Question

Express cross product of two vectors in cartesian co-ordinate.

Answer 1

Let \(\vec A\) and \(\vec B\) be the two vectors. 

\(\vec A\) =\(\vec{(A_x}\hat{i}+\vec{A_y}\hat{j}+\vec{A_z}\hat{k})\)

 and \(\vec R\) = \(\vec{(B_x}\hat{i}+\vec{B_y}\hat{j}+\vec{B_z}\hat{k})\)

\(\vec A\) × \(\vec B\) =\(\vec{(A_x}\hat{i}+\vec{B_y}\hat{j}+\vec{B_z}\hat{k})\)

A_xB_x(\(\hat{i}\times\hat{i}\)) + A_xB_y(\(\hat{i}\times\hat{j}\)) + A_xB_y(\(\hat{i}\times\hat{k}\)) + A_yB_x(\(\hat{j}\times\hat{i}\)) + A_yB_y(\(\hat{j}\times\hat{j}\)) + A_yB_z (\(\hat{j}\times\hat{k}\)) + A_zB_x(\(\hat{k}\times\hat{i}\)) + A_zB_x(\(\hat{k}\times\hat{j}\)) 

+ A_zB_y (\(\hat{k}\times\hat{k}\))

or\(\vec{A}\) × \(\vec{B}\) = (A_yB_z − A_zB_y)\(\hat{i}\)– (A_xB_y − A_zB_x)\(\hat{j}\)+ (A_xB_y − A_yB_x)\(\hat{k}\)

or \(\vec{A}\) × \(\vec{B}\) = A_xB_x(0) + A_xB_y(\(\hat{k}\)) + A_xB_z (\(-\hat{j}\)) + ̂ A_yB_x(\(-\hat{k}\)) + A_yB_y(0) + A_yB_z (\(\hat{i}\)) + A_zB_x(\(\hat{j}\)) + A_zB_y(\(-\hat{i}\)) + A_zB_z(0)

It can be written in determinant from as

 \(\vec{A}\) ×\(\vec{B}\) = \(\begin {vmatrix} \hat{i} & \hat{j} &\hat{k} \\ A_x &A_y&A_z \\ A_x&B_y&B_z \end{vmatrix}\)