Given the vectors $\\mathbf a$, $\\mathbf b$ and $\\mathbf c$, calculate $(\\mathbf a \\times \\mathbf b) \\times \\mathbf c$ and $\\mathbf a \\times (\\mathbf b \\times \\mathbf c)$.
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\n\\[ \\mathbf a = \\var{a},\\quad \\mathbf b = \\var{b}, \\quad \\mathbf c = \\var{c}; \\]
", "advice": "To find $(\\simplify{cross(mathbf:a,mathbf:b)})\\times \\mathbf c$, we want to first find the vector product of $\\mathbf a$ and $\\mathbf b$, then find the vector product between that vector and $\\mathbf c$.
\n\\[ \\begin{split}\\simplify{cross(mathbf:a,mathbf:b)} &\\,= \\begin{vmatrix} \\mathbf i &\\mathbf j &\\mathbf k \\\\ \\var{a[0]} &\\var{a[1]} &\\var{a[2]} \\\\ \\var{b[0]} &\\var{b[1]} &\\var{b[2]} \\end{vmatrix}\\\\ \\\\&\\,= (\\simplify[!collectNumbers]{{a[1]}*{b[2]}-{a[2]}*{b[1]}}) \\mathbf i - (\\simplify[!collectNumbers]{{a[0]}*{b[2]}-{a[2]}*{b[0]}}) \\mathbf j + (\\simplify[!collectNumbers]{{a[0]}*{b[1]}-{a[1]}*{b[0]}}) \\mathbf k \\\\\\\\ &\\,= \\simplify[all,!noLeadingMinus]{{d1[0]}*mathbf:i+ {d1[1]}*mathbf:j + {d1[2]}*mathbf:k}.\\end{split} \\]
\nSo, $(\\simplify{cross(mathbf:a,mathbf:b)})\\times \\mathbf c$ is equal to the vector product of $\\var{d1}$ and $\\var{c}$:
\n\\[ \\begin{split}(\\simplify{cross(mathbf:a,mathbf:b)})\\times \\mathbf c &\\,=\\begin{vmatrix} \\mathbf i &\\mathbf j &\\mathbf k \\\\ \\var{d1[0]} &\\var{d1[1]} &\\var{d1[2]} \\\\ \\var{c[0]} &\\var{c[1]} &\\var{c[2]} \\end{vmatrix}\\\\ \\\\&\\,= (\\simplify[!collectNumbers]{{d1[1]}*{c[2]}-{d1[2]}*{c[1]}}) \\mathbf i - (\\simplify[!collectNumbers]{{d1[0]}*{c[2]}-{d1[2]}*{c[0]}}) \\mathbf j + (\\simplify[!collectNumbers]{{d1[0]}*{c[1]}-{d1[1]}*{c[0]}}) \\mathbf k \\\\\\\\ &\\,= \\simplify[all,!noLeadingMinus]{{sol1[0]}*mathbf:i+ {sol1[1]}*mathbf:j + {sol1[2]}*mathbf:k}.\\end{split} \\]
\n\nTherefore,
\n\\[ (\\simplify{cross(mathbf:a,mathbf:b)}) \\times \\mathbf c = \\simplify{{sol1}} .\\]
\n\\[ \\\\ \\]
\n\nSimilarly, to find $\\mathbf a \\times (\\simplify{cross(mathbf:b,mathbf:c)})$, we want to first find the vector product of $\\mathbf b$ and $\\mathbf c$, then find the vector product between $\\mathbf a$ and that vector.
\n\\[ \\begin{split}\\simplify{cross(mathbf:b,mathbf:c)} &\\,= \\begin{vmatrix} \\mathbf i &\\mathbf j &\\mathbf k \\\\ \\var{b[0]} &\\var{b[1]} &\\var{b[2]} \\\\ \\var{c[0]} &\\var{c[1]} &\\var{c[2]} \\end{vmatrix}\\\\ \\\\&\\,= (\\simplify[!collectNumbers]{{b[1]}*{c[2]}-{b[2]}*{c[1]}}) \\mathbf i - (\\simplify[!collectNumbers]{{b[0]}*{c[2]}-{b[2]}*{c[0]}}) \\mathbf j + (\\simplify[!collectNumbers]{{b[0]}*{c[1]}-{b[1]}*{c[0]}}) \\mathbf k \\\\\\\\ &\\,= \\simplify[all,!noLeadingMinus]{{d2[0]}*mathbf:i+ {d2[1]}*mathbf:j + {d2[2]}*mathbf:k}.\\end{split} \\]
\nSo, $\\mathbf a \\times (\\simplify{cross(mathbf:b,mathbf:c)})$ is equal to the vector product of $\\var{a}$ and $\\var{d2}$:
\n\\[ \\begin{split}(\\simplify{cross(mathbf:a,mathbf:b)})\\times \\mathbf c &\\,=\\begin{vmatrix} \\mathbf i &\\mathbf j &\\mathbf k \\\\ \\var{a[0]} &\\var{a[1]} &\\var{a[2]} \\\\ \\var{d2[0]} &\\var{d2[1]} &\\var{d2[2]} \\end{vmatrix}\\\\ \\\\&\\,= (\\simplify[!collectNumbers]{{a[1]}*{d2[2]}-{a[2]}*{d2[1]}}) \\mathbf i - (\\simplify[!collectNumbers]{{a[0]}*{d2[2]}-{a[2]}*{d2[0]}}) \\mathbf j + (\\simplify[!collectNumbers]{{a[0]}*{d2[1]}-{a[1]}*{d2[0]}}) \\mathbf k \\\\\\\\ &\\,= \\simplify[all,!noLeadingMinus]{{sol2[0]}*mathbf:i+ {sol2[1]}*mathbf:j + {sol2[2]}*mathbf:k}.\\end{split} \\]
\n\nTherefore,
\n\\[\\mathbf a \\times ( \\simplify{cross(mathbf:b,mathbf:c)}) = \\simplify{{sol2}} .\\]
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\n$(\\simplify{cross(mathbf:a,mathbf:b)})\\times \\mathbf c = $[[0]]
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\n$\\mathbf a \\times (\\simplify{cross(mathbf:b,mathbf:c)}) = $[[0]]
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