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# Answer to Question #223884 in Chemical Engineering for Lokika

Question #223884

Find the angle between the surfaces x^{2}+y^{2}+z^{2}=9 and z+3=x^{2}+y^{2} at the point (-2,1,2)

1
2021-09-02T02:32:47-0400

"Suppose \\space \\phi_1=x^{2}+y^{2}+z^{2}-9 \\\\\n\\phi_2= x^{2}+y^{2}-z-3\\\\\nSo, \\Delta \\phi_1 = 2x \\hat{i}+2y \\hat{j}+2z \\hat{k}\\\\\n\\Delta \\phi_2 = 2x \\hat{i}+2y \\hat{j}\\\\\n\n(\\Delta \\phi_1)_{(-2,1,2)} = -4 \\hat{i}+2 \\hat{j}+4 \\hat{k}\\\\\n(\\Delta \\phi_2)_{(-2,1,2)} = -4\\hat{i}+2 \\hat{j}\\\\\n\n\\mathrm{Computing\\:the\\:angle\\:between\\:the\\:vectors}:\\quad \\cos \\left(\\theta \\right)\\:=\\frac{\\vec{a\\:}\\cdot \\vec{b\\:}}{\\left|\\vec{a\\:}\\right|\\cdot \\left|\\vec{b\\:}\\right|}\\\\\n\\theta = \\arccos [\\frac{\\Delta \\phi_1*\\Delta \\phi_2}{|\\Delta \\phi_1||\\Delta \\phi_2|}]\\\\\n\u03b8=\\arccos \\left(\\cos \\left(\u03b8\\right)\\right)=\\arccos \\left(\\frac{\\sqrt{5}}{3}\\right)\\\\\n\u03b8=41.81031^{\\circ \\:}"

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