1. To find the component form of the vector, we subtract the coordinates of the initial point from the coordinates of the terminal point:
π β π = (β6, 7) β (2, β3) = (β8, 10)
So the component form of the vector is γβ8, 10γ. To find the magnitude, we use the formula:
|π β π| = β((-8)^2 + 10^2) = β(164) β 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: γ0, β1γ). 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: γβ1, 0γ). So we have:
v_{airplane} = 440γ0, β1γ
v_{crosswind} = 35γβ1, 0γ
To find the resultant velocity, we add these two vectors:
v_{resultant} = v_{airplane} + v_{crosswind} = 440γ0, β1γ + 35γβ1, 0γ
= γ0, β440γ + γβ35, 0γ
= γβ35, β440γ
Therefore, the resultant velocity is γβ35, β440γ, 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: tanβ»ΒΉ(440/35) β 85.6 degrees).
3. To find u β 2v, we first need to find 2v:
2v = 2(6i β 4j) = 12i β 8j
Then we can subtract 2v from u:
u β 2v = (β2, 5) β (12i β 8j) = (β2 β 12, 5 β (β8)) = (β14, 13)
So u β 2v has the component form γβ14, 13γ.
4. To find the direction angle of a vector, we use the formula:
ΞΈ = tanβ»ΒΉ(y/x)
where y and x are the y- and x-components of the vector, respectively. In this case, the vector has components γ5, 3γ, so we have:
ΞΈ = tanβ»ΒΉ(3/5) β 31 degrees
Therefore, the direction angle for the vector to the nearest degree is 31 degrees.
1. Find the component form and magnitude of a vector with the initial point π(2, β3) and π(β6, 7)
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 β 2v given that u =(β2, 5) and v = 6i β 4j
4. Vector v has a direction of γ5, 3γ. Find the direction angle for v to the nearest degree.
1 answer