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Question: If\[\bar a\],\[\bar b\],\[\bar c\]are unit vectors such that \[\bar a + \bar b + \bar c = 0\]. Find ...

Ifaˉ\bar a,bˉ\bar b,cˉ\bar care unit vectors such that aˉ+bˉ+cˉ=0\bar a + \bar b + \bar c = 0. Find aˉ.bˉ+bˉ.cˉ+cˉ.aˉ\bar a.\bar b + \bar b.\bar c + \bar c.\bar a

Explanation

Solution

Hint : In this question three unit vectors are given and the values of their sum is given and we need to find their sum of the product of two vectors at a time so we will first find the square of the mod of the given vector equation and then we will substitute the values of the unit vector to find the value of the desired vector equation.

Complete step-by-step answer :
Given aˉ\bar a,bˉ\bar b,cˉ\bar c are unit vectors
Vector aˉ+bˉ+cˉ=0\bar a + \bar b + \bar c = 0
Take mod on both side of the given vector
⇒∣aˉ+bˉ+cˉ∣=∣0∣−−(i)\Rightarrow \left| {\bar a + \bar b + \bar c} \right| = \left| 0 \right| - - (i)
Now since we know that the mod of number 0 is always 0, hence we can write the vector equation (i) as
⇒∣aˉ+bˉ+cˉ∣=0−−(ii)[∵∣0∣=0]\Rightarrow \left| {\bar a + \bar b + \bar c} \right| = 0 - - (ii)\left[ {\because \left| 0 \right| = 0} \right]
Now we take square on the both sides of the vector equation (ii), so we will get

⇒∣aˉ+bˉ+cˉ∣2=0 ⇒∣a∣+∣b∣+∣c∣+2a⃗b⃗+2b⃗c⃗+2c⃗a⃗=0−−(iii)  \Rightarrow {\left| {\bar a + \bar b + \bar c} \right|^2} = 0 \\\ \Rightarrow \left| a \right| + \left| b \right| + \left| c \right| + 2\vec a\vec b + 2\vec b\vec c + 2\vec c\vec a = 0 - - (iii) \;

Now since it is already said that the vectors aˉ\bar a,bˉ\bar b,cˉ\bar c are unit vectors, hence we can say ∣a∣=∣b∣=∣c∣=1\left| a \right| = \left| b \right| = \left| c \right| = 1
Now by substituting the values of unit vector in the vector equation (iii), we get
⇒1+1+1+2a⃗b⃗+2b⃗c⃗+2c⃗a⃗=0\Rightarrow 1 + 1 + 1 + 2\vec a\vec b + 2\vec b\vec c + 2\vec c\vec a = 0
By further solving this equation, we get

3+2a⃗b⃗+2b⃗c⃗+2c⃗a⃗=0 2(a⃗b⃗+b⃗c⃗+c⃗a⃗)=−3 a⃗b⃗+b⃗c⃗+c⃗a⃗=−32   3 + 2\vec a\vec b + 2\vec b\vec c + 2\vec c\vec a = 0 \\\ 2\left( {\vec a\vec b + \vec b\vec c + \vec c\vec a} \right) = - 3 \\\ \vec a\vec b + \vec b\vec c + \vec c\vec a = \dfrac{{ - 3}}{2} \;

Hence the value of the vector aˉ.bˉ+bˉ.cˉ+cˉ.aˉ\bar a.\bar b + \bar b.\bar c + \bar c.\bar a is equal to −32\dfrac{{ - 3}}{2}

Note : Students must note that the modulus of 0 is always 0 and the modulus of any positive number is the same number whose modulus is to be found and in the case of any negative number its modulus is the number itself ignoring the negative sign.

∣0∣=0 ∣a∣=a ∣−b∣=b   \left| 0 \right| = 0 \\\ \left| a \right| = a \\\ \left| { - b} \right| = b \;