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 \(\frac{1}{2}i - j + \frac{3}{2}k.\)

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\({\rm{a}} = {\rm{i}} - {\rm{j}} - {\rm{k}}\)and\({\rm{b}} = \frac{1}{2}{\bf{i}} + {\rm{j}} + \frac{1}{2}{\rm{k}}\).

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

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

Thus, the cross product of vectors\(a\)and\(b\)is\(\frac{1}{2}i - j + \frac{3}{2}k.\)

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

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

Substitute\({\rm{a}} \times {\rm{b}} = \frac{1}{2}{\rm{i}} - {\rm{j}} + \frac{3}{2}{\rm{k}}\)and\({\rm{a}} = {\rm{i}} - {\rm{j}} - {\rm{k in }}({\rm{a}} \times {\rm{b}}) \cdot {\rm{a}}\)and obtain the dot product as follows

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

Substitute\({\rm{a}} \times {\rm{b}} = 11{\rm{i}} + 14{\rm{j}} - 2{\rm{k}}\)and\({\rm{b}} = \frac{1}{2}{\bf{i}} + {\rm{j}} + \frac{1}{2}{\rm{k in }}({\rm{a}} \times {\rm{b}}) \cdot {\rm{b}}\)and obtain the dot product as follows.

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