Question

Part A: /2 (Subtract initial from terminal - work) /2 (Correct linear forms)

Part B: /2 (Magnitude) /2 (Reference Angle) /2 (Correct angle)12 (Trig form)
Part C /1(Correct vector) 12 (work shown)
Vector u= PQ has initial point P (2, 14) and terminal point Q (7, 3). Vector v=RS has initial point R (29, 8) and terminal point S (12, 17).
Part A: Write u and v in linear form. Show all necessary work. (4 points)

Part B: Write u and v in trigonometric form. Show all necessary work. (8 points)

Part C: Find 7u-4v. Show all necessary calculations.

Answer 1

To write vectors u and v in linear form, the difference between terminal and initial points is taken. For trigonometric form, magnitudes and reference angles are found. Finally, to find 7u-4v, scalar multiplication and subtraction of components are performed.

Step-by-step explanation:

Writing Vectors in Linear and Trigonometric Forms

The vector u has initial point P (2, 14) and terminal point Q (7, 3). The linear form can be found by subtracting the initial point from the terminal point: u = ⟨7 - 2, 3 - 14⟩ = ⟨5, -11⟩.

The vector v has initial point R (29, 8) and terminal point S (12, 17). Its linear form is: v = ⟨12 - 29, 17 - 8⟩ = ⟨-17, 9⟩.

To write vectors u and v in trigonometric form, we need to calculate their magnitudes and reference angles. The magnitude of u is |u| = √(5² + (-11)²) ≈ 11.66, and the reference angle θ is found using tan-1(-11/5). For vector v, calculate its magnitude as |v| = √((-17)² + 9²) ≈ 19.24, and find its reference angle with tan-1(9/-17). Remember to adjust the angles to their correct quadrants.

For part C, to find 7u - 4v, multiply each component of u and v by 7 and -4, respectively, and then subtract: (7 X u) - (4 X v) = (7 X ⟨5,-11⟩) - (4 X ⟨-17, 9⟩) = ⟨35, -77⟩ - ⟨-68, 36⟩ = ⟨35 + 68, -77 - 36⟩ = ⟨103, -113⟩.