4. Diketahui vektor a= (1 x 2) vektor b= (2 1 -1) dan panjang proyeksi vektor a pada vektor b=2/√6. Jika sudut antara vektor a dan vektor b adalah alpha maka co

a = (1, x, 2)
b = (2, 1, -1)
|c| = 2/√6

[tex] |c| = \frac{ab}{ |b| } [/tex]
ab = (1, x, 2) (2, 1, -1) = (2 + x -2) = x
|b|² = (2)² + (1)² + (-1)²
|b|² = 4 + 1 + 1
|b|² = 6
|b| = √6
[tex] \frac{2}{ \sqrt{6} } = \frac{x}{ \sqrt{6} } \\ \frac{2}{\sqrt{6} } \times \sqrt{6} = x \\ 2 = x[/tex]

maka vektor a = (1, 2, 2) dan b = (2, 1, -1)
ab = (1, 2, 2) (2, 1, -1) = 2
|b| = √6
|a|² = (1)² + (2)² + (2)²
|a|² = 9
|a| = 3
masukan ke rumus
[tex]cos \: \alpha = \frac{ab}{ |a| |b| } [/tex]

[tex] cos \: \alpha = \frac{2}{3 \sqrt{6} } \\ cos \: \alpha = \frac{1}{ 9 } \sqrt{6} [/tex]