The three vectors have magnitude a =3m, b= 4m and c=10m . On the x axis is vector a at 0 degrees then 30 degrees is vector b then 120 degrees is vector c. If vector c =p*vector a + q*vector b .What are the values of p? and q?
I am expose to get -6.67 for p and then 4.33 for q. I don't know what i am doing wrong.
Let me see your work on a, b, c on converting them to s,y coordinates as in
a= something x + something y.
I suspect that is your error.
To find the values of p and q in the equation c = p * a + q * b, we need to convert the given vectors a and b into x and y components.
First, let's convert vector a:
a = 3m at 0 degrees
Since the vector lies on the x-axis, its y-component will be zero:
a_x = magnitude of a * cos(angle of a)
= 3m * cos(0 degrees)
= 3m
a_y = magnitude of a * sin(angle of a)
= 3m * sin(0 degrees)
= 0m
So, a = 3m * (1 x + 0 y)
Next, let's convert vector b:
b = 4m at 30 degrees
To find the x and y components, we can use the following trigonometric formulas:
b_x = magnitude of b * cos(angle of b)
= 4m * cos(30 degrees)
= 4m * (√3/2)
= 2√3m
b_y = magnitude of b * sin(angle of b)
= 4m * sin(30 degrees)
= 4m * (1/2)
= 2m
So, b = 2√3m * (1 x + 2m y)
Now, let's substitute the values of a and b in the equation c = p * a + q * b:
c = p * (3m * (1 x + 0 y)) + q * (2√3m * (1 x + 2m y))
Simplifying this equation, we obtain:
c = (3p + 2q√3) m x + (2q) m y
Comparing the x and y components on both sides of the equation, we get:
3p + 2q√3 = 10m (1)
2q = 0m (2) <--- This equation implies q = 0 (since 2q cannot be equal to 0)
From equation (2), we find that q = 0.
Substituting q = 0 in equation (1), we get:
3p + 2(0√3) = 10m
3p = 10m
p = 10m/3
Therefore, the value of p is p = 10/3 or approximately p = 3.33.
It seems that there was a minor calculation error on your part. The correct value of p is 10/3 or approximately 3.33, not -6.67. The value of q is indeed 0 since 2q = 0.
To recap, the values of p and q are approximately p = 3.33 and q = 0, respectively.