Question

If $ \mathop a\limits^ \to = \hat i + 2\hat j + \hat k,\mathop b\limits^ \to = 2\hat i + \hat j $ and $ \mathop c\limits^ \to = 3\hat i - 4\hat j - 5\hat k $ then find a unit vector perpendicular to both of the vectors $ (\mathop a\limits^ \to - \mathop b\limits^ \to ) $ and $ (\mathop c\limits^ \to - \mathop b\limits^ \to ) $ .

Answer 1

Hint: In this question we have been given three vectors $ \mathop a\limits^ \to ,\mathop b\limits^ \to ,\mathop c\limits^ \to $ we need to find a unit vector which is perpendicular to $ (\mathop a\limits^ \to - \mathop b\limits^ \to ) $ and $ (\mathop c\limits^ \to - \mathop b\limits^ \to ) $ . For that we will find the value of the vector $ (\mathop a\limits^ \to - \mathop b\limits^ \to ) $ and $ (\mathop c\limits^ \to - \mathop b\limits^ \to ) $ as they are perpendicular we will find their cross product, and then we will use the formula: $ \left( {\dfrac{{\left( {\mathop a\limits^ \to - \mathop b\limits^ \to } \right) \times \left( {\mathop c\limits^ \to - \mathop b\limits^ \to } \right)}}{{\left| {\left( {\mathop a\limits^ \to - \mathop b\limits^ \to } \right) \times \left. {\left( {\mathop c\limits^ \to - \mathop b\limits^ \to } \right)} \right|} \right.}}} \right) $ to find the unit vector.

Complete step-by-step answer:
We have been provided with three vectors $ \mathop a\limits^ \to = \hat i + 2\hat j + \hat k,\mathop b\limits^ \to = 2\hat i + \hat j $ and $ \mathop c\limits^ \to = 3\hat i - 4\hat j - 5\hat k $ ,
To find a unit vector which is perpendicular to $ (\mathop a\limits^ \to - \mathop b\limits^ \to ) $ and $ (\mathop c\limits^ \to - \mathop b\limits^ \to ) $ , firstly we need to find $ (\mathop a\limits^ \to - \mathop b\limits^ \to ) $
So, we can compute as $ (\mathop a\limits^ \to - \mathop b\limits^ \to ) = \hat i + 2\hat j + \hat k - (2\hat i + \hat j) $
Simplifying, we will get $ (\mathop a\limits^ \to - \mathop b\limits^ \to ) = - \hat i + \hat j + \hat k $
Similarly, we will find the value of $ (\mathop c\limits^ \to - \mathop b\limits^ \to ) $ ,
So, we get $ (\mathop c\limits^ \to - \mathop b\limits^ \to ) = 3\hat i - 4\hat j - 5\hat k - (2\hat i + \hat j) $
Simplifying, we have $ (\mathop c\limits^ \to - \mathop b\limits^ \to ) = \hat i - 5\hat j - 5\hat k $
Now as we need to find the unit vector which is perpendicular to both $ (\mathop a\limits^ \to - \mathop b\limits^ \to ) $ and $ (\mathop c\limits^ \to - \mathop b\limits^ \to ) $ ,
For that we need to find their cross products,
So, we can write it as
 $ (\mathop a\limits^ \to - \mathop b\limits^ \to ) \times (\mathop c\limits^ \to - \mathop b\limits^ \to ) = \left| {\left( {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  { - 1}&1&1 \\
  1&{ - 5}&{ - 5}
\end{array}} \right)} \right| $
Now we will solve the determinant: $ (\mathop a\limits^ \to - \mathop b\limits^ \to ) \times (\mathop c\limits^ \to - \mathop b\limits^ \to ) = ( - 5 + 5)\hat i - (5 - 1)\hat j + (5 - 1)\hat k $
From this we will get $ (\mathop a\limits^ \to - \mathop b\limits^ \to ) \times (\mathop c\limits^ \to - \mathop b\limits^ \to ) = - 4\hat j + 4\hat k $
Now, for finding the unit vector we will be using the formula: $ \left( {\dfrac{{\left( {\mathop a\limits^ \to - \mathop b\limits^ \to } \right) \times \left( {\mathop c\limits^ \to - \mathop b\limits^ \to } \right)}}{{\left| {\left( {\mathop a\limits^ \to - \mathop b\limits^ \to } \right) \times \left. {\left( {\mathop c\limits^ \to - \mathop b\limits^ \to } \right)} \right|} \right.}}} \right) $
Now, we will be keeping the values then it becomes: $ \dfrac{{ - 4\hat j + 4\hat k}}{{4\sqrt 2 }} = \dfrac{{ - \hat j + \hat k}}{{\sqrt 2 }} $
So, $ \dfrac{{ - \hat j + \hat k}}{{\sqrt 2 }} $ is the required vector.

Note: In this question, be careful while calculating the determinant, about the negative and positive signs. As we need to find a unit vector which is perpendicular to $ (\mathop a\limits^ \to - \mathop b\limits^ \to ) $ and $ (\mathop c\limits^ \to - \mathop b\limits^ \to ) $ , we should calculate the cross product and not the dot product.