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

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.