Question

The unit vector perpendicular to the vector \[\widehat i - \widehat j\] and\[\widehat i + \widehat j\] forming a right-handed system is
A. \[\widehat k\]
B. \[ - \widehat k\]
C. \[\dfrac{{\widehat i - \widehat j}}{{\sqrt 2 }}\]
D. \[\dfrac{{\widehat i + \widehat j}}{{\sqrt 2 }}\]

Answer 1

Hint: The given unit vector is perpendicular to both the vectors. therefore, use the fact that it will be equal to their cross product divided by the magnitude of the cross product.

Complete step-by-step answer:
Given, the unit vector perpendicular to the vector \[\widehat i - \widehat j\] and\[\widehat i + \widehat j\] forming a right-handed system
Let us consider the given unit vector to be equal to \[\overrightarrow c \]
Also let \[\overrightarrow a = \widehat i - \widehat j\] and \[\overrightarrow b = \widehat i + \widehat j\]
Now, since \[\overrightarrow c \] is a unit vector perpendicular to both \[\overrightarrow a \] and \[\overrightarrow b \]
Therefore, \[\overrightarrow c = \dfrac{{\overrightarrow a \times \overrightarrow b }}{{\left| {\overrightarrow a \times \overrightarrow b } \right|}}\]
Now, \[\overrightarrow a \times \overrightarrow b = \left( {\widehat i - \widehat j} \right) \times \left( {\widehat i + \widehat j} \right)\]
\[
   = 2\widehat k \\
   \Rightarrow \left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {{{(2)}^2}} \\
   \Rightarrow \left| {\overrightarrow a \times \overrightarrow b } \right| = 2 \\
\]
Therefore, \[\overrightarrow c = \dfrac{{\overrightarrow a \times \overrightarrow b }}{{\left| {\overrightarrow a \times \overrightarrow b } \right|}}\] is given by
\[
  \overrightarrow c = \dfrac{{2\widehat k}}{2} \\
   \Rightarrow \overrightarrow c = \widehat k \\
\]
Therefore, option (a) \[\widehat k\] is correct.

Note: Whenever a vector is given to be perpendicular to any 2 vectors then it is always equal to their cross product but in case a vector is perpendicular to any one vector then the dot product of the two vectors is 0.