inequality for $\mathbf{u}=\langle 3,\,4\rangle$ and $\mathbf{v}=\langle -3,\,4\rangle$.
View Solution
$\mathbf{u}+\mathbf{v} = \langle 0,\,8\rangle$, so $|\mathbf{u}+\mathbf{v}| = 8$. Also $|\mathbf{u}| + |\mathbf{v}| = 5 + 5 = 10$, so $$|\mathbf{u}+\mathbf{v}| = 8 \le 10 = |\mathbf{u}| + |\mathbf{v}|$$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¶
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 of Theorem 5¶
The proof is very similar to the proof for the triangle inequality.
$$\begin{align*} \lvert\mathbf{u} + \mathbf{v}\rvert^2 &= \left(\mathbf{u} + \mathbf{v}\right)\cdot\left(\mathbf{u} + \mathbf{v}\right) \\ \\ &= \left(\mathbf{u} + \mathbf{v}\right)\cdot\mathbf{u} + \left(\mathbf{u} + \mathbf{v}\right)\cdot\mathbf{v} \\ \\ &= \mathbf{u}\cdot\mathbf{u} + \mathbf{v}\cdot\mathbf{u} + \mathbf{u}\cdot\mathbf{v} + \mathbf{v}\cdot\mathbf{v} \\ \\ &= \lvert \mathbf{u} \rvert^2 + 2\mathbf{u}\cdot\mathbf{v} + \lvert \mathbf{v} \rvert^2 \\ \end{align*}$$
Now
$$ \left|\mathbf{u} + \mathbf{v}\right|^2 = \lvert \mathbf{u} \rvert^2 + \lvert \mathbf{v} \rvert^2$$
if and only if $2\mathbf{u}\cdot\mathbf{v}=0$ if and only if $\mathbf{u}\cdot\mathbf{v}=0$. $\qquad\blacksquare$
Exercise 11¶
Are $\langle 3,\,-2\rangle$ and $\langle 4,\,6\rangle$ orthogonal?
View Solution
$\langle 3,\,-2\rangle\cdot\langle 4,\,6\rangle = 12 - 12 = 0$. Since the dot product is zero, the vectors are orthogonal.Exercise 12¶
Are $\langle 1,\,2,\,2\rangle$ and $\langle 2,\,-2,\,1\rangle$ orthogonal?
View Solution
$\langle 1,\,2,\,2\rangle\cdot\langle 2,\,-2,\,1\rangle = 2 - 4 + 2 = 0$. The vectors are orthogonal.Exercise 13¶
Give a nonzero vector orthogonal to $\langle 4,\,-3\rangle$.
View Solution
Any $\langle a,\,b\rangle$ with $4a - 3b = 0$ works; for example $\langle 3,\,4\rangle$, since $\langle 4,\,-3\rangle\cdot\langle 3,\,4\rangle = 12 - 12 = 0$.Exercise 14¶
Verify the Pythagorean theorem for the orthogonal pair $\mathbf{u}=\langle 3,\,4\rangle$ and $\mathbf{v}=\langle 8,\,-6\rangle$.
View Solution
$\mathbf{u}\cdot\mathbf{v} = 24 - 24 = 0$, so the vectors are orthogonal. Then $$|\mathbf{u}|^2 + |\mathbf{v}|^2 = 25 + 100 = 125, \qquad |\mathbf{u}+\mathbf{v}|^2 = |\langle 11,\,-2\rangle|^2 = 121 + 4 = 125$$ The two agree, as the Pythagorean theorem predicts.Example 6¶
Find the distance from the point $P = (3,1)$ to the line $y=-\frac{3}{4}x$.
The line has the form $\mathbf{y} = \mathbf{m}t + \mathbf{b}$ in vector form, where direction vector $\mathbf{m} = \langle 4, -3 \rangle$, and $\mathbf{b} = \mathbf{0}$.
Compute $\mathbf{m}^\perp = \langle 3,4 \rangle$. The line $\mathbf{y} = \mathbf{m^{\perp}}t$ is perpendicular to the original direction vector $\mathbf{m} = \langle 4, -3\rangle$, so this line meets the given line at a right angle.
The scalar component of vector $\overrightarrow{OP}$ onto $\mathbf{m^{\perp}}$ is the signed distance from the line to point $P$.
The distance $d = \left|\, \text{Comp}_{\mathbf{m^{\perp}}} \overrightarrow{OP} \,\right|$
$$ d = \frac{ \left|\, \mathbf{m^{\perp}}\cdot\overrightarrow{OP} \,\right| }{\lvert\mathbf{m^{\perp}}\rvert} = \frac{ \left|\, 3\cdot 3 + 4\cdot 1 \,\right| }{\sqrt{9 + 16}} = \frac{13}{5} $$
In the next few sections try not to memorize formulas. Think like a Linear Algebra Master. Ask yourself what you are asked to determine. Use linear algebra to describe your solution. Place the given information into the linear algebra model. Then solve the linear algebra problem.
Exercise 15¶
Find the distance from the point $P=(1,\,-2)$ to the line through the origin with direction $\mathbf{m}=\langle 3,\,4\rangle$.
View Solution
A perpendicular direction is $\mathbf{m}^\perp = \langle 4,\,-3\rangle$. Then $$d = \frac{\left|\,\mathbf{m}^\perp\cdot\overrightarrow{OP}\,\right|}{|\mathbf{m}^\perp|} = \frac{|(4)(1) + (-3)(-2)|}{\sqrt{16+9}} = \frac{|4 + 6|}{5} = \frac{10}{5} = 2$$Exercise 16¶
Find the distance from the point $P=(3,\,2)$ to the line through the origin with direction $\mathbf{m}=\langle 5,\,12\rangle$.
View Solution
A perpendicular direction is $\mathbf{m}^\perp = \langle 12,\,-5\rangle$. Then $$d = \frac{\left|\,\mathbf{m}^\perp\cdot\overrightarrow{OP}\,\right|}{|\mathbf{m}^\perp|} = \frac{|(12)(3) + (-5)(2)|}{\sqrt{144+25}} = \frac{|36 - 10|}{13} = \frac{26}{13} = 2$$Example 7¶
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 $$\overrightarrow{P_0P_1} = \mathbf{P_1} - \mathbf{P_0} = \begin{bmatrix} 0-1 \\ 1-0 \\ 0-0 \end{bmatrix} = \begin{bmatrix} -1\ \\ \ \ 1\ \\ \ \ 0\ \end{bmatrix}$$
$$ \overrightarrow{P_0P_2} = \mathbf{P_2} - \mathbf{P_0} = \begin{bmatrix} 0-1 \\ 0-0 \\ 1-0 \end{bmatrix} = \begin{bmatrix} -1\ \\ \ \ 0\ \\ \ \ 1\ \end{bmatrix}$$
- 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,
$$ \mathbf{n} = \overrightarrow{P_0P_1} \times \overrightarrow{P_0P_2} = \begin{vmatrix} \mathbf{\hat{\imath}}\ & -1\ & -1\ \\ \mathbf{\hat{\jmath}}\ &\ \ 1\ &\ \ 0\ \\ \mathbf{\hat{k}}\ &\ \ 0\ &\ \ 1\ \end{vmatrix} = 1\mathbf{\hat{\imath}} + 1\mathbf{\hat{\jmath}} + 1\mathbf{\hat{k}} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} $$
- We denote any point on the plane by $P = (x,y,z)$ and the vector equation of the plane is
$$ \mathbf{n}\cdot\overrightarrow{P_0P} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\cdot\begin{bmatrix} x-1 \\ y-0 \\ z-0 \end{bmatrix} = (x-1) + y + z = 0 $$
- The scalar equation of the plane is then $$ x + y + z = 1 $$
Exercise 17¶
Find the equation of the plane through the points $(1,\,1,\,1)$, $(2,\,3,\,1)$, and $(1,\,2,\,3)$.
View Solution
Using $P_0=(1,1,1)$, form two vectors in the plane: $$\overrightarrow{P_0P_1} = \langle 1,\,2,\,0\rangle, \qquad \overrightarrow{P_0P_2} = \langle 0,\,1,\,2\rangle$$ A normal vector is their cross product: $$\mathbf{n} = \overrightarrow{P_0P_1}\times\overrightarrow{P_0P_2} = \langle\, (2)(2)-(0)(1),\ (0)(0)-(1)(2),\ (1)(1)-(2)(0)\,\rangle = \langle 4,\,-2,\,1\rangle$$ The plane $\mathbf{n}\cdot(\mathbf{p}-\mathbf{p}_0)=0$ gives $4(x-1) - 2(y-1) + (z-1) = 0$, or $$4x - 2y + z = 3$$Example 8¶
Find the distance from point $P = (1,1,2)$ to the plane $2x + 2y + z = 0$.
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 perpendicular distance 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 in fact $\lvert\overrightarrow{QP}\rvert$. The problem is that we do not know where $Q$ lies on the plane. In general, we can pick any convenient point on the plane, say $R$. The component of vector $\overrightarrow{RP}$ onto $\mathbf{n}$ will give us the distance from the point to the plane. For instance, set $x=y=0$ and solve for $z$. The plane in this example is $2x + 2y + z=0$. When $x=y=0$, then $z=0$ too. In this particular problem $\overrightarrow{RP}=\overrightarrow{OP}$ because the plane passes through the origin.
- The distance from the point $P$ to the plane is the absolute value of the scalar component of vector $\overrightarrow{OP}$ onto the normal vector $\mathbf{n}$.
$$ d = \left|\, \text{Comp}_{\mathbf{n}}\overrightarrow{OP} \,\right| = \dfrac{ \lvert\,\overrightarrow{OP}\cdot\mathbf{n} \,\rvert }{\left| \mathbf{n} \right|} = \dfrac{ \left|\, (2)(1) + (2)(1) + (1)(2) \,\right| }{\sqrt{1^2 + 2^2 + 2^2}} = \dfrac{6}{\sqrt{9}} = 2 $$
Exercise 18¶
Find the distance from the point $P=(3,\,3,\,3)$ to the plane $x + 2y + 2z = 6$.
View Solution
The normal is $\mathbf{n} = \langle 1,\,2,\,2\rangle$ with $|\mathbf{n}| = 3$. The distance is the absolute scalar component of $\overrightarrow{RP}$ onto $\mathbf{n}$, which reduces to $$d = \frac{|\,\mathbf{n}\cdot\overrightarrow{OP} - 6\,|}{|\mathbf{n}|} = \frac{|(3 + 6 + 6) - 6|}{3} = \frac{9}{3} = 3$$Exercise 19¶
Find the distance from the point $P=(2,\,-1,\,1)$ to the plane $2x + 3y + 6z = 0$.
View Solution
The normal is $\mathbf{n} = \langle 2,\,3,\,6\rangle$ with $|\mathbf{n}| = 7$. Since the plane passes through the origin, $$d = \frac{|\,\mathbf{n}\cdot\overrightarrow{OP}\,|}{|\mathbf{n}|} = \frac{|4 - 3 + 6|}{7} = \frac{7}{7} = 1$$These questions test the definitions and theorems of this section. In a true/false question, true means always true — a statement that is only sometimes true is false.
True/False Questions¶
For each statement, decide whether it is always true. If true, give a general argument; if false, give a counterexample with specific numbers.
- For all vectors $\mathbf{x}$ and $\mathbf{y}$ in $\mathbb{R}^n$, $\;\mathbf{x}\cdot\mathbf{y} = \mathbf{y}\cdot\mathbf{x}$.
View Solution
TRUE. The dot product is a scalar, so it equals its own transpose: $\mathbf{x}\cdot\mathbf{y} = \mathbf{y}^\mathsf{T}\mathbf{x} = \left(\mathbf{y}^\mathsf{T}\mathbf{x}\right)^\mathsf{T} = \mathbf{x}^\mathsf{T}\mathbf{y} = \mathbf{y}\cdot\mathbf{x}$. (Geometrically, $|\mathbf{x}||\mathbf{y}|\cos\theta = |\mathbf{y}||\mathbf{x}|\cos\theta$.)
- For all vectors $\mathbf{x}$ and $\mathbf{y}$ in $\mathbb{R}^n$, $\;\left|\mathbf{x}\cdot\mathbf{y}\right| = |\mathbf{x}|\,|\mathbf{y}|$.
View Solution
FALSE. The Cauchy-Schwarz inequality gives $\left|\mathbf{x}\cdot\mathbf{y}\right| \le |\mathbf{x}|\,|\mathbf{y}|$, with equality only when the vectors are parallel. CounterExample51: $\mathbf{x} = \langle 1,\,0\rangle$ and $\mathbf{y} = \langle 0,\,1\rangle$. Then $\left|\mathbf{x}\cdot\mathbf{y}\right| = 0$, but $|\mathbf{x}|\,|\mathbf{y}| = (1)(1) = 1$, and $0 \neq 1$.
- If $\mathbf{x}$ and $\mathbf{y}$ are orthogonal vectors in $\mathbb{R}^n$, then $\;|\mathbf{x}+\mathbf{y}|^2 = |\mathbf{x}|^2 + |\mathbf{y}|^2$.
View Solution
TRUE. This is the Pythagorean theorem. Expanding, $$|\mathbf{x}+\mathbf{y}|^2 = (\mathbf{x}+\mathbf{y})\cdot(\mathbf{x}+\mathbf{y}) = |\mathbf{x}|^2 + 2\,\mathbf{x}\cdot\mathbf{y} + |\mathbf{y}|^2.$$ Because $\mathbf{x}$ and $\mathbf{y}$ are orthogonal, $\mathbf{x}\cdot\mathbf{y} = 0$, leaving $|\mathbf{x}+\mathbf{y}|^2 = |\mathbf{x}|^2 + |\mathbf{y}|^2$.
- For a fixed nonzero vector $\mathbf{x}$ in $\mathbb{R}^n$, if $\;\mathbf{x}\cdot\mathbf{y} = \mathbf{x}\cdot\mathbf{z}$, then $\mathbf{y} = \mathbf{z}$.
View Solution
FALSE. The dot product cannot be cancelled. CounterExample51: $\mathbf{x} = \langle 1,\,0\rangle$, $\mathbf{y} = \langle 1,\,2\rangle$, $\mathbf{z} = \langle 1,\,5\rangle$. Then $\mathbf{x}\cdot\mathbf{y} = 1$ and $\mathbf{x}\cdot\mathbf{z} = 1$, so $\mathbf{x}\cdot\mathbf{y} = \mathbf{x}\cdot\mathbf{z}$, yet $\mathbf{y} \neq \mathbf{z}$. (Any $\mathbf{y}$ and $\mathbf{z}$ differing only in a direction orthogonal to $\mathbf{x}$ give the same dot product.)
- The zero vector in $\mathbb{R}^n$ is orthogonal to every vector in $\mathbb{R}^n$.
View Solution
TRUE. For every $\mathbf{y}$ in $\mathbb{R}^n$, $\;\mathbf{0}\cdot\mathbf{y} = 0$. Since a zero scalar product means the vectors are orthogonal, the zero vector is orthogonal to every vector.
Multiple Choice Questions¶
Check the box of each statement that must be true, and leave false statements unchecked. No justification is required.
- If $\mathbf{x}$ and $\mathbf{y}$ are vectors in $\mathbb{R}^n$ and $c$ is a scalar, then check each that must be true.
$\square\ $ $\mathbf{x}\cdot\mathbf{x} = |\mathbf{x}|^2$.
$\square\ $ If $\mathbf{x}\cdot\mathbf{y} = 0$, then $\mathbf{x} = \mathbf{0}$ or $\mathbf{y} = \mathbf{0}$.
$\square\ $ $|c\mathbf{x}| = |c|\,|\mathbf{x}|$.
$\square\ $ $\mathbf{x}\cdot\mathbf{y} \le |\mathbf{x}|\,|\mathbf{y}|$.
$\square\ $ If $|\mathbf{x}| = |\mathbf{y}|$, then $\mathbf{x} = \mathbf{y}$.
View Solution
Check the first, third, and fourth boxes; leave the second and fifth unchecked. (Nonzero orthogonal vectors give $\mathbf{x}\cdot\mathbf{y}=0$ without either being zero, and equal lengths do not force equal vectors.)
- If $\mathbf{u}$ and $\mathbf{v}$ are vectors in $\mathbb{R}^n$, then check each that must be true.
$\square\ $ The projection of $\mathbf{v}$ onto $\mathbf{u}$ is a scalar multiple of $\mathbf{u}$.
$\square\ $ $\mathbf{u}$ and $\mathbf{v}$ are orthogonal if and only if $\mathbf{u}\cdot\mathbf{v} = 0$.
$\square\ $ If $\mathbf{u}$ is a unit vector, then $|\mathbf{u}| = 1$.
$\square\ $ The angle between $\mathbf{u}$ and $\mathbf{v}$ is always between $0$ and $\dfrac{\pi}{2}$.
$\square\ $ $|\mathbf{u} + \mathbf{v}| = |\mathbf{u}| + |\mathbf{v}|$ for all $\mathbf{u}$ and $\mathbf{v}$.
View Solution
Check the first, second, and third boxes; leave the fourth and fifth unchecked. (The angle ranges over $[0,\pi]$, and $|\mathbf{u}+\mathbf{v}| = |\mathbf{u}|+|\mathbf{v}|$ only when the vectors point the same way.)
- Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in $\mathbb{R}^n$. Check each that must be true.
$\square\ $ $\mathbf{u}\cdot\mathbf{v} = \mathbf{v}^\mathsf{T}\mathbf{u}$.
$\square\ $ Every unit vector in $\mathbb{R}^n$ can be mapped to $\mathbf{e}_1$ by some rotation matrix.
$\square\ $ If $\mathbf{u}\cdot\mathbf{v} > 0$, then the angle between $\mathbf{u}$ and $\mathbf{v}$ is acute.
$\square\ $ The distance from a point to a line equals the length of the projection of $\overrightarrow{OP}$ onto the line's direction vector.
$\square\ $ For any nonzero $\mathbf{u}$, the vectors $\mathbf{u}$ and $-\mathbf{u}$ are orthogonal.
View Solution
Check the first, second, and third boxes; leave the fourth and fifth unchecked. (The distance to a line uses the component onto the perpendicular direction, not the direction vector; and $\mathbf{u}\cdot(-\mathbf{u}) = -|\mathbf{u}|^2 \neq 0$, so $\mathbf{u}$ and $-\mathbf{u}$ are not orthogonal.)