VC1.8

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Solution If a=\sin \theta i+\cos \theta j+\theta k,b=\cos \theta i-\sin \theta j-3k,c=2i+3j-k\,.Find {\frac  {d}{d\theta }}[a\times (b\times c)]\, at \theta =0\,

b\times c={\begin{vmatrix}i&j&k\\\cos \theta &-\sin \theta &-3\\2&3&-1\end{vmatrix}}\,

b\times c=(\sin \theta +9)i+(\cos \theta -6)j+(3\cos \theta +2\sin \theta )k\,

Therefore, a\times (b\times c)={\begin{vmatrix}i&j&k\\\sin \theta &\cos \theta &\theta \\\sin \theta +9&\cos \theta -9&3\cos \theta +2\sin \theta \end{vmatrix}}\,

or a\times (b\times c)=(3\cos ^{2}\theta +2\sin \theta \cos \theta -\theta \cos \theta +6\theta )i-3(\sin \theta \cos theta+2\sin ^{2}\theta -\theta \sin \theta -9\theta )j+(\sin \theta \cos \theta -6\sin \theta -\sin \theta \cos \theta -9\cos \theta )k\,

Therefore,{\frac  {d}{d\theta }}(a\times (b\times c)={\frac  {d}{d\theta }}(3\cos ^{2}\theta +2\sin \theta \cos \theta -\theta \cos \theta +6\theta )i-{\frac  {d}{d\theta }}[{\frac  {3}{2}}\sin 2\theta +2\sin ^{2}\theta -\theta \sin \theta -9\theta ]j-{\frac  {d}{d\theta }}(6\sin \theta +9\cos \theta )k\,

[-6\cos \theta \sin \theta +2\cos 2\theta -(\cos \theta -\theta \sin \theta )+6]i-[3\cos 2\theta +4\sin \theta \cos \theta -(\sin \theta +\theta \cos \theta )-9]j-(6\cos \theta -9\sin \theta )k\,

When \theta =0\,, {\frac  {d}{d\theta }}(a\times (b\times c)=(2-1+6)i-(3-9)j-6k=7i+6j-6k\,

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