// Numbas version: finer_feedback_settings {"name": "Vectors: Vector Product 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Vectors: Vector Product 2", "tags": [], "metadata": {"description": "

Find a perpendicular vector to a pair of vectors.

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Find a vector perpendicular to both $\\mathbf p$ and $\\mathbf q$, where \\[\\mathbf p = \\var{p},\\,\\mathbf q = \\var{q}.\\]

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If we want to find a vector perpendicular to both $\\mathbf p$ and $\\mathbf q$, this is done by the taking the vector product between $\\mathbf p$ and $\\mathbf q$.

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(Note: This can be either $\\simplify{cross(mathbf:p,mathbf:q)}$ or $\\simplify{cross(mathbf:q,mathbf:p)}$, both will produce a vector perpendicular to $\\mathbf p$ and $\\mathbf q$, they will just be in opposite directions.)

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Taking the vector product between $\\mathbf p$ and $\\mathbf q$:

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\\[ \\begin{split}\\simplify{cross(mathbf:p,mathbf:q)} &\\,= \\begin{vmatrix} \\mathbf i &\\mathbf j &\\mathbf k \\\\ \\var{p[0]} &\\var{p[1]} &\\var{p[2]} \\\\ \\var{q[0]} &\\var{q[1]} &\\var{q[2]} \\end{vmatrix}\\\\ \\\\&\\,= (\\simplify[!collectNumbers]{{p[1]}*{q[2]}-{p[2]}*{q[1]}}) \\mathbf i - (\\simplify[!collectNumbers]{{p[0]}*{q[2]}-{p[2]}*{q[0]}}) \\mathbf j + (\\simplify[!collectNumbers]{{p[0]}*{q[1]}-{p[1]}*{q[0]}}) \\mathbf k \\\\\\\\ &\\,= \\simplify[all,!noLeadingMinus]{{c1[0]}*mathbf:i+ {c1[1]}*mathbf:j + {c1[2]}*mathbf:k}\\end{split} \\]

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Therefore,

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\\[ \\simplify{cross(mathbf:p,mathbf:q) = {c1}} .\\]

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(NOTE: The vector calculated here is one solution, but any scalar multiple of $\\simplify{cross(mathbf:p,mathbf:q)}$ will also be perpendicular to both $\\mathbf p$ and $\\mathbf q$.)

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