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Show that the points A,B,C,D with position vectors 6{\bar  {i}}-7{\bar  {j}},16{\bar  {i}}-19{\bar  {j}}-4{\bar  {k}},3{\bar  {j}}-6{\bar  {k}},2{\bar  {i}}+5{\bar  {j}}+10{\bar  {k}}\, are not coplanar.

We have {\bar  {AB}}=(16{\bar  {i}}-19{\bar  {j}}-4{\bar  {k}})-(6{\bar  {i}}-7{\bar  {j}})=10{\bar  {i}}-12{\bar  {j}}-4{\bar  {k}}\,

{\bar  {AC}}=(3{\bar  {j}}-6{\bar  {k}})-(6{\bar  {i}}-7{\bar  {j}})=-6{\bar  {i}}+10{\bar  {j}}-6{\bar  {k}}\,

{\bar  {AD}}=(2{\bar  {i}}+5{\bar  {j}}+10{\bar  {k}})-(6{\bar  {i}}-7{\bar  {j}})=-4{\bar  {i}}+12{\bar  {j}}-10{\bar  {k}}\,

Suppose the above three are coplanar. Then there exist two scalars x,y\, such that {\bar  {AB}}=x{\bar  {AC}}+y{\bar  {AD}}\,

-12{\bar  {j}}-4{\bar  {k}}=x(-6{\bar  {i}}+10{\bar  {j}}-6{\bar  {k}})+y(-4{\bar  {i}}+12{\bar  {j}}-10{\bar  {k}})=(-6x-4y){\bar  {i}}+(10x+12y){\bar  {j}}+(10y-6x){\bar  {k}}\,

Since i,j,k are non-coplanar -6x-4y=10,10x+12y=-12,10y-6x=-4\,

Solving the first two we get x=-{\frac  {9}{4}},y={\frac  {7}{8}}\, which donot satisfy the third equation.

Hence the points A,B,C,D are not coplanar.

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