Answer:
uβ= β¨1/(7*β3),β5/(7*β3) ,β11/(7*β3)β©
uβ= β¨-1/(7*β3),5/(7*β3) ,11/(7*β3)β©
Step-by-step explanation:
for Β a=β¨β2,β4,β2β© and b=β¨β3,5,2β©
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 cβ=β¨2,β10,β22β© and cβ= - cβ= β¨-2,10,22β© are orthogonal to a and b
the corresponding unit vectors are
uβ=cβ/|cβ| = β¨2,β10,β22β© / β(2Β²+(β10)Β²+(β22)Β²) = β¨2,β10,β22β©/(14*β3) = Β β¨1/(7*β3),β5/(7*β3) ,β11/(7*β3)β©
then uβ= - uβ= Β β¨-1/(7*β3),5/(7*β3) ,11/(7*β3)β©
then the unit vectors are
uβ= β¨1/(7*β3),β5/(7*β3) ,β11/(7*β3)β©
uβ= β¨-1/(7*β3),5/(7*β3) ,11/(7*β3)β©