Let vector A = 4i^ + 4j^, vector B = -2i^ - 5j^, and vector F = vector A - 5(vector B).
a) Write vector F in component form.
vector F = ?
b) What is the magnitude of vector F? F = ?
c) What is the direction of vector F?
theta = ?
Thank you!
To find vector F, we first need to find the scalar product of vector B with -5, and then subtract this from vector A.
a) The scalar product of a vector with a scalar is obtained by multiplying each component of the vector by the scalar. So, -5(vector B) is calculated as follows:
-5(vector B) = -5 * (-2i^ - 5j^) = 10i^ + 25j^
Now, subtract this from vector A to find vector F:
vector F = vector A - 5(vector B) = 4i^ + 4j^ - (10i^ + 25j^)
To perform the subtraction, combine like terms. Subtract the coefficients of i^ and j^ separately:
vector F = (4 - 10)i^ + (4 - 25)j^
= -6i^ - 21j^
So, vector F in component form is -6i^ - 21j^.
b) The magnitude of a vector can be found using the formula: magnitude of vector F = √(Fx^2 + Fy^2), where Fx and Fy are the x and y components of the vector respectively.
Using the values we obtained for vector F from part a:
Fx = -6, Fy = -21
Plug these values into the formula and calculate the magnitude of vector F:
magnitude of vector F = √((-6)^2 + (-21)^2)
= √(36 + 441)
= √477
≈ 21.82
So, the magnitude of vector F is approximately 21.82.
c) The direction of a vector can be found by calculating the angle it makes with the positive x-axis. This angle, denoted by θ, can be found using the formula: θ = arctan(Fy / Fx).
Using the values we obtained for vector F from part a:
Fx = -6, Fy = -21
Plug these values into the formula and calculate the direction of vector F:
θ = arctan((-21) / (-6))
= arctan(3.5)
≈ 74.92 degrees
So, the direction of vector F is approximately 74.92 degrees.