Question

Find the resultant vector of \((j - k) \times (k - i)\) using cross product.

Short answer

The cross product \((j - k) \times (k - i)\) is\(i + j + k\).

Step-by-step solution

Formula used

Consider the following property

\(\begin{array}{l}{\rm{j}} \times {\rm{k}} = {\rm{i}}\\{\rm{k}} \times {\rm{k}} = 0\\{\rm{j}} \times {\rm{i}} = - {\rm{k}}\\{\rm{k}} \times {\rm{i}} = {\rm{j}}\end{array}\)

Find the resultant vector

As given,\((j - k) \times (k - i) \ldots \ldots \ldots (1)\)

Modify equation (1)

\(\begin{array}{l}({\rm{j}} - {\rm{k}}) \times (k - i) = ({\rm{j}} - {\rm{k}}) \times {\rm{k}} + ({\rm{j}} - {\rm{k}}) + ({\rm{j}} - {\rm{k}}) \times ( - {\rm{i}})\\({\rm{j}} - {\rm{k}}) \times (k - i) = ({\rm{j}} \times {\rm{k}}) + ( - {\rm{k}}) \times {\rm{k}} + {\rm{j}} \times ( - {\rm{i}}) + ( - {\rm{k}}) \times ( - {\rm{i}})\\({\rm{j}} - {\rm{k}}) \times (k - i) = ({\rm{j}} \times {\rm{k}}) + ( - 1)({\rm{k}} \times {\rm{k}}) + ( - 1)({\rm{j}} \times {\rm{i}}) + {( - 1)^2}({\rm{k}} \times {\rm{i}})\end{array}\)

Substitute \(i\) for \(j \times k,0\) for \(k \times k, - k\) for \(j \times i\) and \(j\) for \(k \times i\)

\(\begin{array}{l}({\rm{j}} - {\rm{k}}) \times ({\rm{k}} - {\rm{i}}) = {\rm{i}} + ( - 1)(0) + ( - 1)( - {\rm{k}}) + {( - 1)^2}({\rm{j}})\\({\rm{j}} - {\rm{k}}) \times ({\rm{k}} - {\rm{i}}) = {\rm{i}} - 0 + {\rm{k}} + {\rm{j}}\\({\rm{j}} - {\rm{k}}) \times ({\rm{k}} - {\rm{i}}) = {\rm{i}} + {\rm{j}} + {\rm{k}}\end{array}\)

Thus, the cross product \((j - k) \times (k - i)\) is \(i + j + k\).