Final answer:
a. The magnitude of v₁ is approximately 7.211, and the magnitude of v₂ is approximately 6.708. b. The angle between v₁ and v₂ is approximately 63.535°. c. The scalar projection of v₁ onto v₂ is approximately 3.155. d. The projection of v₁ onto v₂ is approximately (0.470, -0.940).
Step-by-step explanation:
a. To find the magnitude of a vector, we use the formula: |v| = √(x² + y²), where v = (x, y).
For v₁ = (-6, 4):
|v₁| = √((-6)² + 4²) = √(36 + 16) = √52 ≈ 7.211.
For v₂ = (-3, 6):
|v₂| = √((-3)² + 6²) = √(9 + 36) = √45 ≈ 6.708.
b. To find the angle between two vectors, we use the formula: cosθ = (v₁ · v₂) / (|v₁| |v₂|), where θ is the angle between the vectors and · represents the dot product.
For v₁ and v₂:
θ = cos⁻¹((v₁ · v₂) / (|v₁| |v₂|)) = cos⁻¹((-6 * -3 + 4 * 6) / (7.211 * 6.708)) ≈ cos⁻¹(22 / 48.409) ≈ 63.535°.
c. To find the scalar projection of v₁ onto v₂, we use the formula: projᵥ₁ᵥ₂ = |v₁| * cosθ.
For v₁ and v₂:
projᵥ₁ᵥ₂ = |v₁| * cosθ = 7.211 * cos(63.535°) ≈ 3.155.
d. To find the projection of v₁ onto v₂, we use the formula: projv₁v₂ = (projᵥ₁ᵥ₂ / |v₂|) * v₂.
For v₁ and v₂:
projv₁v₂ = (projᵥ₁ᵥ₂ / |v₂|) * v₂ = (3.155 / 6.708) * (-3, 6) ≈ (0.470, -0.940).