Find the angle $ \\theta $ between the following pairs of vectors.
", "advice": "Note that in this advice, the full calculator display is used in the calculation of each step; any rounding is purely for display clarity.
\nThe dot product of two vectors $\\boldsymbol{a}=\\pmatrix{a_1,a_2,a_3}$ and $\\boldsymbol{b}=\\pmatrix{b_1,b_2,b_3}$ is given by
\n\\[\\boldsymbol{a\\cdot b}=a_1b_1+a_2b_2+a_3b_3.\\]
\nIt is also given by
\n\\[\\boldsymbol{a\\cdot b}=ab\\cos(\\theta)\\]
\nwhere $a=\\lvert\\boldsymbol{a}\\rvert=\\sqrt{a_1^2+a_2^2+a_3^2}$ and $b=\\lvert\\boldsymbol{b}\\rvert=\\sqrt{b_1^2+b_2^2+b_3^2}$ are the lengths of the vectors $\\boldsymbol{a}$ and $\\boldsymbol{b}$.
\nEquating the two expressions gives
\n\\[a_1b_1+a_2b_2+a_3b_3=ab\\cos(\\theta)\\]
\nand so
\n\\[\\cos(\\theta)=\\frac{a_1b_1+a_2b_2+a_3b_3}{ab}.\\]
\nIn part a) therefore, we have
\n\\[\\cos(\\theta)=\\frac{\\simplify[std]{{a[0]*b[0]}+{a[1]*b[1]}+{a[2]*b[2]}}}{\\var{precround(lena,2)}\\times\\var{precround(lenb,2)}}=\\frac{\\var{dot(a,b)}}{\\var{precround(lena*lenb,2)}}=\\var{ans1} \\; \\text{to 2d.p.,}\\]
\nand in part b) we have
\n\\[\\cos(\\theta)=\\frac{\\simplify[std]{{c[0]*d[0]}+{c[1]*d[1]}+{c[2]*d[2]}+{c[3]*d[3]}}}{\\var{precround(lenc,2)}\\times\\var{precround(lend,2)}}=\\frac{\\var{dot(c,d)}}{\\var{precround(lenc*lend,2)}}=\\var{ans2} \\; \\text{to 2d.p.}\\]
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\n$\\boldsymbol{a} \\cdot \\boldsymbol{b}=$ [[2]]
\n$\\cos({\\theta})=$ [[0]] (Give your answer to 2d.p.)
\n$\\theta=$ [[1]](Give your answer, in radians, to 1d.p.)
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\n$\\boldsymbol{c} \\cdot \\boldsymbol{d}=$ [[2]]
\n$\\cos({\\theta})=$ [[0]] (Give your answer to 2d.p.)
\n$\\theta=$ [[1]] (Give your answer, in radians, to 1d.p.)
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