1. Find the component form and magnitude of a vector with the initial point 𝑃(2, −3) and 𝑄(−6, 7)

Question

2.an airplane is flying south going 440 mph when it hits a crosswind going west at 35 mph. What is the resultant velocity?

3. Find u – 2v given that u =(−2, 5) and v = 6i – 4j

4. Vector v has a direction of 〈5, 3〉. Find the direction angle for v to the nearest degree.

Answer 1

1. To find the component form of the vector, we subtract the coordinates of the initial point from the coordinates of the terminal point:

𝑄 − 𝑃 = (−6, 7) − (2, −3) = (−8, 10)

So the component form of the vector is 〈−8, 10〉. To find the magnitude, we use the formula:

|𝑄 − 𝑃| = √((-8)^2 + 10^2) = √(164) ≈ 12.806

Therefore, the magnitude of the vector is approximately 12.806.

2. We can think of the airplane's velocity and the velocity of the crosswind as two vectors that are being added together to get the resultant velocity. The southward velocity can be represented by a vector with a magnitude of 440 mph and a direction of due south (which we can represent with a unit vector in the negative y-direction: 〈0, −1〉). The westward velocity can be represented by a vector with a magnitude of 35 mph and a direction of due west (which we can represent with a unit vector in the negative x-direction: 〈−1, 0〉). So we have:

v_{airplane} = 440〈0, −1〉
v_{crosswind} = 35〈−1, 0〉

To find the resultant velocity, we add these two vectors:

v_{resultant} = v_{airplane} + v_{crosswind} = 440〈0, −1〉 + 35〈−1, 0〉
= 〈0, −440〉 + 〈−35, 0〉
= 〈−35, −440〉

Therefore, the resultant velocity is 〈−35, −440〉, which means the airplane is flying in a direction that is 85.6 degrees south of west (using a calculator to find the inverse tangent: tan⁻¹(440/35) ≈ 85.6 degrees).

3. To find u – 2v, we first need to find 2v:

2v = 2(6i − 4j) = 12i − 8j

Then we can subtract 2v from u:

u − 2v = (−2, 5) − (12i − 8j) = (−2 − 12, 5 − (−8)) = (−14, 13)

So u – 2v has the component form 〈−14, 13〉.

4. To find the direction angle of a vector, we use the formula:

θ = tan⁻¹(y/x)

where y and x are the y- and x-components of the vector, respectively. In this case, the vector has components 〈5, 3〉, so we have:

θ = tan⁻¹(3/5) ≈ 31 degrees

Therefore, the direction angle for the vector to the nearest degree is 31 degrees.