F 5 .Maths

(a) |a|2 = (1/2)2 + (√3/2)2 = 1
Therefore a is a unit vector.
(b) Slope of a = √3
So the equation of the required line is:
y - 2 = √3 (x - 1)
√3 x - y + (2 - √3) = 0
(c) The components of u in the direction along a can be found as:
(u . a)a = [6√3/2 + 6(√3/2)] (i/2 + √3j/2)
= (3√3 + 3√3) (i/2 + √3j/2)
= 6√3 (i/2 + √3j/2)
= 3√3 i + 9j
Hence, the components of u in the direction perpendicular to a can be found as:
u - (3√3 i + 9j) = (6√3 i + 6j) - (3√3 i + 9j)
= 3√3 i - 3j

eelyw: 你第三part好似計錯wor!!!
話明係u嘅components, 嗰兩個components一定係要垂直, 但你計到嗰兩個答案嘅dot product唔等於0, 咁即係有問題......

1.
Magnitude of a = sqrt [ (1/2)^2 + ( sqrt 3)^2/4 ] = sqrt [ 1/4 + 3/4 ] = sqrt 1 = 1.
Therefore, a is a unit vector.
2.
Slope of the required line = [sqrt(3)/2]/[(1/2)] = sqrt 3.
Therefore equation of line is
y - 2 = sqrt 3(x -1)
y = (sqrt 3)x + 2 - sqrt 3.
3.
u . a = (6 sqrt 3)(1/2) + (6)(sqrt 3)/2 = 6 sqrt 3.
Magnitude of u = sqrt [ 36 x 3 + 36] = sqrt ( 36 x 4) = 6 x 2 = 12.
Let angle between a and u is y, therefore cos y = (6 sqrt 3)/12 = (sqrt 3)/2
So y = 30 degree.
Therefore, vector in same direction of a is u . cos y =( sqrt 3)/2 [ 6 sqrt 3 i + 6j]
= 9i + 3 sqrt 3 j.
vector perpendicular to a is u. sin y = (1/2)[6 sqrt 3 i + 6j] = 3 sqrt 3 i + 3 j.