Answer:
u = |u|(cos∝+cosβ+cosγ)
Step-by-step explanation:
Explanation
Proof:-
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∝ =[tex]\frac{x}{r}[/tex]
x₁ = rcos∝
InΔOBP , ∠B =90° , cosβ =[tex]\frac{y}{r}[/tex]
y₁ = rcosβ
InΔOCP , ∠C =90° , cosγ =[tex]\frac{z}{r}[/tex]
z₁ = rcosγ
Given u = x₁ i + y₁j +z₁k
|u| = [tex]\sqrt{(x_{1})^{2} +(x_{2} )^{2} +(x_{3} )^{2} }[/tex]
Therefore u = x₁ i + y₁j +z₁k
u = |u|(cos∝+cosβ+cosγ)