Diketahui [x] = √5 dan y = (3,4) jika sudut antara x dan y lancip dan panjang proyeksi x pada y = 2 tentukan x ? tolong ya

Misal x = (a, b) => |x| = √(a^2 + b^2) = √5 => a^2 + b^2 = 5
y = (3, 4) => |y| = √(3^2 + 4^2) = √(9 + 16) = √25 = 5
x.y = 3a + 4b

Proyeksi x pada y = 2
x.y / |y| = 2
(3a + 4b) / 5 = 2
3a + 4b = 10
3a = 10 - 4b
a = (10 - 4b)/3

a^2 + b^2 = 5
((10 - 4b)/3)^2 + b^2 = 5
(100 - 80b + 16b^2)/9 + b^2 = 5 ==> kali 9
100 - 80b + 16b^2 + 9b^2 = 45
25b^2 - 80b + 55 = 0
5b^2 - 16b + 11 = 0
(5b - 11)(b - 1) = 0
b = 11/5 atau b = 1
b = 11/5 => a = (10 - 4(11/5))/3 . 5/5 = (50 - 44)/15 = 6/15 = 2/5
b = 1 => a = (10 - 4(1))/3 = 6/3 = 2
Jadi
x = (2/5, 11/5) atau x = (2, 1)