Question

You are given two vectors A= -3.00i^+7.00j^ and B⃗ =7.00ι^+2.00j^. Let the counterclockwise angles be positive.

1)What angle does A⃗ make with the ?
2)What angle does B⃗ make with the +x-axis?
3)Vector C⃗ is the sum of A⃗ and B⃗ , so C⃗ =A⃗ +B⃗ . What angle does C⃗ make with the +x-axis

Answer 1

The angles vectors A and B make with the +x-axis are found using the arctan function with their respective components, and vector C, being their sum, also has its angle calculated in the same way.

Step-by-step explanation:

To find the angles that vectors α and β make with the +x-axis, we use inverse trigonometric functions based on their components. The angle θ made by a vector with the +x-axis is given by θ = arctan(Δy/Δx), where Δy and Δx are the components of the vector along the y-axis and x-axis respectively.

• For vector A = -3.00i^ + 7.00j^, we have Δx = -3.00 and Δy = 7.00. The angle it makes with the +x-axis is θₐ = arctan(7.00/-3.00).
• For vector B = 7.00i^ + 2.00j^, Δx = 7.00 and Δy = 2.00. The angle it makes with the +x-axis is θₑ = arctan(2.00/7.00).
• Vector C is the sum of A and B, so C = A + B = (-3.00 + 7.00)i^ + (7.00 + 2.00)j^ = 4.00i^ + 9.00j^. The angle θC it makes with the +x-axis is θC = arctan(9.00/4.00).

Note that in practice, we need to consider the signs of the components to determine the correct quadrant for the angle.

Answer 2

1)


First we go found the value of
`\vec{A}`:


a² = (-3)² + 7²

a² = 9 + 49

a² = 58

a = √59

a = 7,68 u


Now we can found the angle with the +x-axis


ax = a · cos θ

3 = 7,68 · cos θ

3 / 7,68 = cos θ

0,39 = cos θ

0,39 ·
`\mathsf{cos^(-1)}` = θ

θ = 67º <----- This is the answer


2)


First we go found the value of
`\vec{B}`:


b² = 7² + 2²

b² = 49 + 4

b² = 53

b = √53

b = 7,28 u


Now we can found the angle with the +x-axis


bx = b · cos θ

7 = 7,28 · cos θ

7 / 7,28 = cos θ

0,96 ·
`\mathsf{cos^(-1)}` = θ

θ = 16º


For counterclockwise:


180 - 16 = 164º <----- This is the answer


3)



`\vec{C}=\vec{A}+\vec{B}`


`\vec{C}=(-3\vec{i}+7\vec{j})+(7\vec{i}+2\vec{j})`


`\vec{C}=(-3\vec{i}+7\vec{i})+(7\vec{j}+2\vec{j})`


`\vec{C}= 4\vec{i}+9\vec{j}`


Now we going to found the module of C vector:


c² = 9² + 4²

c² = 81 + 16

c² = 97

c = √97

c = 9,84 u


The angle is:


Cx = C · cos θ

4 = 9,84 · cos θ

4 / 9,84 = cos θ

0,40 ·
`\mathsf{cos^(-1)}` = θ

θ = 66º


For counterclockwise:


180 - 66 = 114º <----- This is the answer


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