Question

Find the two unit vectors orthogonal to both \(\langle 3,2,1\rangle \) and \(\langle - 1,1,0\rangle \).

Short answer

The two unit vectors are \(\left\langle { - \frac{1}{{3\sqrt 3 }}, - \frac{1}{{3\sqrt 3 }},\frac{5}{{3\sqrt 3 }}} \right\rangle \) and\(\left\langle {\frac{1}{{3\sqrt 3 }},\frac{1}{{3\sqrt 3 }}, - \frac{5}{{3\sqrt 3 }}} \right\rangle \).

Step-by-step solution

Theorem used

The cross product of two vectors is orthogonal to both vectors, when their dot product is equal to zero.

Write the expression to find unit vector of vector\({\bf{a}}\)

Unit vector\( = \frac{{\rm{a}}}{{|{\rm{a}}|}}\)

Find the cross product between \(\langle 3,2,1\rangle \) and \(\langle  - 1,1,0\rangle \)

As given, \(\langle 3,2,1\rangle \) and \(\langle - 1,1,0\rangle \).

Find the cross product between \(\langle 3,2,1\rangle \) and \(\langle - 1,1,0\rangle \)

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

Find unit vectors

\(\begin{array}{l}{\rm{Unit vectors }} = \frac{{ - {\rm{i}} - {\rm{j}} + 5{\rm{k}}}}{{| - {\rm{i}} - {\rm{j}} + 5{\rm{k}}|}}\\{\rm{Unit vectors }} = \pm \frac{{\langle - 1, - 1,5\rangle }}{{\sqrt {{{( - 1)}^2} + {{( - 1)}^2} + {{(5)}^2}} }}\\{\rm{Unit vectors }} = \pm \frac{{\langle - 1, - 1,5\rangle }}{{\sqrt {1 + 1 + 25} }}\\{\rm{Unit vectors }} = \pm \frac{{\langle - 1, - 1,5\rangle }}{{\sqrt {27} }}\end{array}\)

\(\begin{array}{l}{\rm{Unit vectors }} = \pm \frac{{\langle - 1, - 1,5\rangle }}{{3\sqrt 3 }}\\{\rm{Unit vectors}} = \pm \left\langle {\frac{{ - 1}}{{3\sqrt 3 }},\frac{{ - 1}}{{3\sqrt 3 }},\frac{5}{{3\sqrt 3 }}} \right\rangle \end{array}\)

The unit vectors are \(\left\langle { - \frac{1}{{3\sqrt 3 }}, - \frac{1}{{3\sqrt 3 }},\frac{5}{{3\sqrt 3 }}} \right\rangle \) and\(\left\langle {\frac{1}{{3\sqrt 3 }},\frac{1}{{3\sqrt 3 }}, - \frac{5}{{3\sqrt 3 }}} \right\rangle \).

Thus, the two unit vectors are \(\left\langle { - \frac{1}{{3\sqrt 3 }}, - \frac{1}{{3\sqrt 3 }},\frac{5}{{3\sqrt 3 }}} \right\rangle \) and\(\left\langle {\frac{1}{{3\sqrt 3 }},\frac{1}{{3\sqrt 3 }}, - \frac{5}{{3\sqrt 3 }}} \right\rangle \).