Question

If $\left| a \right|=\left| b \right|=1$ and $\left| a+b \right|=\sqrt{3}$, then the value of $\left( 3a-4b \right)\left( 2a+5b \right)$ is
A. $-21$
B. $-\dfrac{21}{2}$
C. 21
D. $\dfrac{21}{2}$

Answer 1

We first need to remove the modulus functions. We take the square value to get rid of the modulus function. We get the values of ${{a}^{2}},{{b}^{2}},{{\left( a+b \right)}^{2}}$. We use the square formula of ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. We put the values to get the value of $ab$. We complete the multiplication of $\left( 3a-4b \right)\left( 2a+5b \right)$ and get the final solution.

Complete step by step answer:
The given condition is $\left| a \right|=\left| b \right|=1$. We take the square value to get rid of the modulus function.
We get ${{a}^{2}}={{b}^{2}}=1$. Applying similar thing for $\left| a+b \right|=\sqrt{3}$, we get ${{\left( a+b \right)}^{2}}=3$.
Now we break the square using ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
Putting the values, we get
$\begin{aligned} & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\\ & \Rightarrow 3=1+1+2ab \\\ & \Rightarrow ab=\dfrac{1}{2} \\\ \end{aligned}$
We now complete the multiplication of $\left( 3a-4b \right)\left( 2a+5b \right)$.
$\left( 3a-4b \right)\left( 2a+5b \right)=6{{a}^{2}}-20{{b}^{2}}+7ab$
We put the values to get
$\begin{aligned} & \left( 3a-4b \right)\left( 2a+5b \right) \\\ & =6{{a}^{2}}-20{{b}^{2}}+7ab \\\ & =6\times 1-20\times 1+7\times \dfrac{1}{2} \\\ & =\dfrac{7}{2}-14 \\\ & =-\dfrac{21}{2} \\\ \end{aligned}$
Therefore, the value of $\left( 3a-4b \right)\left( 2a+5b \right)$ is $-\dfrac{21}{2}$. The correct option is option (B).

Note:
We use the modulus function to get the distance or length in general. That becomes without the sign. The vector form changes it to scalar form. Therefore, without taking square value we cannot use the modulus function in any simplification.