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Q11E

Expert-verifiedFound in: Page 564

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

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

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

**Consider the following property**

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

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 \(i + j + k\).

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