Question
Answers

Find $\overrightarrow a \times \overrightarrow b $ and \[\left| {\overrightarrow a \times \overrightarrow b } \right|\] if \[\overrightarrow a = \widehat i + 3\widehat j - 2\widehat k\] and \[\overrightarrow b = 2\widehat i + \widehat j - 3\widehat k\].

Answer
VerifiedVerified
132k+ views
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.