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**Hint:**Use the concept of cross product or vector product of two vectors and find their product then take the modulus of resultant vector to find the value of \[\left| {\overrightarrow a \times \overrightarrow b } \right|\].

**Complete step by step solution:**Given two vectors,

\[\overrightarrow a = \widehat i + 3\widehat j - 2\widehat k\] and \[\overrightarrow b = 2\widehat i + \widehat j - 3\widehat k\]

To find the cross product and their magnitude

We know that cross product of \[\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\] and \[\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k\] is equal to

\[\overrightarrow a \times \overrightarrow b = \left( {\begin{array}{*{20}{c}}

{\widehat i}&{\widehat j}&{\widehat k} \\

{{a_1}}&{{a_2}}&{{a_3}} \\

{{b_1}}&{{b_2}}&{{b_3}}

\end{array}} \right)\]

On opening the determinant we get,

\[\overrightarrow a \times \overrightarrow b = \widehat i({a_2}{b_3} - {b_2}{a_3}) - \widehat j({a_1}{b_3} - {b_1}{a_3}) + \widehat k({a_1}{b_2} - {b_1}{a_2})\]

Now put the value of vector a and vector b in above formula

We get,

\[\overrightarrow a \times \overrightarrow b = \left( {\begin{array}{*{20}{c}}

{\widehat i}&{\widehat j}&{\widehat k} \\

1&3&{ - 2} \\

2&1&{ - 3}

\end{array}} \right)\]

\[\overrightarrow a \times \overrightarrow b = \widehat i[3 \bullet \left( { - 3} \right) - 1 \bullet \left( { - 2} \right)] - \widehat j[1 \bullet \left( { - 3} \right) - 2 \bullet \left( { - 2} \right)] + \widehat k[1 \bullet 1 - 2 \bullet 3]\]

On simplification,

\[\overrightarrow a \times \overrightarrow b = - 7i - \widehat j - 5\widehat k\]

Now we have to find the modulus or magnitude of $\overrightarrow a \times \overrightarrow b $

We know that \[\left| {\overrightarrow a } \right| = \sqrt {{a^2}_1 + {a^2}_2 + {a^2}_3} \] of a vector \[\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\]

So modulus of $\overrightarrow a \times \overrightarrow b $

On putting the value

\[\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {{{\left( { - 7} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 5} \right)}^2}} \]

On simplifying the above we get

\[\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {49 + 1 + 25} \]

\[\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {75} \]

Or \[\left| {\overrightarrow a \times \overrightarrow b } \right| = 5\sqrt 3 \]

Hence \[\overrightarrow a \times \overrightarrow b = - 7i - \widehat j - 5\widehat k\] and \[\left| {\overrightarrow a \times \overrightarrow b } \right| = 5\sqrt 3 \]

Additional information: Cross product or vector product of two vectors is always a vector quantity which has direction as well as the magnitude, while the dot product or scalar product of two vectors is always a scalar quantity which has no direction only magnitude. The modulus of any vector gives its magnitude.

**Note:**Magnitude or modulus of any vector can not be negative in any condition because it is always a distance from origin.

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