Find two unit vectors orthogonal to a=βŸ¨βˆ’2,βˆ’4,βˆ’2⟩a=βŸ¨βˆ’2,βˆ’4,βˆ’2⟩ and b=βŸ¨βˆ’3,5,2⟩b=βŸ¨βˆ’3,5,2⟩ Enter your answer so that the first non-zero coordinate of the first vector is positive. First Vector: ⟨⟨ , , ⟩⟩ Second Vector: ⟨⟨ , , ⟩⟩ Note: You can earn partial credit on this problem.

Respuesta :

Answer:

u₁= ⟨1/(7*√3),βˆ’5/(7*√3) ,βˆ’11/(7*√3)⟩

uβ‚‚= ⟨-1/(7*√3),5/(7*√3) ,11/(7*√3)⟩

Step-by-step explanation:

for Β a=βŸ¨βˆ’2,βˆ’4,βˆ’2⟩ and b=βŸ¨βˆ’3,5,2⟩

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 c₁=⟨2,βˆ’10,βˆ’22⟩ and cβ‚‚= - c₁= ⟨-2,10,22⟩ are orthogonal to a and b

the corresponding unit vectors are

u₁=c₁/|c₁| = ⟨2,βˆ’10,βˆ’22⟩ / √(2Β²+(βˆ’10)Β²+(βˆ’22)Β²) = ⟨2,βˆ’10,βˆ’22⟩/(14*√3) =  ⟨1/(7*√3),βˆ’5/(7*√3) ,βˆ’11/(7*√3)⟩

then uβ‚‚= - u₁=  ⟨-1/(7*√3),5/(7*√3) ,11/(7*√3)⟩

then the unit vectors are

u₁= ⟨1/(7*√3),βˆ’5/(7*√3) ,βˆ’11/(7*√3)⟩

uβ‚‚= ⟨-1/(7*√3),5/(7*√3) ,11/(7*√3)⟩