Question

For two vectors $\vec A$ and $\vec B$ which of the following relations are not commutative?
A. $\vec P + \vec Q$
B. $\vec P \times \vec Q$
C. $\vec P.\vec Q$
D. None of these

Answer 1

In order to solve this question, we should know that a commutative property of any two quantities is such that when their order is reversed under any mathematical operation, their result is same irrespective of the order so, here we will check for each given operation between two vector and then figure out which relation is not commutative.

Complete step by step answer:
Considering option first we have addition of two vectors say $\vec P + \vec Q$ now let us check the commutative property as, let vector P is written as $\vec P = {p_1}\hat i + {p_2}\hat j + {p_3}\hat k$ and vector Q be $\vec Q = {q_1}\hat i + {q_2}\hat j + {q_3}\hat k$ Now finding $\vec P + \vec Q$ so we have,
$\vec P + \vec Q = {p_1}\hat i + {p_2}\hat j + {p_3}\hat k + ({q_1}\hat i + {q_2}\hat j + {q_3}\hat k)$
$\Rightarrow \vec P + \vec Q = ({p_1} + {q_1})\hat i + ({p_2} + {q_2})\hat j + ({p_3} + {q_3})\hat k$
Now similar if we add $\vec Q + \vec P$ we get,
$\vec Q + \vec P = ({q_1} + {p_1})\hat i + ({q_2} + {p_2})\hat j + ({q_3} + {p_3})\hat k$

Since, addition of scalar quantities are equal irrespective of the order so,
${p_1} + {q_1} = {q_1} + {p_1}$ , ${p_2} + {q_2} = {q_2} + {p_2}$ and ${p_3} + {q_3} = {q_3} + {p_3}$ hence, all components of these sum of vector P and Q are equal, $\vec P + \vec Q = \vec Q + \vec P$ hence, this operation is commutative. Now, for second option $\vec P \times \vec Q$ which is a vector product between two vectors now since vector product between two vectors is given by,
\vec P \times \vec Q = \left| {\left( {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\\ {{p_1}}&{{p_2}}&{{p_3}} \\\ {{q_1}}&{{q_2}}&{{q_3}} \end{array}} \right)} \right| which is the determinant of the shown matrix.

Now, similarly,
\vec Q \times \vec P = \left| {\left( {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\\ {{q_1}}&{{q_2}}&{{q_3}} \\\ {{p_1}}&{{p_2}}&{{p_3}} \end{array}} \right)} \right|
So, we see that this determinant is obtained when rows of matrix \vec P \times \vec Q = \left| {\left( {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\\ {{p_1}}&{{p_2}}&{{p_3}} \\\ {{q_1}}&{{q_2}}&{{q_3}} \end{array}} \right)} \right| are interchanged. Hence, we know from the properties of determinants if rows are interchanged then, determinant will change its sign hence,
$\vec Q \times \vec P = - \vec P \times \vec Q$ hence, this relation is not equal to each other but negative of one another, hence this vector product is not commutative as $\vec Q \times \vec P \ne \vec P \times \vec Q$

Hence, the correct option is (B).

Note: It should be remembered that, the scalar product is also commutative as scalar product of real numbers is not affected by the orders in which they are multiplied such as $a \times b = b \times a$ where, $a$ and $b$ are real numbers hence, vector product is only operation which is not commutative as it comprise of determinant matrix of its components which depends upon the order in which they are written.