VC2.10

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a\cdot b\times c={\begin{vmatrix}t&-3&2t\\1&-2&2\\3&t&-1\end{vmatrix}}\,

Performing C_{2}\longrightarrow C_{2}+2C_{1},C_{3}=C_{3}-2C_{1}\, we have

{\begin{vmatrix}t&2t-3&0\\1&0&0\\3&6+t&-7\end{vmatrix}}\,

-7[0-(2t-3)]=7(2t-3)\,

Hence \int _{1}^{2}a\cdot (b\times c)\,dt=7\int _{1}^{2}(2t-3)\,dt=7[t^{2}-3t]_{1}^{2}\,

=7[(4-6)-(1-3)]=0\,

Now b\times c={\begin{vmatrix}i&j&k\\1&-2&2\\3&t&1\end{vmatrix}}\,

=-(2+2t)i+5j+(t+6)k\,

Now a\times (b\times c)={\begin{vmatrix}i&j&k\\t&-3&2t\\-(2+2t)&5&t+6\end{vmatrix}}\,

=-[(13t+18)i+5(t^{2}+2t)j+(t+6)k]\,

Therefore,\int _{1}^{2}[a\cdot (b\times c)+a\times (b\times c)]\,dt=-\int _{1}^{2}[(13t+18)i+5(t^{2}+2t)j+(t+6)k]\,dt\,

=-i\int _{1}^{2}(13t+18)\,dt-5j\int _{1}^{2}(t^{2}+2t)\,dt-k\int _{1}^{2}(t+6)\,dt\,

=-i[{\frac  {13}{2}}t^{2}+18t]_{1}^{2}-5j[{\frac  {1}{3}}t^{3}+t^{2}]_{1}^{2}-k[{\frac  {1}{2}}t^{2}+6t]_{1}^{2}\,

=-{\frac  {55}{2}}i-{\frac  {80}{3}}j-{\frac  {15}{2}}k\, on simplification.

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