Question

\[\vec{a,} \vec{b} \text { and } \vec{c}\]  are the position vectors of points A, B and C respectively, prove that: \[\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}\]is a vector perpendicular to the plane of triangle ABC.

Answer 1

We know that if any vector is perpendicular to all three sides of ∆ ABC, it must be perpendicular to the plane of ∆ ABC .

Now, 

\[ \overrightarrow{AB} = \vec{b} - \vec{a} , \overrightarrow{BC} = \vec{c} - \vec{b} , \overrightarrow{CA} = \vec{a} - \vec{c} \left( Position vectors of A, B and C are \vec{a} , \vec{b} , \vec{c} \right)\]

We have

\[ \overrightarrow{AB} . ( \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} )\]

\[ = \left( \vec{b} - \vec{a} \right) . \left( \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} \right) \]

\[ = \vec{b} . \left( \vec{a} \times \vec{b} \right) + \vec{b} . \left( \vec{b} \times \vec{c} \right) + \vec{b} . \left( \vec{c} \times \vec{a} \right) - \vec{a} . \left( \vec{a} \times \vec{b} \right) - \vec{a} . \left( \vec{b} \times \vec{c} \right) - \vec{a} . \left( \vec{c} \times \vec{a} \right) \left( By distributive law \right)\]

\[ = \left[ \vec{b} \vec{a} \vec{b} \right] + \left[ \vec{b} \vec{b} \vec{c} \right] + \left[ \vec{b} \vec{c} \vec{a} \right] - \left[ \vec{a} \vec{a} \vec{b} \right] - \left[ \vec{a} \vec{b} \vec{c} \right] - \left[ \vec{a} \vec{c} \vec{a} \right]\]

\[ = 0 + 0 + \left[ \vec{b} \vec{c} \vec{a} \right] - 0 - \left[ \vec{a} \vec{b} \vec{c} \right] - 0\]

\[ = 0 \left( \because \left[ \vec{b} \vec{c} \vec{a} \right] = \left[ \vec{a} \vec{b} \vec{c} \right] \right) \]

\[ \overrightarrow{BC} . ( \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} )\]

\[ = \left( \vec{c} - \vec{b} \right) . \left( \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} \right) \]

\[ = \vec{c} . \left( \vec{a} \times \vec{b} \right) + \vec{c} . \left( \vec{b} \times \vec{c} \right) + \vec{c} . \left( \vec{c} \times \vec{a} \right) - \vec{b} . \left( \vec{a} \times \vec{b} \right) - \vec{b} . \left( \vec{b} \times \vec{c} \right) - \vec{b} . \left( \vec{c} \times \vec{a} \right) \left( By distributive law \right)\]

\[ = \left[ \vec{c} \vec{a} \vec{b} \right] + \left[ \vec{c} \vec{b} \vec{c} \right] + \left[ \vec{c} \vec{c} \vec{a} \right] - \left[ \vec{b} a^\to \vec{b} \right] - \left[ \vec{b} \vec{b} \vec{c} \right] - \left[ \vec{b} \vec{c} \vec{a} \right]\]

\[ = \left[ \vec{c} \vec{a} \vec{b} \right] + 0 + 0 - 0 - 0 - \left[ \vec{b} \vec{c} \vec{a} \right]\]

\[ = 0 \left( \because \left[ \vec{c} \vec{a} \vec{b} \right] = \left[ \vec{b} \vec{c} \vec{a} \right] \right) \]

Similarly, 

\[ \overrightarrow{CA} . ( \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} )\]

\[ = \left( \vec{a} - \vec{c} \right) . \left( \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} \right) \]

\[ = \vec{a} . \left( \vec{a} \times \vec{b} \right) + \vec{a} . \left( \vec{b} \times \vec{c} \right) + \vec{a} \left( \vec{c} \times \vec{a} \right) - \vec{c} . \left( \vec{a} \times \vec{b} \right) - \vec{c} . \left( \vec{b} \times \vec{c} \right) - \vec{c} . \left( \vec{c} \times \vec{a} \right) \left( By distributive law \right)\]

\[ = \left[ \vec{a} \vec{a} \vec{b} \right] + \left[ \vec{a} \vec{b} \vec{c} \right] + \left[ \vec{a} \vec{c} \vec{a} \right] - \left[ \vec{c} \vec{a} \vec{b} \right] - \left[ \vec{c} \vec{b} \vec{c} \right] - \left[ \vec{c} \vec{c} \vec{a} \right]\]

\[ = 0 + \left[ \vec{a} \vec{b} \vec{c} \right] + 0 - \left[ \vec{c} \vec{a} \vec{b} \right] - 0 - 0\]

\[ = 0 \left( \because \left[ \vec{c} \vec{a} \vec{b} \right] = \left[ \vec{a} \vec{b} \vec{c} \right] \right) \]

\[\text {Hence, vector } \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}\text { is perpendicular to all sides of ∆ ABC and also perpendicular to the plane of ∆ ABC } .\]