Question

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

Answer 1

Hint: First find the product $\left( {3\overline a - 4\overline b } \right) \cdot \left( {2\overline a + 5\overline b } \right)$ .
Now, \[\left| {\overline a } \right| = \left| {\overline b } \right| = 1 \Rightarrow {\left| {\overline a } \right|^2} = {\left| {\overline b } \right|^2} = 1\] .
Thus, using the square of the equation $\left| {\overline a + \overline b } \right| = \sqrt 3 $ , find \[\left| {\overline a } \right|\left| {\overline b } \right|\] .
Now, when we get all values of \[\left| {\overline a } \right|\left| {\overline b } \right|\] , \[{\left| {\overline a } \right|^2},{\left| {\overline b } \right|^2}\] , we will substitute them in the product of $\left( {3\overline a - 4\overline b } \right) \cdot \left( {2\overline a + 5\overline b } \right)$ and get the correct answer.

Complete step-by-step answer:
It is known that, \[\left| {\overline a } \right| = \left| {\overline b } \right| = 1\] and $\left| {\overline a + \overline b } \right| = \sqrt 3 $ .
Now, expanding the product $\left( {3\overline a - 4\overline b } \right) \cdot \left( {2\overline a + 5\overline b } \right)$ will give
 $\left( {3\overline a - 4\overline b } \right) \cdot \left( {2\overline a + 5\overline b } \right) = 6{\left| {\overline a } \right|^2} - 8\left| {\overline a \overline b } \right| + 15\left| {\overline a \overline b } \right| - 20{\left| {\overline b } \right|^2}$
                    $ = 6{\left| {\overline a } \right|^2} + 7\left| {\overline a \overline b } \right| - 20{\left| {\overline b } \right|^2}$ ... (1)
Also, $\left| {\overline a } \right| = 1$ and $\left| {\overline b } \right| = 1$
 $\therefore {\left| {\overline a } \right|^2} = 1$ and ${\left| {\overline b } \right|^2} = 1$ .
And $\left| {\overline a + \overline b } \right| = \sqrt 3 $
 \[
  \therefore {\left| {\overline a + \overline b } \right|^2} = {\left( {\sqrt 3 } \right)^2} \\
  \therefore {\left| {\overline a } \right|^2} + {\left| {\overline b } \right|^2} + 2\left| {\overline a } \right|\left| {\overline b } \right| = 3 \\
  \therefore 1 + 1 + 2\left| {\overline a } \right|\left| {\overline b } \right| = 3 \\
  \therefore 2\left| {\overline a } \right|\left| {\overline b } \right| = 3 - 2 \\
  \therefore \left| {\overline a } \right|\left| {\overline b } \right| = \dfrac{1}{2} \\
 \]
Now, substituting the values ${\left| {\overline a } \right|^2} = 1$ , ${\left| {\overline b } \right|^2} = 1$ and \[\left| {\overline a } \right|\left| {\overline b } \right| = \dfrac{1}{2}\] in equation (1)
 $
   = 6\left( 1 \right) + 7\left( {\dfrac{1}{2}} \right) - 20\left( 1 \right) \\
   = \dfrac{{12 + 7 - 40}}{2} \\
   = - \dfrac{{21}}{2} \\
 $
Thus, $\left( {3\overline a - 4\overline b } \right) \cdot \left( {2\overline a + 5\overline b } \right)$ \[ = \dfrac{{ - 21}}{2}\] .
Option (B) is correct.

Note: The dot product of any two vectors can be found using normal multiplication method.
For example, here $\left( {3\overline a - 4\overline b } \right) \cdot \left( {2\overline a + 5\overline b } \right) = 6{\left| {\overline a } \right|^2} - 8\left| {\overline a \overline b } \right| + 15\left| {\overline a \overline b } \right| - 20{\left| {\overline b } \right|^2}$ .
This is similar to finding $\left( {3a - 4b} \right)\left( {2a + 5b} \right) = 6{a^2} + 15ab - 8ab - 20{b^2}$