Question

Determine the dot product of the vector\(a\)and\(b\)and verify\(a \times b\) is orthogonal on both\(a\)and \(b.\)

Short answer

The cross product of vectors \(a\)and \(b\)is \(15{\rm{i}} - 3{\rm{j}} + 3{\rm{k}}.\)

Step-by-step solution

The cross product formula.

If the vectors\({\bf{a}} = \left\langle {{a_1},{a_2},{a_3}} \right\rangle \)and\({\bf{b}} = \left\langle {{b_1},{b_2},{b_3}} \right\rangle {\rm{,}}\)then the cross product of\({\bf{a}}\)and\({\bf{b}}\)is\({\rm{a}} \times {\rm{b}} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right|\).

Use the cross product formula for calculation.

The two given vectors are \({\rm{a}} = {\bf{i}} + 3{\rm{j}} - 2{\rm{k}}\) and \({\rm{b}} = - {\bf{i}} + 5{\rm{k}}\).

Apply the formula with \({a_1} = 1,{a_2} = 3,{a_3} = - 2,{b_1} = - 1,{b_2} = 0,\) and \({b_3} = 5\)and obtain the cross product of two vectors.

\(\begin{array}{l}{\rm{a}} \times {\rm{b}} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\1&3&{ - 2}\\{ - 1}&0&5\end{array}} \right|\\ = \left| {\begin{array}{*{20}{c}}3&{ - 2}\\0&5\end{array}} \right|{\rm{i}} - \left| {\begin{array}{*{20}{c}}1&{ - 2}\\{ - 1}&5\end{array}} \right|{\rm{j}} + \left| {\begin{array}{*{20}{c}}1&3\\{ - 1}&0\end{array}} \right|{\rm{k}}\\ = (15 + 0){\rm{i}} - (5 - 2){\rm{j}} + (1 + 3){\rm{k}}\\ = 15{\rm{i}} - 3{\rm{j}} + 3{\rm{k}}\end{array}\)

Thus, the cross product of vectors \(a\)and \(b\)is \(15{\rm{i}} - 3{\rm{j}} + 3{\rm{k}}.\)

Note that the cross product \(a \times b\) is orthogonal to both \(a\) and \(b\) if \((a \times b) \cdot a = 0\) and\((a \times b) \cdot b = 0\).

Rewrite the vector \(a = \langle 1,1, - 1\rangle \) in terms of unit vector as \({\rm{a}} = {\bf{i}} + {\rm{j}} - {\rm{k}}\)

Calculate the vector value\((a \times b) \cdot a\) as follows.

Substitute \({\rm{a}} \times {\rm{b}} = 15{\rm{i}} - 3{\rm{j}} + 3{\rm{k}}\) and \({\rm{a}} = {\bf{i}} + 3{\rm{j}} - 2{\rm{k in }}({\rm{a}} \times {\rm{b}}) \cdot {\rm{a}}\)and obtain the dot product as follows

Calculate the vector value\((a \times b) \cdot b\) as follows.

Substitute \({\rm{a}} \times {\rm{b}} = 15{\rm{i}} - 3{\rm{j}} + 3{\rm{k}}\)and \({\rm{b}} = - {\bf{i}} + 5{\rm{k in }}({\rm{a}} \times {\rm{b}}) \cdot {\rm{b}}\) and obtain the dot product as follows.

Thus, the cross product\(a \times b\)is orthogonal on both\(a\) and\(b.\)