Question

Find the direction cosines and direction angles of the vector (3, 1, 2). Give the direction angles correct to the nearest tenth of a degree.

Answer 1

The direction cosines of the vector (3, 1, 2) are cos(α) = 3 / √14, cos(β) = 1 / √14, and cos(γ) = 2 / √14. The direction angles are α ≈ 53.1°, β ≈ 29.3°, and γ ≈ 67.8°.

Step-by-step explanation:

To find the direction cosines and direction angles of a vector, we use the formulas:

Direction cosines:

cos(α) = x / √(x^2 + y^2 + z^2)

cos(β) = y / √(x^2 + y^2 + z^2)

cos(γ) = z / √(x^2 + y^2 + z^2)

Direction angles:

α = arccos(x / √(x^2 + y^2 + z^2))

β = arccos(y / √(x^2 + y^2 + z^2))

γ = arccos(z / √(x^2 + y^2 + z^2))

For the vector (3, 1, 2), we have:

x = 3, y = 1, z = 2

cos(α) = 3 / √(3^2 + 1^2 + 2^2) = 3 / √14

cos(β) = 1 / √(3^2 + 1^2 + 2^2) = 1 / √14

cos(γ) = 2 / √(3^2 + 1^2 + 2^2) = 2 / √14

α = arccos(3 / √(3^2 + 1^2 + 2^2)) ≈ 53.1°

β = arccos(1 / √(3^2 + 1^2 + 2^2)) ≈ 29.3°

γ = arccos(2 / √(3^2 + 1^2 + 2^2)) ≈ 67.8°