Answer:
Let the vectors be
a = [0, 1, 2] and
b = [1, -2, 3]
( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.
Let the cross product be another vector c.
To find the cross product (c) of a and b, we have
[tex]\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right][/tex]
c = i(3 + 4) - j(0 - 2) + k(0 - 1)
c = 7i + 2j - k
c = [7, 2, -1]
( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:
c / | c |
Where | c | = √ (7)² + (2)² + (-1)² = 3√6
Therefore, the unit vector is
[tex]\frac{[7,2,-1]}{3\sqrt{6} }[/tex]
or
[ [tex]\frac{7}{3\sqrt{6} }[/tex] , [tex]\frac{2}{3\sqrt{6} }[/tex] , [tex]\frac{-1}{3\sqrt{6} }[/tex] ]
The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:
[ [tex]\frac{-7}{3\sqrt{6} }[/tex] , [tex]\frac{-2}{3\sqrt{6} }[/tex] , [tex]\frac{1}{3\sqrt{6} }[/tex] ]
In conclusion, the two unit vectors are;
[ [tex]\frac{7}{3\sqrt{6} }[/tex] , [tex]\frac{2}{3\sqrt{6} }[/tex] , [tex]\frac{-1}{3\sqrt{6} }[/tex] ]
and
[ [tex]\frac{-7}{3\sqrt{6} }[/tex] , [tex]\frac{-2}{3\sqrt{6} }[/tex] , [tex]\frac{1}{3\sqrt{6} }[/tex] ]
Hope this helps!
