Question

Find out the value of the dot product of $\left( {3\vec a + 2\vec b} \right).\left( {5\vec a - 6\vec b} \right)$ if $\vec a$ and $\vec b$ are mutually perpendicular unit vectors.
$
  {\text{a}}{\text{. 5}} \\
  {\text{b}}{\text{. 3}} \\
  {\text{c}}{\text{. 6}} \\
  {\text{d}}{\text{. 12}} \\
$

Answer 1

Hint: For mutually perpendicular vectors the dot product is always zero. Using this concept we can find out the solution of this problem.

Complete step-by-step answer:
As we know if two unit vectors are mutually perpendicular then their dot product should be zero.
I.e. the dot product of vector $\vec a$ with vector $\vec b$ is zero.
And the dot product of vector $\vec a$ with itself is one.
$
   \Rightarrow \vec a.\vec b = 0{\text{, }}\vec a.\vec a = \vec b.\vec b = 1 \\
   \Rightarrow \left( {3\vec a + 2\vec b} \right).\left( {5\vec a - 6\vec b} \right) = 15\left( {\vec a.\vec a} \right) - 18\left( {\vec a.\vec b} \right) + 10\left( {\vec a.\vec b} \right) - 12\left( {\vec b.\vec b} \right) \\
   \Rightarrow \left( {3\vec a + 2\vec b} \right).\left( {5\vec a - 6\vec b} \right) = 15 + 0 - 0 - 12 \\
   \Rightarrow \left( {3\vec a + 2\vec b} \right).\left( {5\vec a - 6\vec b} \right) = 3 \\
$
Hence option (c) is correct.

Note: In such types of questions the key concept we have to remember is that the dot product of two mutually perpendicular vectors is always zero and the modules of vectors is always 1, then according to these properties simplify the given equation we will get the required answer.