LinAlg4.2.22

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Find the vector equation of the line passing through three non-collinear points -2{\bar  {i}}+6{\bar  {j}}-6{\bar  {k}},-3{\bar  {i}}+10{\bar  {j}}-9{\bar  {k}},-5{\bar  {i}}-6{\bar  {k}}\,.Also find its cartesian equation.

Let P={\bar  {r}}\, be any point in the plane ABC.

Now the vector equation of the plane is ({\bar  {r}}-{\bar  {a}})\cdot ({\bar  {b}}-{\bar  {a}})\times ({\bar  {c}}-{\bar  {a}})=0\,

Now {\bar  {b}}-{\bar  {a}}=-{\bar  {i}}+4{\bar  {j}}-3{\bar  {k}},{\bar  {c}}-{\bar  {a}}=-3{\bar  {i}}-6{\bar  {j}}\,

Therefore ({\bar  {b}}-{\bar  {a}})\times ({\bar  {c}}-{\bar  {a}})={\begin{vmatrix}{\bar  {i}}&{\bar  {j}}&{\bar  {k}}\\-1&4&-3\\-3&-6&0\end{vmatrix}}\, which is equal to

{\bar  {i}}(0-18)-{\bar  {j}}(0-9)+{\bar  {k}}(6+12)=-18{\bar  {i}}+9{\bar  {j}}+18{\bar  {k}}\,

Hence the vector equation is [{\bar  {r}}-(-2{\bar  {i}}+6{\bar  {j}}-6{\bar  {k}})]\cdot (-18{\bar  {i}}+9{\bar  {j}}+18{\bar  {k}})=0\,

{\bar  {r}}\cdot (-18{\bar  {i}}+9{\bar  {j}}+18{\bar  {k}})=(-2{\bar  {i}}+6{\bar  {j}}-6{\bar  {k}})\cdot (-18{\bar  {i}}+9{\bar  {j}}+18{\bar  {k}})\,

{\bar  {r}}\cdot (-18{\bar  {i}}+9{\bar  {j}}+18{\bar  {k}})=2\,

Cartesian equation is (x{\bar  {i}}+y{\bar  {j}}+z{\bar  {k}})\cdot (2{\bar  {i}}-{\bar  {j}}-2{\bar  {k}})=2\, which is equal to 2x-y-2z=2\,


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