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Show that the points whose position vectors are 2{\bar  {i}}-{\bar  {j}}+{\bar  {k}},{\bar  {i}}-3{\bar  {j}}-5{\bar  {k}},3{\bar  {i}}-4{\bar  {j}}-4{\bar  {k}}\, are the vertices of a right angled triangle.

Let A=2{\bar  {i}}-{\bar  {j}}+{\bar  {k}},B={\bar  {i}}-3{\bar  {j}}-5{\bar  {k}},C=3{\bar  {i}}-4{\bar  {j}}-4{\bar  {k}}\, be the given points.

Now {\bar  {AB}}=-{\bar  {i}}-2{\bar  {j}}-6{\bar  {k}},{\bar  {BC}}=2{\bar  {i}}-{\bar  {j}}+{\bar  {k}},{\bar  {AC}}={\bar  {i}}-3{\bar  {j}}-5{\bar  {k}}\,

Now {\bar  {AB}}+{\bar  {BC}}=(-{\bar  {i}}-2{\bar  {j}}-6{\bar  {k}})+(2{\bar  {i}}-{\bar  {j}}+{\bar  {k}})={\bar  {i}}-3{\bar  {j}}-5{\bar  {k}}={\bar  {AC}}=-{\bar  {CA}}\,

{\bar  {AB}}+{\bar  {BC}}+{\bar  {CA}}={\bar  {O}}\,

This shows that the the given points are the vertices of a triangle.

{\bar  {BC}}\cdot {\bar  {CA}}=(2{\bar  {i}}-{\bar  {j}}+{\bar  {k}})\cdot (-{\bar  {i}}+3{\bar  {j}}+5{\bar  {k}})=-2-3+5=0\,

Hence the two sides are perpendicular to each other.Therefore,ABC is a right angled triangle.

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