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Question: If \(\overrightarrow a ,\overrightarrow b ,\overrightarrow c \) are unit vectors such that \(\overri...

If a→,b→,c→\overrightarrow a ,\overrightarrow b ,\overrightarrow c are unit vectors such that a→+b→+c→=0→\overrightarrow a + \overrightarrow b + \overrightarrow c = \overrightarrow 0 , find the value of vector a→.b→+b→.c→+c→.a→\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a .
A) 12\dfrac{1}{2}
B) 00
C) 11
D) −32\dfrac{-3}{2}

Explanation

Solution

In the question it is given that a→,b→,c→\overrightarrow a ,\overrightarrow b ,\overrightarrow c are unit vectors that means its magnitude is 1. So to solve the question we will do square of equation given in the question above and then we will solve it step by step and at necessary places we will replace ∣a→∣,∣b→∣,∣c→∣\left| {\overrightarrow a } \right|,\left| {\overrightarrow b } \right|,\left| {\overrightarrow c } \right|with 1 to find the answer.

Complete step by step answer:
According to the question,
a→,b→,c→\overrightarrow a ,\overrightarrow b ,\overrightarrow c are unit vectors and which means its magnitude is 1.
Given: ∣a→∣=∣b→∣=∣c→∣=1\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right| = 1------(1)
a→+b→+c→=0\overrightarrow a + \overrightarrow b + \overrightarrow c = 0
So, ∣a→+b→+c→∣=∣0→∣=0\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right| = \left| {\overrightarrow 0 } \right| = 0------(2)
Now, we will square this equation, and we get
=∣a→+b→+c→∣2= {\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2}
=(a→+b→+c→).(a→+b→+c→)= \left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right).\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)
=a→.a→+a→.b→+a→.c→+b→.a→+b→.b→+b→.c→+c→.a→+b→.c→+c→.c→= \overrightarrow a .\overrightarrow a + \overrightarrow a .\overrightarrow b + \overrightarrow a .\overrightarrow c + \overrightarrow b .\overrightarrow a + \overrightarrow b .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow c
Now, we know that a→.b→=b→.a→\overrightarrow a .\overrightarrow b = \overrightarrow b .\overrightarrow a so, similarly we will replace it in the above equation
=a→.a→+a→.b→+c→.a→+a→.b→+b→.b→+b→.c→+c→.a→+b→.c→+c→.c→= \overrightarrow a .\overrightarrow a + \overrightarrow a .\overrightarrow b + \overrightarrow c .\overrightarrow a + \overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow c
=a→.a→+b→.b→+c→.c→+2a→.b→+2b→.c→+2c→.a→= \overrightarrow a .\overrightarrow a + \overrightarrow b .\overrightarrow b + \overrightarrow c .\overrightarrow c + 2\overrightarrow a .\overrightarrow b + 2\overrightarrow b .\overrightarrow c + 2\overrightarrow c .\overrightarrow a
Now, we know that a→.a→=∣a→∣2,b→.b→=∣b→∣2,c→.c→=∣c→∣2\overrightarrow a .\overrightarrow a = {\left| {\overrightarrow a } \right|^2},\overrightarrow b .\overrightarrow b = {\left| {\overrightarrow b } \right|^2},\overrightarrow c .\overrightarrow c = {\left| {\overrightarrow c } \right|^2}so, we will replace it in above equation
=∣a→∣2+∣b→∣2+∣c→∣2+2(a→.b→+b→.c→+c→.a→)= {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} + 2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right)
Now, using equation (1) we will replace value of ∣a→∣,∣b→∣,∣c→∣\left| {\overrightarrow a } \right|,\left| {\overrightarrow b } \right|,\left| {\overrightarrow c } \right|with 1
=12+12+12+2(a→.b→+b→.c→+c→.a→)= {1^2} + {1^2} + {1^2} + 2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right)
=3+2(a→.b→+b.→c→+c→.a→)= 3 + 2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow {b.} \overrightarrow c + \overrightarrow c .\overrightarrow a } \right)
So, ∣a→+b→+c→∣2=3+2(a→.b→+b→.c→+c→.a→){\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = 3 + 2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right)
Using equation (2) we get,
3+2(a→.b→+b→.c→+c→.a→)=03 + 2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right) = 0
2(a→.b→+b→.c→+c→.a→)=−32\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right) = - 3
(a→.b→+b→.c→+c→.a→)=−32\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right) = \dfrac{{ - 3}}{2}
Therefore, the value of vector a→.b→+b→.c→+c→.a→\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a is −32\dfrac{-3}{2}. So, option (D) is correct.

Additional information:
A vector is an object that has both magnitude and direction. the students must know the difference between dot product and cross product. the dot product results in a scalar quantity and the cross product results in a vector quantity. In these kinds of questions we have to use the equation provided in the question to calculate the answer.

Note:
This question can also be solved by another method also. We can solve by multiplying a→,b→,c→\overrightarrow a ,\overrightarrow b ,\overrightarrow c one by one with (a→+b→+c→)\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)then, we will get some value from all of these and then after adding all three of them together and after solving we will get the answer.