Final answer:
The magnitude of vector a is 5 and the magnitude of vector b is √6. The cosine of the angle between a and b is 7/(5*√(6)). A unit vector perpendicular to both a and b is (8/√65)î - (1/√65)ĵ. Vectors u₁ and u₂ can be found as -2î - 4ĵ - 2k and 5î + 4ĵ + 6k respectively.
Step-by-step explanation:
(a) To find the magnitude of vector a, we can use the formula |a| = √(a_x² + a_y² + a_z²). Plugging in the values, we get |a| = √(3² + 0² + 4²) = √(9 + 16) = √(25) = 5. Similarly, |b| = √(1² + 2² + 1²) = √(1 + 4 + 1) = √(6).
(b) To find cos(θ), we can use the dot product formula: a · b = |a| |b| cos(θ). Plugging in the given values and the magnitudes from part (a), we get 3*1 + 0*2 + 4*1 = 5*√(6)*cos(θ). Solving for cos(θ), we get cos(θ) = (3 + 4)/(5*√(6)) = 7/(5*√(6)).
(c) To find a unit vector perpendicular to both a and b, we can use the cross-product formula: a x b = |a| |b| sin(θ) * n. Plugging in the given values and taking the cross product, we get (2(4) - 1(0))î + (1(3) - 1(4))ĵ + (3(0) - 2(0))k = 8î - 1ĵ. To find the unit vector, we divide by the magnitude: (8/√65)î - (1/√65)ĵ.
(d) To find vectors u₁ and u₂, we can use the fact that u₁ is parallel to b and u₂ is perpendicular to b. Since u₁ is parallel to b, we can write u₁ as a multiple of b: u₁ = kb, where k is a scalar. Since u₂ is perpendicular to b, we can write u₂ as the difference of a and u₁: u₂ = a - u₁. Let's solve for k: u₁ = kb = 1(1î + 2ĵ + 1k) = 1î + 2ĵ + 1k. Since u₂ is perpendicular to b, the dot product of u₂ and b is 0: u₂ · b = (3 - 1k)(1) + (0 - 2k)(2) + (4 - 1k)(1) = 0. Solving for k, we get k = -2. Substituting k into u₁, we get u₁ = -2î - 4ĵ - 2k. Finally, substituting k into u₂, we get u₂ = a - u₁ = (3î + 0ĵ + 4k) - (-2î - 4ĵ - 2k) = 5î + 4ĵ + 6k.