Answer:
u = |u|(cos∝+cosβ+cosγ)
Step-by-step explanation:
<u>Explanation</u>
<u>Proof:-</u>
Given a vector u = x₁ i + y₁j +z₁k
let O X, OY, O Z be the positive co-ordinate axes
P(x₁,y₁,z₁) be any point in the space
Let OP makes angles α,β,γ with co-ordinate axes OX , OY ,OZ .
The angle α,β,γ are known as direction angles and cosine of the angle
l =cosα , m= cosβ , n=cosγ
The perpendicular PA,PB,PC are drawn co-ordinate axes OX,OY,OZ respecctively
InΔOAP , ∠A =90° , cos∝ =
x₁ = rcos∝
InΔOBP , ∠B =90° , cosβ =
y₁ = rcosβ
InΔOCP , ∠C =90° , cosγ =
z₁ = rcosγ
Given
u = x₁ i + y₁j +z₁k
|u| = 
Therefore u = x₁ i + y₁j +z₁k
u = |u|(cos∝+cosβ+cosγ)