Question: Demonstrate the property of BAC-CAB: ((mathbf{a} times mathbf{b}) times mathbf{c}=mathbf{b}(mathbf{a} cdot mathbf{c})-mathbf{c}(mathbf{a} cdot mathbf{b})), verifying it by direct development after arranging the vector (mathbf{c}) along
Demonstrate the property of BAC-CAB: \((\mathbf{a} \times \mathbf{b}) \times \mathbf{c}=\mathbf{b}(\mathbf{a} \cdot \mathbf{c})-\mathbf{c}(\mathbf{a} \cdot \mathbf{b})\), verifying it by direct development after arranging the vector \(\mathbf{c}\) along the \(x\) axis and the vector \(\mathbf{b}\) in the plane \((x, y)\).
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To demonstrate the property of the vector triple product mathbfa times mathbfb times mathbfc mathbfbmathbfa cdot mathbfc mathbfcmathbfa cdot mathbfb we can perform the direct development of both sides of the equation Given that mathbfc is aligned along the x axis and mathbfb lies in the xy plane we can express the vectors as mathbfa ax mathbfi ay mathbfj az mathbfk mathbfb bx mathbfi by mathbfj mathbfc cx mathbfi Now lets compute the lefthand side LHS of the equation mathbfa times mathbfb times mathbfc First lets find mathbfa times mathbfb mathbfa times mathbfb beginvmatrix mathbfi mathbfj mathbfk ax ay az bx by 0 endvmatrix ay bz az by mathbfi ax bz az bx mathbfj ax by ay bx mathbfk Now take the cross product of mathbfa times mathbfb and mathbfc mathbfa times mathbfb times mathbfc beginvmatrix mathbfi mathbfj mathbfk ay bz az by ax bz az bx ax by ay bx cx 0 0 endvmatrix ax by ay bx cx mathbfj Now lets compute the righthand side RHS of the equation mathbfb mathbfa cdot mathbfc mathbfc mathbfa cdot mathbfb First find mathbfa cdot mathbfc mathbfa cdot mathbfc ax cx ay cdot 0 az cdot 0 ax cx Now compute mathbfb mathbfa cdot mathbfc mathbfb ... View full answer
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