Answer:
uâ= â¨1/(7*â3),â5/(7*â3) ,â11/(7*â3)âİ
uâ= â¨-1/(7*â3),5/(7*â3) ,11/(7*â3)âİ
Step-by-step explanation:
for  a=â¨â2,â4,â2âİ and b=â¨â3,5,2âİ
a vector orthogonal to a and b can be found through the vectorial product of a and b. Thus
c= a x b[tex]c=\left[\begin{array}{ccc}i&j&k\\-2&-4&-2\\-3&5&2\end{array}\right] = \left[\begin{array}{ccc}-4&-2\\5&2\end{array}\right]*i+\left[\begin{array}{ccc}-2&-2\\-3&2\end{array}\right]*j+\left[\begin{array}{ccc}-2&-4\\-3&5\end{array}\right]*k = 2*i -10*j - 22*k[/tex]
then câ=â¨2,â10,â22âİ and câ= - câ= â¨-2,10,22âİ are orthogonal to a and b
the corresponding unit vectors are
uâ=câ/|câ| = â¨2,â10,â22âİ / â(2²+(â10)²+(â22)²) = â¨2,â10,â22âİ/(14*â3) =  â¨1/(7*â3),â5/(7*â3) ,â11/(7*â3)âİ
then uâ= - uâ=  â¨-1/(7*â3),5/(7*â3) ,11/(7*â3)âİ
then the unit vectors are
uâ= â¨1/(7*â3),â5/(7*â3) ,â11/(7*â3)âİ
uâ= â¨-1/(7*â3),5/(7*â3) ,11/(7*â3)âİ