Question

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

Answer 1

Hint: First we will find a vector that is mutually perpendicular to both the given vectors i.e. \[\hat i - \hat j\] and \[\hat i + \hat j\] using the cross product of the two vectors, we get a vector perpendicular to both the vectors. Now, we will find the unit vector corresponding to that vector by dividing it by the modulus of that vector.

Complete step by step answer:

Given data: the vector \[\hat i - \hat j\] and \[\hat i + \hat j\]
We know that the cross product of two vectors results in the mutual perpendicular vector to both the vectors.
The vector perpendicular to the vectors \[\hat i - \hat j\] and \[\hat i + \hat j\]let say $\vec p$
$\vec p = \left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  1&{ - 1}&0 \\
  1&1&0
\end{array}} \right|$
 $ = (0 - 0)\hat i - (0 - 0)\hat j + \left( {1 - ( - 1)} \right)\hat k$
$ = 2\hat k$
Therefore, $\vec p = 2\hat k$
But we have to find the unit vector, and if \[a\hat i + b\hat j + c\hat k\] is a vector, then a unit vector proportional to this vector is given by
\[\dfrac{{a\hat i + b\hat j + c\hat k}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]
Therefore, unit vector along $\vec p$ i.e. $\hat p$is given by
\[ \Rightarrow \hat p = \dfrac{{2\hat k}}{{\sqrt {{0^2} + {0^2} + {2^2}} }}\]
\[ \Rightarrow \hat p = \hat k\]
Therefore, the required unit vector is \[\hat p = \hat k\]
Option(A) is correct

Note: While finding the cross product of the two vectors we have to find the determinant of $3 \times 3$ matrix and while taking the component of \[\hat j\], most of the students forget to put a negative sign and takes positive sign which is wrong as in this it does not matter as it is resulting zero but for other cases learn to remember this.