Given the vectors 𝐴⃗=5𝑖̂+3𝑗̂+𝑘̂ and 𝐵⃗⃗=−3𝑖̂+2𝑗̂−4𝑘̂, find the following:
• 𝐴⃗ ∙ 𝐵⃗
• Angle between vector A and B
• 𝐴⃗ 𝑥 𝐵⃗
(a)
"{\\vec A}\\cdot {\\vec B}=(5\ud835\udc56\u0302+3\ud835\udc57\u0302+\ud835\udc58\u0302)\\cdot(\u22123\ud835\udc56\u0302+2\ud835\udc57\u0302\u22124\ud835\udc58\u0302)\\\\\n=-15+6-4=-13"(b)
"\\theta=\\cos^{-1}\\frac{-13}{\\sqrt{5^2+3^2+1^2}\\sqrt{(-3)^2+2^2+(-4)^2}}\\\\\n=114^\\circ"(c)
"{\\vec A}\\times {\\vec B}=(5\ud835\udc56\u0302+3\ud835\udc57\u0302+\ud835\udc58\u0302)\\times(\u22123\ud835\udc56\u0302+2\ud835\udc57\u0302\u22124\ud835\udc58\u0302)\\\\\n=10\ud835\udc58\u0302+20\ud835\udc57\u0302+9\ud835\udc58\u0302-12\ud835\udc56\u0302-3\ud835\udc57\u0302-2\ud835\udc56\u0302\\\\\n=-14\ud835\udc56\u0302+17\ud835\udc57\u0302+19\ud835\udc58\u0302"
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