Question

The unit vector orthogonal to \[ - \hat i + \hat k,2\hat j - \hat k\] and forming a right handed system with them is
A. \[2\hat i + \hat j + 2\hat k\]
B. \[\dfrac{{2\hat i + \hat j + 2\hat k}}{3}\]
C. \[ - \dfrac{{2\hat i + \hat j + 2\hat k}}{3}\]
D. \[\dfrac{{2\hat i + \hat j + 2\hat k}}{9}\]

Answer 1

Hint: We will begin by finding the vector which is orthogonal to given vectors by finding the cross product of the given vectors. Then, we will find the magnitude of the resultant vector. Divide the resultant vector by the magnitude to find the corresponding unit vector. If the vector is $a\hat i + b\hat j + c\hat k$, then the magnitude of the vector is $\sqrt {{a^2} + {b^2} + {c^2}} $

Complete step by step Answer:

We have to find the unit vector orthogonal to \[ - \hat i + \hat k,2\hat j - \hat k\].
First, we will find the vector which is orthogonal to the given vectors.
That is, we will find the cross product of \[ - \hat i + \hat k,2\hat j - \hat k\]
The cross product of the vectors will give the corresponding value of the orthogonal vector.
If $a\hat i + b\hat j + c\hat k$ and $p\hat i + q\hat j + r\hat k$ are two vectors, then the cross product is given as
$\left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  a&b&c \\
  p&q&r
\end{array}} \right|$
Then, the cross product of \[ - \hat i + \hat k\] and \[2\hat j - \hat k\] is
$\left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  { - 1}&0&1 \\
  0&2&{ - 1}
\end{array}} \right|$
On solving the determinant along the first row, we will get,
$
  \left( {0\left( { - 1} \right) - 1\left( 2 \right)} \right)\hat i - \left( {\left( { - 1} \right)\left( { - 1} \right) - \left( 1 \right)0} \right)\hat j + \left( {\left( { - 1} \right)\left( 2 \right) - 0} \right)\hat k \\
   \Rightarrow - 2\hat i - \hat j - 2\hat k \\
$
We now have to find the unit vector corresponding to $ - 2\hat i - \hat j - 2\hat k$
We will first find the magnitude of the vector $ - 2\hat i - \hat j - 2\hat k$
If the vector is $a\hat i + b\hat j + c\hat k$, then the magnitude of the vector is $\sqrt {{a^2} + {b^2} + {c^2}} $
Then, the magnitude of $ - 2\hat i - \hat j - 2\hat k$ is,
$
  \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 2} \right)}^2}} \\
   = \sqrt {4 + 1 + 4} \\
   = \sqrt 9 \\
   = 3 \\
$
We can now write the unit vector by dividing the vector by its magnitude.
$\dfrac{{ - 2\hat i - \hat j - 2\hat k}}{3}$ which can be rewritten as $ - \dfrac{{2\hat i + \hat j + 2\hat k}}{3}$
Hence, the unit vector orthogonal to \[ - \hat i + \hat k,2\hat j - \hat k\] is $ - \dfrac{{2\hat i + \hat j + 2\hat k}}{3}$
Thus, option C is correct.

Note: If a vector is orthogonal to given vectors, it means the vector is perpendicular to both the given vectors. The cross product of any two vectors gives the vector which is perpendicular to both the vectors. One makes mistakes by not dividing the resultant orthogonal vector by its magnitude which will not give us the unit vector. And, the magnitude of the unit vector is always one.