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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 $t\\in\\mathbb{R}$.

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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])}. \$

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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 $\\var{va}$:

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

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

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

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(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

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$t=$[[0]]

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(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

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$t=$[[0]]

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

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