Question

Is A cross B is equal to B cross A?

Answer 1

Hint: Suppose that A and B are two vectors, where \[\overrightarrow A = a\widehat i + b\widehat j + c\widehat k{\rm{ and }}\overrightarrow B = p\widehat i + q\widehat j + r\widehat k\], then find the cross product of A and B, then find the cross product of B and A and observe that whether \[A \times B = B \times A\] or \[A \times B \ne B \times A\] .



Formula Used:The formula of cross product is \[\overrightarrow A \times \overrightarrow B = \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\a&b&c\\p&q&r\end{array}} \right|\] ,
where \[\overrightarrow A = a\widehat i + b\widehat j + c\widehat k{\rm{ and }}\overrightarrow B = p\widehat i + q\widehat j + r\widehat k\].



Complete step by step solution:We have,
\[\overrightarrow A \times \overrightarrow B = \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\a&b&c\\p&q&r\end{array}} \right|\]
             \[ = \widehat i\left( {br - cq} \right) - \widehat j\left( {ar - cp} \right) + \widehat k\left( {aq - pb} \right)\]
And,
\[\overrightarrow B \times \overrightarrow A = \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\p&q&r\\a&b&c\end{array}} \right|\]
           \[ = \widehat i\left( {cq - br} \right) - \widehat j\left( {cp - ar} \right) + \widehat k\left( {pb - aq} \right)\]
           \[ = - \widehat i\left( {br - cq} \right) + \widehat j\left( {ar - cp} \right) - \widehat k\left( {aq - pb} \right)\]
           \[ = - \left[ {\widehat i\left( {br - cq} \right) - \widehat j\left( {ar - cp} \right) + \widehat k\left( {aq - pb} \right)} \right]\]
           \[ = - \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\a&b&c\\p&q&r\end{array}} \right|\]
           \[ = - \left( {\overrightarrow A \times \overrightarrow B } \right)\]

Therefore, A cross B is not equal to B cross A, as \[\overrightarrow B \times \overrightarrow A = - \left( {\overrightarrow A \times \overrightarrow B } \right)\].





Additional Information:Cross product of two vectors represents a vector that is perpendicular to the plane where the vectors lie.



Note: Sometime students get confused when they calculate the value of the determinant, so calculate the value of the determinant with expanding the first row as \[\widehat i\left( {br - cq} \right) - \widehat j\left( {ar - cp} \right) + \widehat k\left( {aq - pb} \right)\] .