Question

$ABC$ is an equilateral triangle of sides $a$. The value of \[\overrightarrow {AB} \cdot \overrightarrow {BC} + \overrightarrow {BC} \cdot \overrightarrow {CA} + \overrightarrow {CA} \cdot \overrightarrow {AB} \] is equal to
(A) $\dfrac{{3{a^2}}}{2}$
(B) $3{a^2}$
(C) $ - \dfrac{{3{a^2}}}{2}$
(D) None of these

Answer 1

Hint: We have to calculate the value of \[\overrightarrow {AB} \cdot \overrightarrow {BC} + \overrightarrow {BC} \cdot \overrightarrow {CA} + \overrightarrow {CA} \cdot \overrightarrow {AB} \] where $AB$, $BC$ and $CA$ are the sides of an equilateral triangle \[ABC\]. First of all we have to find the dot product of two adjacent sides. We have studied in chapter “vector” that the dot product of two vectors $\overrightarrow X $ and $\overrightarrow Y $is \[\overrightarrow {X} \overrightarrow{Y} = |X||Y|\cos \theta \]. Since the given triangle is an equilateral triangle, the angle between the two sides is ${60^ \circ }$. Then we can find the dot product of two adjacent sides and then add them to get the required value.

Complete step by step solution:Here, the given triangle is an equilateral triangle and we have to find the value of \[\overrightarrow {AB} \cdot \overrightarrow {BC} + \overrightarrow {BC} \cdot \overrightarrow {CA} + \overrightarrow {CA} \cdot \overrightarrow {AB} \].
It is clearly visible that we have to find the dot product of two adjacent sides. We also know that each angle of the equilateral triangle is equal and of ${60^ \circ }$ each.
Now, find the value of \[A{B’}B{C’}\] by using the formula of dot products of two vectors.
\[\overrightarrow {AB} \cdot \overrightarrow {BC} = \overrightarrow {|AB|} |\overrightarrow {BC|} \cos {60^ \circ }\]
It is given that the sides of the equilateral triangle is $a$. And $\left| {\overrightarrow {AB} } \right|$,$\left| {\overrightarrow {BC} } \right|$ are the length of sides and $\cos {60^ \circ } = \dfrac{1}{2}$.
\[\overrightarrow {AB} \cdot \overrightarrow {BC} = \frac{1}{2}{a^2}\]
Similarly,
\[\overrightarrow {BC} \cdot \overrightarrow {CA} = \overrightarrow {|BC|} |\overrightarrow {CA|} \cos {60^ \circ }\]
Similarly,
\[\overrightarrow {CA} \cdot \overrightarrow {AB} = \overrightarrow {|CA|} |\overrightarrow {AB|} \cos {60^ \circ }\]
Now, all these terms to get the value of \[\overrightarrow {AB} \cdot \overrightarrow {BC} + \overrightarrow {BC} \cdot \overrightarrow {CA} + \overrightarrow {CA} \cdot \overrightarrow {AB} \].
$ = \dfrac{1}{2}{a^2} + \dfrac{1}{2}{a^2} + \dfrac{1}{2}{a^2}$
$ = \dfrac{3}{2}{a^2}$
Thus, the required value of is $\dfrac{3}{2}{a^2}$.

Hence, option (A) is correct.

Note: Similar concept is applied if we have to calculate the value of $\overrightarrow {AB} \times \overrightarrow {BC} + \overrightarrow {BC} \times \overrightarrow {CA} + \overrightarrow {CA} \times \overrightarrow {AB} $ remaining all condition is same as above. In spite of the dot product of two adjacent sides we have to find the cross product. The formula for the magnitude of cross product as $\left| {\overrightarrow {AB} \times \overrightarrow {BC} } \right| = \left| {\overrightarrow {AB} } \right|\left| {\overrightarrow {BC} } \right|\sin \theta $ . And all other steps are the same as above.