Consider the vector 𝐴⃑ in the first quadrant, with magnitude

Question

Consider the vector 𝐴⃑ in the first quadrant, with magnitude

given by 𝐴 = 10, that makes an angle of 𝜃 = 25° with the 𝑥-
axis.

(a) Find a vector, 𝐵⃑⃑, such that 𝐴⃑ + 𝐵⃑⃑ = 0. What is the
magnitude of 𝐵⃑⃑? Carefully sketch both 𝐴⃑ and 𝐵⃑⃑, making
sure they are proportional.
(b) Find a unit vector, 𝑢̂, such that 𝐴⃑ ⋅ 𝑢̂ = 0.
(c) Let 𝑉⃑⃑ = 5 x̂ − 7 ŷ. Compute 𝐴⃑ × 𝑉⃑⃑ and 𝑉⃑⃑ × 𝐴⃑. Compare your answers; does this make sense?

Answer 1

(a) clearly, B = -A

(b) u must point in a direction 25±90°, right?
so figure the x and y components
(c) I assume you can figure the cross product. Your calculations should verify that AxV = -(VxA)
(you know that switching rows in a determinant changes its sign, right?)

Answer 2

(a) To find the vector B, we need to consider that the sum of A and B is zero, which means that they have equal magnitudes but opposite directions.

Given that the magnitude of A is 10, we can find the magnitude of B as follows:
|A| = |B| (Equal magnitudes)
10 = |B|

Now, let's find the direction of B. Since A makes an angle of 25° with the x-axis, B should make an angle of 180° - 25° = 155° with the x-axis to have opposite direction.

Therefore, the vector B can be written as:
B⃑ = 10(cos(155°)x̂ + sin(155°)ŷ)

To sketch A and B, we draw two vectors with the same length but opposite direction, starting from the same origin point.

(b) To find a unit vector û such that the dot product of A and û is zero, we know that their dot product (A⃑ ⋅ û) is given by:
A⃑ ⋅ û = |A||û|cos(theta)

Since the dot product equals zero, we have:
A⃑ ⋅ û = 0
|A||û|cos(theta) = 0
|A|cos(theta) = 0

Since the magnitude of A is non-zero, we obtain:
cos(theta) = 0

The angle theta for A is 25°, so the only solution for cos(theta) = 0 is when theta = 90°.

Thus, a unit vector û such that A⃑ ⋅ û = 0 is a vector that makes an angle of 90° with the x-axis. We can write this unit vector as:
û = cos(90°)x̂ + sin(90°)ŷ
û = ŷ

(c) Given V⃑ = 5x̂ - 7ŷ, we can compute the cross product of A⃑ and V⃑ using the formula:
A⃑ × V⃑ = |A||V⃑|sin(theta) * n̂

Here, n̂ is the unit vector perpendicular to the plane formed by A⃑ and V⃑. Since A⃑ lies in the first quadrant, V⃑ should be in the fourth quadrant because the cross product follows the right-hand rule (thumb in the direction of A⃑, fingers in the direction of V⃑, and the palm gives the direction of the cross product).

Therefore, V⃑ = -5x̂ + 7ŷ.

Now, calculating the cross product:
A⃑ × V⃑ = |A||V⃑|sin(theta) * n̂
A⃑ × V⃑ = |A||V⃑|sin(180° - theta) * n̂

Using the given values, we have:
A⃑ × V⃑ = 10 * |-5x̂ + 7ŷ| * sin(180° - 25°) * n̂
A⃑ × V⃑ = 10 * |-5x̂ + 7ŷ| * sin(155°) * n̂

Similarly, calculating the cross product of V⃑ and A⃑:
V⃑ × A⃑ = |V⃑||A|sin(theta) * n̂
V⃑ × A⃑ = |-5x̂ + 7ŷ| * |A|sin(180° - theta) * n̂

Substituting the given values:
V⃑ × A⃑ = |-5x̂ + 7ŷ| * 10 * sin(155°) * n̂

Comparing both cross products, we can observe that they are equal in magnitude but have opposite directions, indicating that the cross product is anti-commutative. This makes sense as the cross product represents a vector orthogonal (perpendicular) to both input vectors, and changing the order of vectors will not affect the direction of the perpendicular vector.