The direction cosines of the vector r with direction angles α, β, γ to the x, y and z axis are
- cosα = x/r = 3√2/10
- cosβ = y/r = 4√2/10 and
- cosγ = z/r = √2/2
So, the direction angles are
- α = cos⁻¹(x/r) = cos⁻¹(3√2/10) = 64.9°
- β = cos⁻¹(y/r) = cos⁻¹(4√2/10) = 55.6° and
- γ = cos⁻¹(z/r) = cos⁻¹(√2/2) = 45°
How to find the direction cosines and direction angles of the vector?
Given a vector r = xi + yj + zk where
- x, y and z are the components of the vector in the x, y and z directions. and
- r = √(x² + y² + z²)
The direction cosines of the vector r with direction angles α, β, γ to the x, y and z axis are
- cosα = x/r,
- cosβ = y/r and
- cosγ = z/r
So, their direction angles are
- α = cos⁻¹(x/r),
- β = cos⁻¹(y/r) and
- γ = cos⁻¹(z/r)
As an example, let r = 3i + 4j + 5k
So, r = √(x² + y² + z²)
= √(3² + 4² + 5²)
= √(9 + 16 + 25)
= √50
= 5√2
The direction cosines of the vector r with direction angles α, β, γ to the x, y and z axis are
- cosα = x/r = 3/5√2 = 3√2/10
- cosβ = y/r = 4/5√2 = 4√2/10 and
- cosγ = z/r = 5/5√2 = 1/√2 = √2/2
So, the direction angles are
- α = cos⁻¹(x/r) = cos⁻¹(3√2/10) = cos⁻¹(0.4243) = 64.9°
- β = cos⁻¹(y/r) = cos⁻¹(4√2/10) = cos⁻¹(0.5657) = 55.6° and
- γ = cos⁻¹(z/r) = cos⁻¹(√2/2) = cos⁻¹(0.5657) = 45°
Learn more about direction cosine here:
https://brainly.com/question/27228226
#SPJ4
