Notebook 3 – Math 2121, Fall 2020

In today's notebook we'll try to get some feeling for vectors in $R^{2}$ and $R^{3}$ and their linear combinations. We'll also think about what a solution to a vector equation means. The code here involves a lot of new plotting functions in Julia.

xxxxxxxxxx
 
begin
    using PlutoUI
    using Plots
    using LinearAlgebra
​
    # Returns true if row has all zeros.
    function is_zero_row(A, row, tolerance=10e-16)
        (m, n) = size(A)
        return all([abs(A[row,col]) < tolerance for col=1:n])
    end
    
    # Returns first index of nonzero entry in row, or 0 if none exists.
    function find_leading(A, row=1, tolerance=10e-16)
        (m, n) = size(A)
        for col=1:n
            if abs(A[row, col]) > tolerance
                return col
            end
        end
        return 0
    end
​
    function find_nonzeros_in_column(A, row_start, col, tol=10e-12)
        (m, n) = size(A)
        return [i for i=row_start:m if abs(A[i,col]) > tol]
    end
    
    function columns(A)
        (m, n) = size(A)
        return n
    end
​
    function rowop_replace(A, source_row, target_row, scalar_factor)
        @assert source_row != target_row
        for j=1:columns(A)
            A[target_row,j] += A[source_row,j] * scalar_factor
        end
        return A
    end
​
    function rowop_scale(A, target_row, scalar_factor)
        @assert scalar_factor != 0
        for j=1:columns(A)
            A[target_row,j] *= scalar_factor
        end
        A
    end
    
    function rowop_swap(A, s, t)
        for j=1:columns(A)
            A[t,j], A[s,j] = A[s,j], A[t,j]
        end
        A
    end
​
    function RREF_with_rowop_count(A, tol=10e-12)
        rowop_count = 0
        A = float(copy(A))
        (m, n) = size(A)
        
        row = 1
        for col=1:n
            nonzeros = find_nonzeros_in_column(A, row, col, tol)
            if length(nonzeros) == 0
                continue
            end
            
            i = nonzeros[1]
            
            if abs(A[i, col] - 1) < tol
                A[i, col] = 1
            else
                A = rowop_scale(A, i, 1 / A[i, col])
                rowop_count += 1
            end
            
            for j=nonzeros[2:end]
                A = rowop_replace(A, i, j, -A[j, col])
                rowop_count += 1
            end
​
            if i != row
                A = rowop_swap(A, row, i)
                rowop_count += 1
            end
            
            row += 1
        end
        
        for row=m:-1:1
            l = find_leading(A, row, tol)
            if l > 0
                for k=1:row - 1
                    if abs(A[k,l]) > tol
                        rowop_replace(A, row, k, -A[k,l])
                        rowop_count += 1
                    end
                end
            end
        end
        
        # round entries to give nicer output; could omit this step
        for row=1:m
            for col=1:n
                if abs(A[row,col] - round(A[row,col])) < tol
                    A[row,col] = round(A[row,col])
                end
            end
        end
        
        return A, rowop_count
    end
    
    function RREF_RowOpCount(A, tol=10e-12)
        A, ops = RREF_with_rowop_count(A, tol)
        return ops
    end
​
    function RREF(A, tol=10e-12)
        A, ops = RREF_with_rowop_count(A, tol)
        return A
    end
    
    md"# Notebook 3 -- Math 2121, Fall 2020
In today's notebook we'll try to get some feeling for vectors in $\mathbb{R}^2$ and $\mathbb{R}^3$ and their linear combinations. We'll also think about what a solution to a vector equation means. The code here involves a lot of new plotting functions in Julia."
end

1.3 ms

Running this notebook (optional)

If you have Pluto up and running, you can access the notebook we are currently viewing by entering this link (right click -> Copy Link) in the Open from file menu in Pluto.

xxxxxxxxxx

2.3 ms

Vectors in $R^{2}$

Let's start out by drawing some vectors in $R^{2}$ .

xxxxxxxxxx
 
md"## Vectors in $\mathbb{R}^2$
​
Let's start out by drawing some vectors in $\mathbb{R}^2$."

15.7 μs

Create some vectors in $R^{2}$

u1 = v1 = w1 =

u2 = v2 = w2 =

x
 
begin   
    u1_slider = @bind u1 Slider(-10:0.1:10, default=4, show_value=false)
    u2_slider = @bind u2 Slider(-10:0.1:10, default=8, show_value=false)
​
    v1_slider = @bind v1 Slider(-10:0.1:10, default=-10, show_value=false)
    v2_slider = @bind v2 Slider(-10:0.1:10, default=6, show_value=false)
​
    w1_slider = @bind w1 Slider(-10:0.1:10, default=2, show_value=false)
    w2_slider = @bind w2 Slider(-10:0.1:10, default=-9, show_value=false)
    
    
    md"""**Create some vectors in $$\mathbb{R}^2$$**
    
    `u1` = $(u1_slider) `v1` = $(v1_slider) `w1` = $(w1_slider)
    
    `u2` = $(u2_slider) `v2` = $(v2_slider) `w2` = $(w2_slider)
    
    """
end

138 μs

The vectors we created:

x
 
md"The vectors we created:"

9.7 μs

u = [ 6.7]     v = [  -5]     w = [  10]
    [-3.7]         [-5.1]         [ -10]

xxxxxxxxxx

57.9 μs

Display: u v w

xxxxxxxxxx
 
md"""**Display**: 
`u` $(@bind uon html"<input type=checkbox >") `v` $(@bind von html"<input type=checkbox >") `w` $(@bind won html"<input type=checkbox >")
"""

87.1 μs

xxxxxxxxxx
 
begin
    lims = (
        -(maximum([1, uon + von + won])) * 10, 
        maximum([1, uon + von + won]) * 10
    )
    
    p1 = scatter([0], [0], xlims = lims, ylims = lims, 
        legend = false, label = "origin",  aspect_ratio=:equal,
        title = "Vectors u, v, w")
    
    if uon
        quiver!(quiver = ([u1],[u2]), [0], [0], color = [:blue],)
    end
    
    if von
        quiver!(quiver = ([v1],[v2]), [0], [0], color = [:orange],)
    end
    
    if won
        quiver!(quiver = ([w1],[w2]), [0], [0], color = [:green],)
    end 
    
    p2 = scatter([0], [0], xlims = lims, ylims = lims, 
        legend = false, label = "origin",  aspect_ratio=:equal,
        title = "Sum of vectors")
    
    quiver!(
        quiver = ([uon*u1 + von*v1 + won*w1],[uon*u2 + von*v2 + won*w2]), 
        [0], [0], color = [:black],
    )
    
    quiver!(
        quiver = ([uon*u1],[uon*u2]), 
        [0], [0], 
        color = [:blue], 
        linestyle = :dashdot,
    )
    quiver!(
        quiver = ([von*v1],[von*v2]), 
        [uon*u1], [uon*u2], 
        color = [:orange], 
        linestyle = :dashdot,
    )
    quiver!(
        quiver = ([won*w1],[won*w2]), 
        [uon*u1 + von*v1], [uon*u2 + von*v2], 
        color = [:green], 
        linestyle = :dashdot,
    )
    
    plot(p1, p2, layout=2)
end
    lims = (-(maximum([1, uon + von + won])) * 10, maximum([1, uon + von + won]) * 10)

7.3 ms

The vector $u$ is shown in blue, $v$ in orange, and $w$ in green.

The right hand graph shows the sum of the displayed vectors on the left as the solid arrow. The dotted arrows are just copies of u, v, and w moved around.

x
 
md"The vector $u$ is shown in blue, $v$ in orange, and $w$ in green.
​
The right hand graph shows the sum of the displayed vectors on the left as the solid arrow. The dotted arrows are just copies of u, v, and w moved around."

11 μs

Linear combinations

Now that we have chosen some vectors $u$ , $v$ , $w$ , let's think about the meaning of the linear combination $a u + b v + c w$ where $a, b, c \in R$ are scalars.

x
 
md"## Linear combinations
​
Now that we have chosen some vectors $u$, $v$, $w$, let's think about the meaning of the linear combination $au + bv + cw$ where $a,b,c \in \mathbb{R}$ are scalars."

9.1 μs

Choose some scalars in $R$

a = b = c =

x
 
begin   
    a_slider = @bind a Slider(-10:0.1:10, default=1, show_value=false)
    b_slider = @bind b Slider(-10:0.1:10, default=1, show_value=false)  
    c_slider = @bind c Slider(-10:0.1:10, default=1, show_value=false)
    
    md"""**Choose some scalars in $$\mathbb{R}$$**
    
    `a` = $(a_slider) `b` = $(b_slider) `c` = $(c_slider)
    """
end

93.8 μs

The scalars we chose:

xxxxxxxxxx
 
md"The scalars we chose:"

8.5 μs

a = -1.4     b = -2.1     c = 0

xxxxxxxxxx

42.7 μs

xxxxxxxxxx
 
begin
    p3 = scatter([0], [0], xlims = lims, ylims = lims, 
        legend = false, label = "origin",  aspect_ratio=:equal,
        title = "a•u, b•v, and c•w")
    
    if uon
        quiver!(quiver = ([a*u1],[a*u2]), [0], [0], color = [:blue],)
    end
    
    if von
        quiver!(quiver = ([b*v1],[b*v2]), [0], [0], color = [:orange],)
    end
    
    if won
        quiver!(quiver = ([c*w1],[c*w2]), [0], [0], color = [:green],)
    end 
    
    p4 = scatter([0], [0], xlims = lims, ylims = lims, 
        legend = false, label = "origin",  aspect_ratio=:equal,
        title = "a•u + b•v + c•w")
    
    quiver!(
        quiver = (
            [a*uon*u1 + b*von*v1 + c*won*w1],
            [a*uon*u2 + b*von*v2 + c*won*w2]
        ), 
        [0], [0], color = [:black],
    )
    
    quiver!(
        quiver = ([a*uon*u1],[a*uon*u2]), 
        [0], [0], 
        color = [:blue], 
        linestyle = :dashdot,
    )
    quiver!(
        quiver = ([b*von*v1],[b*von*v2]), 
        [a*uon*u1], [a*uon*u2], 
        color = [:orange], 
        linestyle = :dashdot,
    )
    quiver!(
        quiver = ([c*won*w1],[c*won*w2]), 
        [a*uon*u1 + b*von*v1], [a*uon*u2 + b*von*v2], 
        color = [:green], 
        linestyle = :dashdot,
    )
    
    plot(p3, p4, layout=2)
end

7 ms

Varying $a$ among positive values scales the length of the arrow that represents the vector $u$ .

If $a$ is negative then this scalaing reverses the direction of $u$ .

The same thing happens to the arrows representing $v$ and $w$ when we change $b$ and $c$ .

xxxxxxxxxx
 
md"Varying $a$ among positive values scales the length of the arrow that represents the vector $u$. 
​
If $a$ is negative then this scalaing reverses the direction of $u$. 
​
The same thing happens to the arrows representing $v$ and $w$ when we change $b$ and $c$."

24.1 μs

Vector equations

Let's now think of $a, b, c$ as the variables $x_{1} = a$ and $x_{2} = b$ and $x_{3} = c$ .

Choose a target vector $t = [\begin{array}{l} t_{1} \\ t_{2} \end{array}] \in R^{2}$ .

What does a solution to the equation $x_{1} u + x_{2} v + x_{3} w = t$ mean?

We know that the solutions $x = [\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}]$ to this equation are the same as the solutions to the linear system whose augmented matrix is $A = [\begin{array}{cccc} u_{1} & v_{1} & w_{1} & t_{1} \\ u_{2} & v_{2} & w_{2} & t_{2} \end{array}]$ .

xxxxxxxxxx
 
md"## Vector equations
​
Let's now think of $a,b,c$ as the variables $x_1=a$ and $x_2=b$ and $x_3=c$.
​
Choose a target vector $t =\left[\begin{array}{l} t_1 \\ t_2\end{array}\right] \in \mathbb{R}^2$.
​
What does a solution to the equation $x_1 u + x_2 v + x_3 w = t$ mean?
​
We know that the solutions $x = \left[\begin{array}{l} x_1 \\ x_2 \\ x_3\end{array}\right]$ to this equation are the same as the solutions to the linear system whose augmented matrix is 
$$A = \left[\begin{array}{ccc|c} u_1 & v_1 & w_1 & t_1 \\ u_2 & v_2 & w_2 & t_2 \end{array}\right]$$."
​

16 μs

Let's create a random target vector $t$ .

xxxxxxxxxx
 
md"Let's create a random target vector $t$."

7.9 μs

target

2×1 Array{Float64,2}:
  1.0
 15.4

xxxxxxxxxx
 
target = rand(-20:0.1:20, 2, 1)

13.6 μs

Here's our coefficient matrix.

xxxxxxxxxx
 
md"Here's our coefficient matrix."

8.3 μs

coefficient_matrix

2×3 Array{Float64,2}:
  6.7  -5.0   10.0
 -3.7  -5.1  -10.0

 
coefficient_matrix = [u1 v1 w1; u2 v2 w2]

52 μs

Here's our augmented matrix.

xxxxxxxxxx
 
md"Here's our augmented matrix."

9.5 μs

augmented_matrix

2×4 Array{Float64,2}:
  6.7  -5.0   10.0   1.0
 -3.7  -5.1  -10.0  15.4

 
augmented_matrix = hcat(coefficient_matrix, target)

8.5 μs

To solve the corresponding linear system we compute RREF(augmented_matrix).

x
 
md"To solve the corresponding linear system we compute `RREF(augmented_matrix)`."

9.9 μs

rref

2×4 Array{Float64,2}:
 1.0  0.0  1.917600151889121   -1.3651034744636414
 0.0  1.0  0.5695842035314223  -2.02923865578128

xxxxxxxxxx
 
rref = RREF(augmented_matrix)

8 μs

For typical values of $t$ , the pivot positions are (1,1) and (2,2).

This means $x_{1}$ and $x_{2}$ are basic variables and $x_{3}$ is a free variable.

To get a solution, choose any value for $x_{3}$ , then solve for $x_{1}$ , $x_{2}$ .

x
 
md"For typical values of $t$, the pivot positions are (1,1) and (2,2). 
​
This means $x_1$ and $x_2$ are basic variables and $x_3$ is a free variable.
​
To get a solution, choose any value for $x_3$, then solve for $x_1$, $x_2$."

9.8 μs

free_variable_value

x
 
free_variable_value = 0

1.5 μs

solution

-1.3651

-2.02924

xxxxxxxxxx
 
solution = (
    rref[1,4] - free_variable_value * rref[1,3], 
    rref[2,4] - free_variable_value * rref[2, 3],
    free_variable_value
)

18.4 μs

The meaninng of this solution is that if we set $x_{1}$ , $x_{2}$ , and $x_{3}$ to the corresponding values, then the linear combination $x_{1} u + x_{2} v + x_{3} w$ should match up with the target vector $t$ .

We can see this visually below.

xxxxxxxxxx
 
md"The meaninng of this solution is that if we set $x_1$, $x_2$, and $x_3$ to the corresponding values, then the linear combination $x_1 u + x_2 v + x_3 w$ should match up with the target vector $t$.
​
We can see this visually below."

11.1 μs

x1 = x2 = x3 =

xxxxxxxxxx
 
begin   
    md"""
    `x1` = $(a_slider) `x2` = $(b_slider) `x3` = $(c_slider)
    """
end

10.9 μs

x1 = -1.4     x2 = -2.1     x3 = 0

xxxxxxxxxx
 
Text(join([string("x1 = ", a), string("x2 = ", b), string("x3 = ", c)], "     "))

33.3 μs

xxxxxxxxxx
 
begin
    p5 = scatter([0 target[1]], [0 target[2]], xlims = (-20,20), ylims = (-20,20), 
        legend = true, label = ["origin" "target"],  
        # aspect_ratio=:equal,
        title = "x1•u + x2•v + x3•w = ?",
    )
    
    quiver!(
        quiver = ([a*uon*u1 + b*von*v1 + c*won*w1],[a*uon*u2 + b*von*v2 + c*won*w2]), 
        [0], [0], color = [:black],
    )
    
    quiver!(
        quiver = ([a*uon*u1],[a*uon*u2]), 
        [0], [0], 
        color = [:blue], 
        linestyle = :dashdot,
    )
    quiver!(
        quiver = ([b*von*v1],[b*von*v2]), 
        [a*uon*u1], [a*uon*u2], 
        color = [:orange], 
        linestyle = :dashdot,
    )
    quiver!(
        quiver = ([c*won*w1],[c*won*w2]), 
        [a*uon*u1 + b*von*v1], [a*uon*u2 + b*von*v2], 
        color = [:green], 
        linestyle = :dashdot,
    )
    
    plot(p5)
end
        # aspect_ratio=:equal,

5.3 ms

Vectors in $R^{3}$

Let's also try to visualize vectors in $R^{3}$ and think about their span.

This is a little more challenging to display.

x
 
md"## Vectors in $\mathbb{R}^3$
​
Let's also try to visualize vectors in $\mathbb{R}^3$ and think about their span. 
​
This is a little more challenging to display."

10.2 μs

Display parameters (changing these rotates our view):"

phi = 0

psi = 0

xxxxxxxxxx
 
begin
    p_slider = @bind _phi Slider(-360:12:360, default=0, show_value=true)
    q_slider = @bind _psi Slider(-360:12:360, default=0, show_value=true)
    
    md"""**Display parameters** (changing these rotates our view):"
    
    `phi` = $(p_slider) 
    
    `psi` = $(q_slider)
    """ 
end

111 μs

Create some vectors in $R^{3}$

v1 = w1 =

v2 = w2 =

v3 = w3 =

xxxxxxxxxx
 
begin
    a1_slider = @bind a1 Slider(0:1:20, default=4, show_value=false)
    a2_slider = @bind a2 Slider(0:1:20, default=8, show_value=false)
    a3_slider = @bind a3 Slider(0:1:20, default=0, show_value=false)
​
​
    b1_slider = @bind b1 Slider(0:1:20, default=10, show_value=false)
    b2_slider = @bind b2 Slider(0:1:20, default=6, show_value=false)
    b3_slider = @bind b3 Slider(0:1:20, default=0, show_value=false)
    
    
    md"""**Create some vectors in $$\mathbb{R}^3$$**
    
    `v1` = $(a1_slider) `w1` = $(b1_slider)
    
    `v2` = $(a2_slider) `w2` = $(b2_slider)
​
    `v3` = $(a3_slider) `w3` = $(b3_slider)
    
    """
end

139 μs

The vectors we created:

xxxxxxxxxx
 
md"The vectors we created:"

10.1 μs

v = [   9]     w = [   0]
    [   0]         [  20]
    [   5]         [   7]

xxxxxxxxxx

63.9 μs

show shadows in $x y$ -plane:

x
 
md"""show shadows in $xy$-plane: 
$(@bind nothide html"<input type=checkbox >")
"""

45.3 μs

draw span:

x
 
md"""draw span: 
$(@bind span html"<input type=checkbox >")
"""

65.2 μs

xxxxxxxxxx
 
begin
    phi = (_phi - 90) / 360 * pi
    psi = (_psi - 270) / 360 * pi
    A = [           sin(psi)           -cos(psi)         0
         cos(phi) * cos(psi) cos(phi) * sin(psi) -sin(phi)]
    md""
end

55.3 μs

x
 
begin
    nlims = (-20,20)
    
    p6 = scatter([0], [0], 
        legend = false,
        label = "origin", 
        aspect_ratio=:equal, 
        grid=nothing, 
        axis=nothing, 
        foreground_color_subplot="white",
    )
    
    for i=6:6:60
        quiver!(
            quiver = ([100 * A[1,2]], [100 * A[2,2]]),
            [i * A[1,1]], [i * A[2,1]], 
            color=[:lightgray],
            linestyle = :dashdot,
        )
        quiver!(
            quiver = ([100 * A[1,1]], [100 * A[2,1]]),
            [i * A[1,2]], [i * A[2,2]], 
            color=[:lightgray],
            linestyle = :dashdot,
        )
    end
    
    quiver!(
        quiver = (100 * A[1:1,1:3], 100 * A[2:2,1:3]),
        [0, 0, 0], 
        [0, 0, 0],
        xlims = nlims,
        ylims = nlims,
        color = [:grey],
    )
    
    if nothide
        x = A * [a1; a2; 0]
        quiver!(
            quiver = ([x[1]], [x[2]]), [0], [0], 
            color = [:blue],
            linestyle = :dashdot,
        )
        y = A * [b1; b2; 0]
        quiver!(
            quiver = ([y[1]], [y[2]]), [0], [0], 
            color = [:orange],
            linestyle = :dashdot,
        )
    end
    
    x = A * [a1; a2; a3]
    y = A * [b1; b2; b3]
    
    if norm(x) != 0
        nx = x / norm(x)
    else
        nx = x
    end
    
    if norm(y) != 0
        ny = y / norm(y)
    else
        ny = y 
    end
    
    if span
        for i=0:20
            for j=0:20
                scatter!(
                    [i * nx[1] + j * ny[1]], [i * nx[2]  + j * ny[2]], 
                    color=[:lightgray],
                )
            end
        end
    end
​
    quiver!(quiver = ([x[1]], [x[2]]), [0], [0], color = [:blue],)
    quiver!(quiver = ([y[1]], [y[2]]), [0], [0], color = [:orange],)
    
    p7 = scatter([0], [0], 
        legend = false,
        label = "origin", 
        aspect_ratio=:equal, 
        grid=nothing, 
        axis=nothing, 
        foreground_color_subplot="white"
    )
    
    for i=6:6:60
        quiver!(
            quiver = ([100 * A[1,2]], [100 * A[2,2]]),
            [i * A[1,1]], [i * A[2,1]], 
            color=[:lightgray],
            linestyle = :dashdot,
        )
        quiver!(
            quiver = ([100 * A[1,1]], [100 * A[2,1]]),
            [i * A[1,2]], [i * A[2,2]], 
            color=[:lightgray],
            linestyle = :dashdot,
        )
    end
    
    quiver!(
        quiver = (100 * A[1:1,1:3], 100 * A[2:2,1:3]),
        [0, 0, 0], 
        [0, 0, 0],
        xlims = nlims,
        ylims = nlims,
        color = [:grey],
    )
​
    if nothide
        x = A * [a1; a2; 0]
        y = A * [b1; b2; 0]
        quiver!(
            quiver = ([x[1]], [x[2]]), [0], [0], 
            color = [:grey], 
            linestyle = :dashdot,)
        quiver!(
            quiver = ([y[1]], [y[2]]), [x[1]], [x[2]], 
            color = [:grey], 
            linestyle = :dashdot,)
        quiver!(
            quiver = ([x[1] + y[1]], [x[2] + y[2]]), [0], [0], 
            color = [:grey])
    end
​
    x = A * [a1; a2; a3]
    y = A * [b1; b2; b3]
        
    quiver!(
        quiver = ([x[1]], [x[2]]), [0], [0], 
        color = [:blue], 
        linestyle = :dashdot)
    quiver!(
        quiver = ([y[1]], [y[2]]), [x[1]], [x[2]], 
        color = [:orange], 
        linestyle = :dashdot)
    quiver!(
        quiver = ([x[1] + y[1]], [x[2] + y[2]]), [0], [0], 
        color = [:black])
    
    plot(p6, p7, layout=2)
end
begin

34.2 ms

The left picture shows that vectors $v$ and $w$ .

The right picture shows their sum.

We can also display the span of the two vectors, denoted $R -span {v, w}$ .

For different inputs, we can get $R -span {v, w}$ to be either a plane, a line, or a single point.

xxxxxxxxxx
 
md"The left picture shows that vectors $v$ and $w$.
​
The right picture shows their sum.
​
We can also display the span of the two vectors, denoted $\mathbb{R}\text{-span}\{v,w\}$.
​
For different inputs, we can get $\mathbb{R}\text{-span}\{v,w\}$ to be either a plane, a line, or a single point."

11.9 μs

Some remarks

Of course, these pictures are not actually in $R^{3}$ !

On a flat surface like this screen, we can only draw vectors in $R^{2}$ .

Secretly, we are drawing many vectors $R^{2}$ to simulate the appearance of $R^{3}$ . If we erase those extra vectors and display the usual axes, the plots below are what we are actually showing.

x
 
md"## Some remarks
​
Of course, these pictures are not actually in $\mathbb{R}^3$!
​
On a flat surface like this screen, we can only draw vectors in $\mathbb{R}^2$.
​
Secretly, we are drawing many vectors $\mathbb{R}^2$ to simulate the appearance of $\mathbb{R}^3$. If we erase those extra vectors and display the usual axes, the plots below are what we are actually showing."

43.8 μs

x
 
begin
    p8 = scatter([0], [0], 
        legend = false,
        label = "origin", 
        aspect_ratio=:equal, 
        xlims = nlims,
        ylims = nlims,
    )
    quiver!(quiver = ([x[1]], [x[2]]), [0], [0], color = [:blue],)
    quiver!(quiver = ([y[1]], [y[2]]), [0], [0], color = [:orange],)
​
    p9 = scatter([0], [0], 
        legend = false,
        label = "origin", 
        aspect_ratio=:equal, 
        xlims = nlims,
        ylims = nlims,
    )       
    quiver!(
        quiver = ([x[1]], [x[2]]), [0], [0], 
        color = [:blue], 
        linestyle = :dashdot)
    quiver!(
        quiver = ([y[1]], [y[2]]), [x[1]], [x[2]], 
        color = [:orange], 
        linestyle = :dashdot)
    quiver!(
        quiver = ([x[1] + y[1]], [x[2] + y[2]]), [0], [0], 
        color = [:black])
    
    plot(p8, p9, layout=2)
end

5.8 ms

How are these vectors in $R^{2}$ computed from $v$ and $w$ , which we chose to be in $R^{3}$ ?

Some kind of transformation is used to change vectors in $R^{3}$ to vectors in $R^{2}$ that we can display. We'll talk about these kinds of transformations next lecture.

x
 
md"How are these vectors in $\mathbb{R}^2$ computed from $v$ and $w$, which we chose to be in $\mathbb{R}^3$? 
​
Some kind of transformation is used to change vectors in $\mathbb{R}^3$ to vectors in $\mathbb{R}^2$ that we can display. We'll talk about these kinds of transformations next lecture."