LinAlg4.2.13

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Find a unit vector perpendiculars to the plane ABC where A=(3,-1,2),B(1,-1,-3),C(4,-3,1)\,

Given A=(3,-1,2),B(1,-1,-3),C(4,-3,1)\,

{\bar  {AB}}=-2{\bar  {i}}-5{\bar  {k}},{\bar  {AC}}={\bar  {i}}-2{\bar  {j}}-{\bar  {k}}\,

Now {\bar  {AB}}\times {\bar  {AC}}={\begin{vmatrix}{\bar  {i}}&{\bar  {j}}&{\bar  {k}}\\-2&0&-5\\1&-2&-1\end{vmatrix}}\, which is equal to

{\bar  {i}}(0-10)-{\bar  {j}}(2+5)+{\bar  {k}}(4-0)=-10{\bar  {i}}-7{\bar  {j}}+4{\bar  {k}}\,

Therefore,a unit vector perpendicular to the plane ABC =\pm {\frac  {{\bar  {AB}}\times {\bar  {AC}}}{|AB\times AC|}}=\pm {\frac  {-10{\bar  {i}}-7{\bar  {j}}+4{\bar  {k}}}{{\sqrt  {(-10)^{2}+(-7)^{2}+4^{2}}}=\pm {\frac  {1}{\sqrt  {165}}}}}(-10{\bar  {i}}-7{\bar  {j}}+4{\bar  {k}})\,


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