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Let $\\var{vi},\\var{vj},\\var{vk}$ form an orthonormal basis. Consider
\\[
    \\var{va}=\\var{disp(a)}, \\qquad \\var{vb}= \\var{disp(b)} , \\qquad \\var{vc}=\\var{{ccoef}*a[0]}\\var{vi} -\\var{lcoef} t \\var{vj}-\\var{abs({ccoef}*a[2])}\\var{vk},
\\]
where $t\\in\\mathbb{R}$.

", "advice": "

a) Let $\\var{vah}$ be the unit vector corresponding to $\\var{va}$, i.e.
\\[
\\var{vah}=\\frac{1}{|\\var{va}|}\\var{va}.
\\]
Since $\\var{vd}$ is opposite to $\\var{va}$, it is also opposite to $\\var{vah}$. Since $|\\var{vd}|=\\simplify{{acoef}*{alen}}$, we conclude that
\\[
\\var{vd} = -\\simplify{{acoef}*{alen}}\\var{vah}=-\\frac{\\simplify{{acoef}*{alen}}}{|\\var{va}|}\\var{va}.
\\]
We have:
\\[
|\\var{va}|=\\sqrt{\\var{abs(a[0])}^2+\\var{abs(a[1])}^2+\\var{abs(a[2])}^2}=\\var{alen}.
\\]
Therefore,
\\[
\\var{vd} =-\\frac{\\simplify{{acoef}*{alen}}}{\\var{alen}}\\bigl(\\var{disp(a)}\\bigr)=\\var{disp(apar)}.
\\]

\n

\n

b) We know that
\\[
\\cos \\theta =\\frac{\\var{va}\\cdot\\var{vb}}{|\\var{va}|\\, |\\var{vb}|}.
\\]
We have found that $|\\var{va}|=\\var{alen}$. Next,
\\[
|\\var{vb}|=\\sqrt{\\var{abs(b[0])}^2+\\var{abs(b[1])}^2+\\var{abs(b[2])}^2}=\\var{blen}.
\\]
We have also that
\\[
\\var{va}\\cdot\\var{vb}=\\bigl(\\var{disp(a)}\\bigr)\\cdot\\bigl(\\var{disp(b)}\\bigr)=\\var{dot(a,b)}
\\]
Therefore,
\\[
\\cos \\theta=\\frac{\\var{dot(a,b)}}{\\var{alen} \\cdot\\var{blen}}=\\var{sfrac(costheta[0],costheta[1])}.
\\]

\n

c) Since vectors $\\var{va}=\\var{disp(a)}$ and $\\var{vc}=\\var{{ccoef}*a[0]}\\var{vi} -\\var{lcoef} t \\var{vj}+\\var{abs({ccoef}*a[2])}\\var{vk}$ are parallel, there exists a number $r\\in\\mathrm{R}$ such that
\\[
\\var{vc}=r\\var{va}.
\\]
Equating coefficients before $\\var{vi},\\var{vj},\\var{vk}$, we get
\\[
\\begin{cases}
\\displaystyle \\var{{ccoef}*a[0]} = \\simplify{{a[0]}*r},\\\\[2mm]
\\displaystyle -\\var{lcoef} t =\\simplify{{a[1]}*r},\\\\[2mm]
\\displaystyle \\var{abs({ccoef}*a[2])} =\\simplify{{a[2]}*r},
\\end{cases}
\\qquad\\qquad
\\begin{cases}
\\displaystyle r=\\var{ccoef},\\\\[3mm]
\\displaystyle t = -\\frac{ \\simplify{{a[1]}*r}}{\\var{lcoef}},
\\end{cases}
\\qquad\\qquad t=\\var{sfrac(l[0],l[1])}.
\\]

\n

d) Vectors $\\var{vb}= \\var{disp(b)}$ and $ \\var{vc}=\\var{{ccoef}*a[0]}\\var{vi} -\\var{lcoef} t \\var{vj}+\\var{abs({ccoef}*a[2])}\\var{vk}$ are orthogonal if and only if
\\[
\\var{vb}\\cdot\\var{vc}=0.
\\]
Since
\\[
\\var{vb}\\cdot\\var{vc}=\\bigl( \\var{disp(b)}\\bigr) \\cdot\\bigl( \\var{{ccoef}*a[0]}\\var{vi} -\\var{lcoef} t \\var{vj}+\\var{abs({ccoef}*a[2])}\\var{vk}\\bigr) =\\simplify{{dop} - { dop1} * t},
\\]
we require
\\[
\\simplify{{dop} - { dop1} * t}=0, \\qquad\\qquad  t=\\var{sfrac(lort[0],lort[1])}.
\\]

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Find the vector $\\var{vd}$ whose magnitude is $\\simplify{{acoef}*{alen}}$ and which is opposite to $\\var{va}$:

\n

$ \\var{vd}=$[[0]]$\\var{vi}+$[[1]]$\\var{vj}+$[[2]]$\\var{vk}$.

\n

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Let $\\theta$ be the angle between vectors $\\var{va}$ and $\\var{vb}$. Find

\n

$\\cos\\theta = $[[0]]

\n

(To type a fraction use the form 'p/q' or '-p/q' where 'p' and 'q' are natural numbers.)

\n

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Let $t\\in\\mathbb{R}$ be such that vectors $\\var{va}$ and $\\var{vc}$ are parallel. Find

\n

$t=$[[0]]

\n

(To type a fraction use the form 'p/q' or '-p/q' where 'p' and 'q' are natural numbers.)

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Let $t\\in\\mathbb{R}$ be such that vectors $\\var{vb}$ and $\\var{vc}$ are orthogonal. Find

\n

$t=$[[0]]

\n

(To type a fraction use the form 'p/q' or '-p/q' where 'p' and 'q' are natural numbers.)

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