Question

Show that a vector

Answer 1

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∝ =$\frac{x}{r}$

                                 x₁ = rcos∝

InΔOBP , ∠B =90° , cosβ =$\frac{y}{r}$

                                 y₁ = rcosβ

InΔOCP , ∠C =90° , cosγ =$\frac{z}{r}$

                                z₁ = rcosγ

Given   u = x₁ i + y₁j +z₁k

      |u| = $\sqrt{(x_{1})^{2} +(x_{2} )^{2} +(x_{3} )^{2} }$

Therefore  u = x₁ i + y₁j +z₁k

               u = |u|(cos∝+cosβ+cosγ)

         

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