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The picture below is made of two cubes.

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For which point $X$ is the following statement true $\\overrightarrow{AE}+\\overrightarrow{ED} =\\overrightarrow{AX}$?

Your answer: $X=$[[1]].

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Calculate the following.

$\\lVert \\boldsymbol{v} \\rVert=$ [[0]]

$\\lVert \\boldsymbol{w} \\rVert = $ [[1]]

$\\lVert \\boldsymbol{v}+\\boldsymbol{w} \\rVert = $ [[2]]

Let $\\boldsymbol{z}=\\boldsymbol{v}+\\boldsymbol{w}$.

Find the unit vector $\\boldsymbol{u_z}$ in the direction of $\\boldsymbol{z}$. Write $\\boldsymbol{u_z}$ as a row vector.

$\\boldsymbol{u_z}= \\big($ [[0]], [[1]], [[2]] $\\big)$

You must enter your answers exactly, using the function sqrt(x) if necessary.

Find

$\\var{a4}\\boldsymbol{v} = $ [[0]]

$\\var{b4}\\boldsymbol{w} = $ [[1]]

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Find the unit vector $\\boldsymbol{u_v}$ parallel to $\\boldsymbol{v}$, and the unit vector $-\\boldsymbol{u_w}$ anti-parallel to $\\boldsymbol{w}$. Write both vectors as row vectors.

$\\boldsymbol{u_v} = \\big($ [[0]], [[1]], [[2]] $\\big)$

$-\\boldsymbol{u_w} = \\big($ [[3]], [[4]], [[5]] $\\big)$

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Elementary operations on vectors; sum, modulus, unit vector, scalar multiple.

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a)

\\[\\boldsymbol{v}+\\boldsymbol{w} = \\var{vector(a,b,g)} + \\var{vector(c,d,f)} = \\var{vector(a+c,b+d,g+f)} \\]

b)

In general for a vector $\\boldsymbol{x}= \\begin{pmatrix}x_1 \\\\ x_2 \\\\ x_3 \\end{pmatrix}$, we have $\\lVert \\boldsymbol{x} \\rVert = \\sqrt{x_1^2+x_2^2+x_3^2}$.

Hence:

\\begin{align}
\\lVert \\boldsymbol{v} \\rVert &= \\sqrt{\\var{a^2}+\\var{b^2}+\\var{g^2}} = \\simplify[all]{ sqrt({a^2+b^2+g^2})} \\\\
\\lVert \\boldsymbol{w} \\rVert &= \\sqrt{\\var{c^2}+\\var{d^2}+\\var{f^2}} = \\simplify[all]{ sqrt({c^2+d^2+f^2})} \\\\
\\lVert \\boldsymbol{v+w} \\rVert &= \\sqrt{\\var{(a+c)^2}+\\var{(b+d)^2}+\\var{(g+f)^2}} = \\simplify[all]{ sqrt({(a+c)^2+(b+d)^2+(f+g)^2})}
\\end{align}

c)

Given a vector $\\boldsymbol{x}= \\begin{pmatrix} x_1 \\\\ x_2 \\\\ x_3 \\end{pmatrix}$, the unit vector parallel to $\\boldsymbol{x}$ is given by:

\\[ \\boldsymbol{u_x} = \\frac{1}{\\lVert \\boldsymbol{x} \\rVert} \\begin{pmatrix} x_1 \\\\ x_2 \\\\ x_3 \\end{pmatrix} = \\begin{pmatrix} \\frac{x_1}{\\lVert \\boldsymbol{x} \\rVert} \\\\ \\frac{x_2}{\\lVert \\boldsymbol{x} \\rVert} \\\\ \\frac{x_3}{\\lVert \\boldsymbol{x} \\rVert} \\end{pmatrix} \\]

For this example we have $\\lVert \\boldsymbol{v+w} \\rVert =\\simplify[std]{sqrt({(a+c)^2+(b+d)^2+(f+g)^2})}$, hence:

\\begin{align}
&&\\boldsymbol{z} = \\boldsymbol{v} + \\boldsymbol{w} &= \\begin{pmatrix} \\var{a+c} \\\\ \\var{b+d} \\\\ \\var{g+f} \\end{pmatrix} \\\\
\\implies && \\boldsymbol{u_z} &= \\frac{1}{\\sqrt{\\var{ssquares}}} \\begin{pmatrix} \\var{a+c} \\\\ \\var{b+d} \\\\ \\var{g+f} \\end{pmatrix} \\\\[1em]
&& &= \\begin{pmatrix} \\simplify[std]{{a+c}/sqrt({ssquares})} \\\\ \\simplify[std]{{b+d}/sqrt({ssquares})} \\\\ \\simplify[std]{{g+f}/sqrt({ssquares})} \\end{pmatrix}
\\end{align}

d)

\\begin{align}
\\var{a4}\\boldsymbol{v} &= \\simplify{vector({a4}*{a}, {a4}*{b}, {a4}*{g})} \\\\[1em]
&= \\var{a4*vector(a,b,g)}
\\end{align}

\\begin{align}
\\var{-b4}\\boldsymbol{v} &= \\simplify{vector({-b4}*{c}, {-b4}*{d}, {-b4}*{f})} \\\\[1em]
&= \\var{-b4*vector(c,d,f)}
\\end{align}

e)

Using the information above, the unit vector parallel to $\\boldsymbol{v}$ is:

\\[ \\boldsymbol{u_v} = \\begin{pmatrix} \\simplify[std]{{a}/sqrt({ssquaresA})} \\\\ \\simplify[std]{{b}/sqrt({ssquaresA})} \\\\ \\simplify[std]{{g}/sqrt({ssquaresA})} \\end{pmatrix} \\]

and the unit vector anti-parallel to $\\boldsymbol{w}$ is:

\\[ -\\boldsymbol{u_w} = \\begin{pmatrix} \\simplify[std]{{-c}/sqrt({ssquaresB})} \\\\ \\simplify[std]{{-d}/sqrt({ssquaresB})} \\\\ \\simplify[std]{{-f}/sqrt({ssquaresB})} \\end{pmatrix} \\]

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