LinAlg4.1.13

From Exampleproblems

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.

Main Page:Linear Algebra

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