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Question: \(ABC\) is an equilateral triangle of sides \(a\). The value of \[\overrightarrow {AB} \cdot \overri...

ABCABC is an equilateral triangle of sides aa. The value of AB→⋅BC→+BC→⋅CA→+CA→⋅AB→\overrightarrow {AB} \cdot \overrightarrow {BC} + \overrightarrow {BC} \cdot \overrightarrow {CA} + \overrightarrow {CA} \cdot \overrightarrow {AB} is equal to
(A) 3a22\dfrac{{3{a^2}}}{2}
(B) 3a23{a^2}
(C) −3a22 - \dfrac{{3{a^2}}}{2}
(D) None of these

Explanation

Solution

We have to calculate the value of AB→⋅BC→+BC→⋅CA→+CA→⋅AB→\overrightarrow {AB} \cdot \overrightarrow {BC} + \overrightarrow {BC} \cdot \overrightarrow {CA} + \overrightarrow {CA} \cdot \overrightarrow {AB} where ABAB, BCBC and CACA are the sides of an equilateral triangle ABCABC. 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 X→\overrightarrow X and Y→\overrightarrow Y is X→Y→=∣X∣∣Y∣cos⁡θ\overrightarrow {X} \overrightarrow{Y} = |X||Y|\cos \theta . Since the given triangle is an equilateral triangle, the angle between the two sides is 60∘{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 AB→⋅BC→+BC→⋅CA→+CA→⋅AB→\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∘{60^ \circ } each.
Now, find the value of AB’BC’A{B’}B{C’} by using the formula of dot products of two vectors.
AB→⋅BC→=∣AB∣→∣BC∣→cos⁡60∘\overrightarrow {AB} \cdot \overrightarrow {BC} = \overrightarrow {|AB|} |\overrightarrow {BC|} \cos {60^ \circ }
It is given that the sides of the equilateral triangle is aa. And ∣AB→∣\left| {\overrightarrow {AB} } \right|,∣BC→∣\left| {\overrightarrow {BC} } \right| are the length of sides and cos⁡60∘=12\cos {60^ \circ } = \dfrac{1}{2}.
AB→⋅BC→=12a2\overrightarrow {AB} \cdot \overrightarrow {BC} = \frac{1}{2}{a^2}
Similarly,
BC→⋅CA→=∣BC∣→∣CA∣→cos⁡60∘\overrightarrow {BC} \cdot \overrightarrow {CA} = \overrightarrow {|BC|} |\overrightarrow {CA|} \cos {60^ \circ }
Similarly,
CA→⋅AB→=∣CA∣→∣AB∣→cos⁡60∘\overrightarrow {CA} \cdot \overrightarrow {AB} = \overrightarrow {|CA|} |\overrightarrow {AB|} \cos {60^ \circ }
Now, all these terms to get the value of AB→⋅BC→+BC→⋅CA→+CA→⋅AB→\overrightarrow {AB} \cdot \overrightarrow {BC} + \overrightarrow {BC} \cdot \overrightarrow {CA} + \overrightarrow {CA} \cdot \overrightarrow {AB} .
=12a2+12a2+12a2= \dfrac{1}{2}{a^2} + \dfrac{1}{2}{a^2} + \dfrac{1}{2}{a^2}
=32a2= \dfrac{3}{2}{a^2}
Thus, the required value of is 32a2\dfrac{3}{2}{a^2}.

Hence, option (A) is correct.

Note: Similar concept is applied if we have to calculate the value of AB→×BC→+BC→×CA→+CA→×AB→\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 ∣AB→×BC→∣=∣AB→∣∣BC→∣sin⁡θ\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.