Find the value of ‘a’ for which the vector $\overrightarrow A = \,3\widehat i + 3\widehat j + 9\widehat k$ and $\overrightarrow B = \,\widehat i + a\widehat j + 3\widehat k$ are parallel.

Hint: Vectors are the quantities that have direction as well as magnitude. In order to answer the question, we must be aware about basic properties of vectors (i.e., addition, multiplication, subtraction and division of vectors).

Complete step by step answer:
If we observe properties of vectors then we will notice that the cross product of two parallel vectors is always zero.
$\overrightarrow A \times \overrightarrow B = 0$
Cross product of the given vectors will give,
$\overrightarrow A \times \overrightarrow B = \left( {3\widehat i + 3\widehat j + 9\widehat k} \right) \times \left( {\widehat i + a\widehat j + 3\widehat k} \right)$
Solving the cross product by converting the given vectors in matrix,
$\overrightarrow A \times \overrightarrow B = \left( {\begin{array}{*{20}{c}}
  i&j&k \\
  3&3&9 \\
  1&a&3
\end{array}} \right)$
Calculating the determinant of the matrix,
$\overrightarrow A \times \overrightarrow B = \left( {\begin{array}{*{20}{c}}
  3&9 \\
  a&3
\end{array}} \right)\widehat i - \left( {\begin{array}{*{20}{c}}
  3&9 \\
  1&3
\end{array}} \right)\widehat j + \left( {\begin{array}{*{20}{c}}
  3&3 \\
  1&a
\end{array}} \right)\widehat k$
Evaluating the above vector will yield,
$\overrightarrow A \times \overrightarrow B = \left( {\left( {3 \times 3} \right) - \left( {9 \times a} \right)} \right)\widehat i - \left( {\left( {3 \times 3} \right) - \left( {9 \times 1} \right)} \right)\widehat j + \left( {\left( {3 \times a} \right) - \left( {3 \times 1} \right)} \right)\widehat k$
Solving the basic maths will yield,
$\overrightarrow A \times \overrightarrow B = \left( {9 - 9a} \right)\widehat i - \left( {9 - 9} \right)\widehat j + \left( {3a - 3} \right)\widehat k$

As we know that the cross product of two vectors yield zero, hence
$0\widehat i + 0\widehat j + 0\widehat k = \left( {9 - 9a} \right)\widehat i - \left( 0 \right)\widehat j + \left( {3a - 3} \right)\widehat k$
We can observe from the above equation that,
$9 – 9a = 0 \\
\Rightarrow 3a – 3 = 0$
Evaluating value of ‘a’ from either of two equations
$9 = 9a \\
\therefore a = 1$
For the given vectors to be parallel to each other $a = 1$.

Note: Multiplication of vectors can be achieved by either their dot product or their cross product. Dot products of the vectors always yield a scalar result while cross product of two vectors results in a vector quantity.