Vector Algebra and Spaces: Learning Objectives 1, 2
Question 1: Determine if $\textbf{v}$ and $\textbf{w}$ are equivalent vectors:
- $\textbf{v}$ has initial point $(3,2)$ and terminal point $(7,2)$, $\textbf{w}$ has initial point $(1,-4)$ and terminal point $(1,0)$.
- $\textbf{v}$ has initial point $(0,0)$ and terminal point $(1,1)$, $\textbf{w}$ has initial point $(-2,-2)$ and terminal point $(-1,-1)$.
Two vectors are equivalent if they have the same component form. Find the component form for each vector by subtracting the coordinates of the initial point from the coordinates of the terminal point (terminal - initial).
- Not equivalent.
- Equivalent.
a)
Component form of $\textbf{v}$: $\langle 7-3, 2-2 \rangle = \langle 4, 0 \rangle$.
Component form of $\textbf{w}$: $\langle 1-1, 0-(-4) \rangle = \langle 0, 4 \rangle$.
Since $\langle 4, 0 \rangle \neq \langle 0, 4 \rangle$, the vectors are not equivalent.
b)
Component form of $\textbf{v}$: $\langle 1-0, 1-0 \rangle = \langle 1, 1 \rangle$.
Component form of $\textbf{w}$: $\langle -1-(-2), -1-(-2) \rangle = \langle 1, 1 \rangle$.
Since the component forms are identical, the vectors are equivalent.
Component form of $\textbf{v}$: $\langle 7-3, 2-2 \rangle = \langle 4, 0 \rangle$.
Component form of $\textbf{w}$: $\langle 1-1, 0-(-4) \rangle = \langle 0, 4 \rangle$.
Since $\langle 4, 0 \rangle \neq \langle 0, 4 \rangle$, the vectors are not equivalent.
b)
Component form of $\textbf{v}$: $\langle 1-0, 1-0 \rangle = \langle 1, 1 \rangle$.
Component form of $\textbf{w}$: $\langle -1-(-2), -1-(-2) \rangle = \langle 1, 1 \rangle$.
Since the component forms are identical, the vectors are equivalent.
Question 2: The component form of a vector $\langle x_2 - x_1,\, y_2 - y_1 \rangle $ describes the change in the x- and y-directions from its initial to terminal points. Given the initial point $(-3, 4)$ and terminal point $(1, 2)$, express the vector $\textbf{v}$ in component form.
The component form is $\langle x_{terminal} - x_{initial}, y_{terminal} - y_{initial} \rangle$.
$\langle 4, -2 \rangle$
Let the initial point be $P_1 = (-3, 4)$ and the terminal point be $P_2 = (1, 2)$.
The component form of $\textbf{v}$ is given by:
\[ \textbf{v} = \langle 1 - (-3), 2 - 4 \rangle = \langle 4, -2 \rangle \]
Question 3: Let $\textbf{v}$ be the vector with initial point $(2,5)$ and terminal point $(8, 13)$ and let $\textbf{w} = \langle -2, 4 \rangle$.
- Express $\textbf{v}$ in component form and find $||\textbf{v}||$, the magnitude of vector $\textbf{v}$.
- Find $\textbf{v} + \textbf{w}$.
- Find $3\textbf{v}$.
- Find $\textbf{v}- 2\textbf{w}$.
First, find the component form of $\textbf{v}$. The magnitude (or norm) is $||\textbf{v}|| = \sqrt{v_1^2 + v_2^2}$. Vector addition and scalar multiplication are performed component-wise.
- $\textbf{v} = \langle 6, 8 \rangle$, $||\textbf{v}|| = 10$.
- $\langle 4, 12\rangle$.
- $\langle 18, 24 \rangle$.
- $\langle 10, 0 \rangle$.
a) $\textbf{v} = \langle 8-2, 13-5 \rangle = \langle 6, 8 \rangle$.
$||\textbf{v}|| = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = 10$.
b) $\textbf{v} + \textbf{w} = \langle 6, 8 \rangle + \langle -2, 4 \rangle = \langle 6-2, 8+4 \rangle = \langle 4, 12 \rangle$.
c) $3\textbf{v} = 3\langle 6, 8 \rangle = \langle 3 \cdot 6, 3 \cdot 8 \rangle = \langle 18, 24 \rangle$.
d) $\textbf{v} - 2\textbf{w} = \langle 6, 8 \rangle - 2\langle -2, 4 \rangle = \langle 6, 8 \rangle - \langle -4, 8 \rangle = \langle 6-(-4), 8-8 \rangle = \langle 10, 0 \rangle$.
$||\textbf{v}|| = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = 10$.
b) $\textbf{v} + \textbf{w} = \langle 6, 8 \rangle + \langle -2, 4 \rangle = \langle 6-2, 8+4 \rangle = \langle 4, 12 \rangle$.
c) $3\textbf{v} = 3\langle 6, 8 \rangle = \langle 3 \cdot 6, 3 \cdot 8 \rangle = \langle 18, 24 \rangle$.
d) $\textbf{v} - 2\textbf{w} = \langle 6, 8 \rangle - 2\langle -2, 4 \rangle = \langle 6, 8 \rangle - \langle -4, 8 \rangle = \langle 6-(-4), 8-8 \rangle = \langle 10, 0 \rangle$.
Question 4: Let $\textbf{v} = \langle 1,2\rangle$.
- Find a unit vector with the same direction as $\textbf{v}$.
- Find a vector $\textbf{w}$ with the same direction as $\textbf{v}$ such that $||\textbf{w}|| = 7$.
A unit vector in the direction of $\textbf{v}$ is found by dividing the vector by its magnitude: $\textbf{u} = \frac{\textbf{v}}{||\textbf{v}||}$. To find $\textbf{w}$, scale this unit vector by the desired magnitude.
- $\langle\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\rangle$
- $\langle\frac{7}{\sqrt{5}}, \frac{14}{\sqrt{5}}\rangle$
a) First, find the magnitude of $\textbf{v}$:
\[ ||\textbf{v}|| = \sqrt{1^2 + 2^2} = \sqrt{1+4} = \sqrt{5} \]
The unit vector $\textbf{u}$ is:
\[ \textbf{u} = \frac{\textbf{v}}{||\textbf{v}||} = \frac{\langle 1,2\rangle}{\sqrt{5}} = \left\langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right\rangle \]
b) To create a vector $\textbf{w}$ with magnitude 7 in the same direction, we scale the unit vector by 7: \[ \textbf{w} = 7\textbf{u} = 7\left\langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right\rangle = \left\langle \frac{7}{\sqrt{5}}, \frac{14}{\sqrt{5}} \right\rangle \]
b) To create a vector $\textbf{w}$ with magnitude 7 in the same direction, we scale the unit vector by 7: \[ \textbf{w} = 7\textbf{u} = 7\left\langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right\rangle = \left\langle \frac{7}{\sqrt{5}}, \frac{14}{\sqrt{5}} \right\rangle \]
Question 5: Let $\textbf{u} = \langle3,5,2\rangle$ and $\textbf{v} = \langle -1, 3, 0 \rangle$.
- Find the dot product of $\textbf{u}$ and $\textbf{v}$.
- Give the geometric interpretation of the dot product.
- Show that the dot product is commutative. That is, for $\textbf{u} = \langle u_1, u_2, u_3 \rangle$ and $\textbf{v}= \langle v_1, v_2, v_3 \rangle$, $\textbf{v} \cdot \textbf{u} = \textbf{u} \cdot \textbf{v}$.
- Explain why the commutative property of the dot product is consistent with the geometric interpretation.
For (a), sum the products of corresponding components. For (b), recall the formula involving the angle between the vectors. For (c), use the commutative property of scalar multiplication. For (d), consider if the magnitudes or the angle change when you swap the order of the vectors.
- 12
- The dot product is a scalar quantity given by $\textbf{u} \cdot \textbf{v} = ||\textbf{u}||\cdot||\textbf{v}||\cos(\theta)$, which measures the extent to which two vectors point in the same direction.
- The proof relies on the commutative property of multiplication for real numbers.
- Geometrically, the dot product depends on the magnitudes of the vectors and the angle between them, none of which depend on the order of the vectors.
a) \[ \textbf{u} \cdot \textbf{v} = (3)(-1) + (5)(3) + (2)(0) = -3 + 15 + 0 = 12 \]
b) The dot product is a scalar quantity. Geometrically, it is the product of the magnitudes of the two vectors and the cosine of the angle between them: $\textbf{u} \cdot \textbf{v} = ||\textbf{u}|| \cdot ||\textbf{v}|| \cos(\theta)$. It can be interpreted as the length of the projection of one vector onto the other, scaled by the magnitude of the other vector.
c) By definition, $\textbf{u} \cdot \textbf{v} = u_1v_1 + u_2v_2 + u_3v_3$. Also, $\textbf{v} \cdot \textbf{u} = v_1u_1 + v_2u_2 + v_3u_3$. Since scalar multiplication is commutative ($u_iv_i = v_iu_i$), the sums are equal. Thus, $\textbf{u} \cdot \textbf{v} = \textbf{v} \cdot \textbf{u}$.
d) The geometric definition involves the magnitudes of the vectors ($||\textbf{u}||, ||\textbf{v}||$) and the angle between them ($\theta$). Since scalar multiplication is commutative, $||\textbf{u}|| \cdot ||\textbf{v}|| = ||\textbf{v}|| \cdot ||\textbf{u}||$. The angle from $\textbf{u}$ to $\textbf{v}$ is the same as the angle from $\textbf{v}$ to $\textbf{u}$. Therefore, the geometric result is the same regardless of the order.
b) The dot product is a scalar quantity. Geometrically, it is the product of the magnitudes of the two vectors and the cosine of the angle between them: $\textbf{u} \cdot \textbf{v} = ||\textbf{u}|| \cdot ||\textbf{v}|| \cos(\theta)$. It can be interpreted as the length of the projection of one vector onto the other, scaled by the magnitude of the other vector.
c) By definition, $\textbf{u} \cdot \textbf{v} = u_1v_1 + u_2v_2 + u_3v_3$. Also, $\textbf{v} \cdot \textbf{u} = v_1u_1 + v_2u_2 + v_3u_3$. Since scalar multiplication is commutative ($u_iv_i = v_iu_i$), the sums are equal. Thus, $\textbf{u} \cdot \textbf{v} = \textbf{v} \cdot \textbf{u}$.
d) The geometric definition involves the magnitudes of the vectors ($||\textbf{u}||, ||\textbf{v}||$) and the angle between them ($\theta$). Since scalar multiplication is commutative, $||\textbf{u}|| \cdot ||\textbf{v}|| = ||\textbf{v}|| \cdot ||\textbf{u}||$. The angle from $\textbf{u}$ to $\textbf{v}$ is the same as the angle from $\textbf{v}$ to $\textbf{u}$. Therefore, the geometric result is the same regardless of the order.
Question 6: Let $\textbf{u} = \langle 0,2,1 \rangle$ and $\textbf{v} = \langle 3, -1, 0\rangle$.
- Calculate $\textbf{u} \times \textbf{v}$ and $\textbf{v} \times \textbf{u}$. Are they equivalent?
- Based on the previous result, is the cross product commutative?
- Give the geometric interpretation of the cross product (in terms of magnitude and direction). Explain why the geometric interpretation is consistent with the cross product not being commutative.
For part (c), consider the right-hand rule for determining the direction of the cross product vector. What happens when you swap the order of the vectors?
- $\textbf{u} \times \textbf{v} = \langle 1, 3, -6\rangle$ and $\textbf{v} \times \textbf{u} = \langle -1, -3, 6\rangle$. They are not equivalent.
- No, the cross product is anti-commutative.
- The magnitude of the cross product is the area of the parallelogram spanned by the vectors, which is independent of order. The direction is given by the right-hand rule, which reverses when the order of the vectors is swapped.
a) \[ \textbf{u} \times \textbf{v} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \\ 0 & 2 & 1 \\ 3 & -1 & 0 \end{vmatrix} = \textbf{i}(0 - (-1)) - \textbf{j}(0 - 3) + \textbf{k}(0 - 6) = \langle 1, 3, -6 \rangle \]
\[ \textbf{v} \times \textbf{u} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \\ 3 & -1 & 0 \\ 0 & 2 & 1 \end{vmatrix} = \textbf{i}(-1 - 0) - \textbf{j}(3 - 0) + \textbf{k}(6 - 0) = \langle -1, -3, 6 \rangle \]
The vectors are not equivalent.
b) No, the cross product is not commutative. It is anti-commutative, since $\textbf{u} \times \textbf{v} = -(\textbf{v} \times \textbf{u})$.
c) Geometrically, the magnitude of the cross product, $||\textbf{u} \times \textbf{v}||$, represents the area of the parallelogram formed by vectors $\textbf{u}$ and $\textbf{v}$. This area is the same as the area of the parallelogram formed by $\textbf{v}$ and $\textbf{u}$. However, the direction of the cross product vector is determined by the right-hand rule. For $\textbf{u} \times \textbf{v}$, your fingers curl from $\textbf{u}$ to $\textbf{v}$, and your thumb points in the direction of the resultant vector. For $\textbf{v} \times \textbf{u}$, your fingers curl from $\textbf{v}$ to $\textbf{u}$, causing your thumb to point in the exact opposite direction. This explains why the cross product is anti-commutative.
b) No, the cross product is not commutative. It is anti-commutative, since $\textbf{u} \times \textbf{v} = -(\textbf{v} \times \textbf{u})$.
c) Geometrically, the magnitude of the cross product, $||\textbf{u} \times \textbf{v}||$, represents the area of the parallelogram formed by vectors $\textbf{u}$ and $\textbf{v}$. This area is the same as the area of the parallelogram formed by $\textbf{v}$ and $\textbf{u}$. However, the direction of the cross product vector is determined by the right-hand rule. For $\textbf{u} \times \textbf{v}$, your fingers curl from $\textbf{u}$ to $\textbf{v}$, and your thumb points in the direction of the resultant vector. For $\textbf{v} \times \textbf{u}$, your fingers curl from $\textbf{v}$ to $\textbf{u}$, causing your thumb to point in the exact opposite direction. This explains why the cross product is anti-commutative.
Question 7: Let $\textbf{x} \in \mathbb{R}^n$ and $c \in \mathbb{R}$. Show that $c\textbf{x} = \mathbf{0}$ if and only if $c=0$ or $\textbf{x}= \mathbf{0}$.
An "if and only if" proof requires proving two directions. First, assume $c=0$ or $\textbf{x}=\mathbf{0}$ and show that $c\textbf{x}=\mathbf{0}$. Second, assume $c\textbf{x}=\mathbf{0}$ and show that it must be true that $c=0$ or $\textbf{x}=\mathbf{0}$.
The proof involves showing the implication in both directions, using the definitions of scalar multiplication and the zero vector.
This proof requires two parts. Let $\textbf{x} = \langle x_1, \dots, x_n \rangle$.
Part 1 ($\implies$): Assume $c=0$ or $\textbf{x}=\mathbf{0}$. Show $c\textbf{x}=\mathbf{0}$.
Part 2 ($\impliedby$): Assume $c\textbf{x}=\mathbf{0}$. Show $c=0$ or $\textbf{x}=\mathbf{0}$. We prove this by contradiction. Assume $c\textbf{x}=\mathbf{0}$ AND $c \neq 0$ AND $\textbf{x} \neq \mathbf{0}$. Since $\textbf{x} \neq \mathbf{0}$, at least one of its components must be non-zero. Let's say $x_i \neq 0$. The equation $c\textbf{x}=\mathbf{0}$ means $\langle cx_1, \dots, cx_n \rangle = \langle 0, \dots, 0 \rangle$. This implies that $cx_i = 0$ for every component $i$. Since we assumed $c \neq 0$, we can divide by $c$: \[ x_i = \frac{0}{c} = 0 \] This contradicts our assumption that at least one component $x_i$ was non-zero. Therefore, the assumption that "$c \neq 0$ and $\textbf{x} \neq \mathbf{0}$" must be false. This leaves only the possibility that if $c\textbf{x}=\mathbf{0}$, then $c=0$ or $\textbf{x}=\mathbf{0}$.
Part 1 ($\implies$): Assume $c=0$ or $\textbf{x}=\mathbf{0}$. Show $c\textbf{x}=\mathbf{0}$.
- If $c=0$, then $c\textbf{x} = 0 \cdot \langle x_1, \dots, x_n \rangle = \langle 0 \cdot x_1, \dots, 0 \cdot x_n \rangle = \langle 0, \dots, 0 \rangle = \mathbf{0}$.
- If $\textbf{x}=\mathbf{0}$, then $\textbf{x} = \langle 0, \dots, 0 \rangle$. So $c\textbf{x} = c\langle 0, \dots, 0 \rangle = \langle c \cdot 0, \dots, c \cdot 0 \rangle = \langle 0, \dots, 0 \rangle = \mathbf{0}$.
Part 2 ($\impliedby$): Assume $c\textbf{x}=\mathbf{0}$. Show $c=0$ or $\textbf{x}=\mathbf{0}$. We prove this by contradiction. Assume $c\textbf{x}=\mathbf{0}$ AND $c \neq 0$ AND $\textbf{x} \neq \mathbf{0}$. Since $\textbf{x} \neq \mathbf{0}$, at least one of its components must be non-zero. Let's say $x_i \neq 0$. The equation $c\textbf{x}=\mathbf{0}$ means $\langle cx_1, \dots, cx_n \rangle = \langle 0, \dots, 0 \rangle$. This implies that $cx_i = 0$ for every component $i$. Since we assumed $c \neq 0$, we can divide by $c$: \[ x_i = \frac{0}{c} = 0 \] This contradicts our assumption that at least one component $x_i$ was non-zero. Therefore, the assumption that "$c \neq 0$ and $\textbf{x} \neq \mathbf{0}$" must be false. This leaves only the possibility that if $c\textbf{x}=\mathbf{0}$, then $c=0$ or $\textbf{x}=\mathbf{0}$.
Question 8: List the conditions for a subset $W$ to be a subspace of a vector space $V$.
There are three main conditions that a non-empty subset must satisfy to be considered a subspace. They involve the zero vector, addition, and scalar multiplication.
- The zero vector of $V$ is in $W$.
- $W$ is closed under vector addition: If $\textbf{u}$ and $\textbf{v}$ are in $W$, then $\textbf{u}+\textbf{v}$ is in $W$.
- $W$ is closed under scalar multiplication: If $\textbf{u}$ is in $W$ and $c$ is a scalar, then $c\textbf{u}$ is in $W$.
Question 9: Prove or disprove that the given set $V$ is a subspace of $\mathbb{R}^2$.
- $V = \{ \begin{bmatrix} x \\ 2x \end{bmatrix} : x \in \mathbb{R} \}$
- $V = \{ \begin{bmatrix} x \\ y \end{bmatrix} \in \mathbb{R}^2: x \geq 0 \}$
For each set, check the three subspace conditions: presence of the zero vector, closure under addition, and closure under scalar multiplication. If any one condition fails, it is not a subspace.
- $V$ is a subspace.
- $V$ is not a subspace.
a) $V = \{ \begin{bmatrix} x \\ 2x \end{bmatrix} : x \in \mathbb{R} \}$ (a line through the origin)
1. Zero vector: If $x=0$, the vector is $\begin{bmatrix} 0 \\ 0 \end{bmatrix}$, which is the zero vector. So it's in $V$. (Pass)
2. Closure under addition: Let $\textbf{u}=\begin{bmatrix} a \\ 2a \end{bmatrix}$ and $\textbf{v}=\begin{bmatrix} b \\ 2b \end{bmatrix}$ be two vectors in $V$. Their sum is $\textbf{u}+\textbf{v} = \begin{bmatrix} a+b \\ 2a+2b \end{bmatrix} = \begin{bmatrix} (a+b) \\ 2(a+b) \end{bmatrix}$. This is of the form $\begin{bmatrix} x \\ 2x \end{bmatrix}$ where $x=a+b$. So it's in $V$. (Pass)
3. Closure under scalar multiplication: Let $\textbf{u}=\begin{bmatrix} a \\ 2a \end{bmatrix}$ be in $V$ and $c$ be a scalar. Then $c\textbf{u} = \begin{bmatrix} ca \\ c(2a) \end{bmatrix} = \begin{bmatrix} (ca) \\ 2(ca) \end{bmatrix}$. This is of the form $\begin{bmatrix} x \\ 2x \end{bmatrix}$ where $x=ca$. So it's in $V$. (Pass)
Since all three conditions hold, $V$ is a subspace.
b) $V = \{ \begin{bmatrix} x \\ y \end{bmatrix} \in \mathbb{R}^2: x \geq 0 \}$ (the right half-plane)
Let's check closure under scalar multiplication. Let $\textbf{u} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}$. Since its x-component (1) is $\ge 0$, $\textbf{u}$ is in $V$. Let $c=-2$.
Then $c\textbf{u} = -2\begin{bmatrix} 1 \\ 3 \end{bmatrix} = \begin{bmatrix} -2 \\ -6 \end{bmatrix}$. The x-component of this new vector is -2, which is not $\ge 0$. So $c\textbf{u}$ is not in $V$.
Since the set is not closed under scalar multiplication, $V$ is not a subspace.
1. Zero vector: If $x=0$, the vector is $\begin{bmatrix} 0 \\ 0 \end{bmatrix}$, which is the zero vector. So it's in $V$. (Pass)
2. Closure under addition: Let $\textbf{u}=\begin{bmatrix} a \\ 2a \end{bmatrix}$ and $\textbf{v}=\begin{bmatrix} b \\ 2b \end{bmatrix}$ be two vectors in $V$. Their sum is $\textbf{u}+\textbf{v} = \begin{bmatrix} a+b \\ 2a+2b \end{bmatrix} = \begin{bmatrix} (a+b) \\ 2(a+b) \end{bmatrix}$. This is of the form $\begin{bmatrix} x \\ 2x \end{bmatrix}$ where $x=a+b$. So it's in $V$. (Pass)
3. Closure under scalar multiplication: Let $\textbf{u}=\begin{bmatrix} a \\ 2a \end{bmatrix}$ be in $V$ and $c$ be a scalar. Then $c\textbf{u} = \begin{bmatrix} ca \\ c(2a) \end{bmatrix} = \begin{bmatrix} (ca) \\ 2(ca) \end{bmatrix}$. This is of the form $\begin{bmatrix} x \\ 2x \end{bmatrix}$ where $x=ca$. So it's in $V$. (Pass)
Since all three conditions hold, $V$ is a subspace.
b) $V = \{ \begin{bmatrix} x \\ y \end{bmatrix} \in \mathbb{R}^2: x \geq 0 \}$ (the right half-plane)
Let's check closure under scalar multiplication. Let $\textbf{u} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}$. Since its x-component (1) is $\ge 0$, $\textbf{u}$ is in $V$. Let $c=-2$.
Then $c\textbf{u} = -2\begin{bmatrix} 1 \\ 3 \end{bmatrix} = \begin{bmatrix} -2 \\ -6 \end{bmatrix}$. The x-component of this new vector is -2, which is not $\ge 0$. So $c\textbf{u}$ is not in $V$.
Since the set is not closed under scalar multiplication, $V$ is not a subspace.