Question

Find the vector of using cross product.

\((i + j) \times (i - j)\)

Short answer

The vector \((i + j) \times (i - j)\) is\( - 2k\).

Step-by-step solution

Formula used

The following property is consider

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

Find the resultant vector

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

Calculate the value of\((i + j) \times (i - j)\)as follows

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

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

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

Thus, the vector \((i + j) \times (i - j)\) is\( - 2k\).

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

Modify equation (1)

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

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 \(i + j + k\).