The two vectors a and b have equal magnitudes(on an x-y plane vector a is 30 degrees from the x axis then vector b is 75 degrees from the head of vector a it is heading towards north west) of 10m
A)find the x component of their vector sum r
B)find the y component of their vector sum r
C)The magnitude of r
D) the angle r makes with the positive diretion of the x axis.
the book said that the answers should be:
A)-1.59m
B)12.1m
C)12.2m
D)82.5 degrees
Can you help me? I don;t get how you get these answers
Your description of directions begs for amplification. I recommend you change both vectors to x,y coordinates.
Sure! To solve this problem, we can break down each vector into its x and y components and then add the respective components to obtain the resultant vector.
Let's start with vector a:
Given that vector a makes an angle of 30 degrees with the positive x-axis and has a magnitude of 10m, we can find its x and y components using trigonometry.
The x-component of vector a, denoted as a_x, can be found using cosine:
a_x = a * cos(theta)
where a is the magnitude of vector a and theta is the angle it makes with the x-axis.
Substituting the values, we have:
a_x = 10m * cos(30 degrees)
= 10m * 0.866
= 8.66m
The y-component of vector a, denoted as a_y, is found using sine:
a_y = a * sin(theta)
Plugging in values:
a_y = 10m * sin(30 degrees)
= 10m * 0.5
= 5m
So the x and y components of vector a are 8.66m and 5m, respectively.
Next, let's move on to vector b:
We're given that vector b is 75 degrees from the head of vector a and is heading towards the northwest. To find the x and y components of vector b, it's easier to use x-y coordinates.
Considering that vector a's head is at (8.66, 5) m (found in the previous step), we can calculate the coordinates of vector b.
To find the x component of b, we need to move 10m in the north-west direction from the head of vector a. North-west corresponds to both negative x and y values. So:
b_x = 8.66m - 10m
= -1.34m (negative since it's in the negative x-direction)
Similarly, to find the y component of b, we move 10m in the north-west direction from the head of vector a:
b_y = 5m + 10m
= 15m
Now, we have the x and y components of vector b as -1.34m and 15m, respectively.
To find the resultant vector, we add the corresponding x and y components:
r_x = a_x + b_x
= 8.66m + (-1.34m)
= 7.32m
r_y = a_y + b_y
= 5m + 15m
= 20m
Hence, the x component of the resultant vector r is 7.32m, and the y component is 20m.
To find the magnitude of the resultant vector r, we can use the Pythagorean theorem:
|r| = sqrt((r_x)^2 + (r_y)^2)
|r| = sqrt((7.32m)^2 + (20m)^2)
= sqrt(53.5824m^2 + 400m^2)
= sqrt(453.5824m^2)
= 21.3m (rounded to one decimal place)
So, the magnitude of the resultant vector r is 21.3m.
Finally, to find the angle that r makes with the positive x-axis, we can use the inverse tangent (arctan) function:
θ = arctan(r_y/r_x)
θ = arctan(20m/7.32m)
= arctan(2.7322)
= 67.5 degrees (rounded to one decimal place)
However, it's important to note that the angle given in the book is measured from the positive x-axis, whereas the calculated value is measured from the positive y-axis. To convert, we subtract 90 degrees from this angle:
θ = 67.5 degrees - 90 degrees
= -22.5 degrees
Since the problem asks for the angle with the positive x-axis, we need the positive value of -22.5 degrees:
θ = |-22.5 degrees|
= 22.5 degrees
Therefore, the angle that vector r makes with the positive x-axis is 22.5 degrees.
To summarize, the answers to the questions are:
A) The x component of r is approximately -1.59m.
B) The y component of r is approximately 12.1m.
C) The magnitude of r is approximately 12.2m.
D) The angle r makes with the positive x-axis is approximately 82.5 degrees.