Let u have a length of 20 and direction of 60 degrees, and v= 5i-3j
what is u?
what is abs(2v-u)?
let u = [a,b]
then a^2 + b^2 = 20 and tan60° = b/a = √3
b = √3 a
back in a^2 + b^2 = 20
a^2 + (√3a)^2 = 20
a^2 + 3a^2 = 20
a^2 = 5
a = ±√5
b = ±√15
u = [√5,√15] or [-√5,-√15] or in your notation
u = ±√5i ± √15
then 2v - u = (10-√5)i + (√15-3)j
abs(2v-u) = √(100 - 20√5 + 5 + 15 - 6√15 + 9)
= etc
repeat for the other value of u
solve the rest because i am not getting correct answer?
If you're not getting the correct answer, I suggest you post how you got your incorrect answer so that we will help you find what went wrong. This is the whole purpose of solving problems.
To find u given its length and direction, we can use trigonometry.
In this particular case, we know that the magnitude of u is 20, so its length is 20. The direction of u is given as 60 degrees.
Using trigonometry, we can express u in terms of its x and y components. Let's call the x-component of u as u_x and the y-component as u_y.
The magnitude of u can be calculated using the Pythagorean theorem:
|u| = sqrt(u_x^2 + u_y^2)
Since |u| = 20, we have:
20 = sqrt(u_x^2 + u_y^2) (1)
The direction of u can be expressed using the tangent function:
tan(theta) = u_y / u_x
Where theta is the direction angle given as 60 degrees. We can rearrange the equation to solve for u_y:
u_y = u_x * tan(theta) (2)
Now, we can substitute equation (2) into equation (1) to solve for u_x:
20 = sqrt(u_x^2 + (u_x * tan(theta))^2)
Squaring both sides:
400 = u_x^2 + (u_x^2 * tan(theta))^2
Combining like terms:
400 = u_x^2 * (1 + tan(theta)^2)
Dividing both sides by (1 + tan(theta)^2):
u_x^2 = 400 / (1 + tan(theta)^2)
Taking the square root:
u_x = sqrt(400 / (1 + tan(theta)^2))
u_x = 20 / sqrt(1 + tan(theta)^2)
Note: Since theta is given as 60 degrees, we can substitute tan(theta) with the exact value sqrt(3).
u_x = 20 / sqrt(1 + (sqrt(3))^2)
u_x = 20 / sqrt(1 + 3)
u_x = 20 / sqrt(4)
u_x = 20 / 2
u_x = 10
Now that we found u_x, we can substitute it back into equation (2) to find u_y:
u_y = 10 * tan(theta)
u_y = 10 * sqrt(3)
Therefore, the vector u is u = 10i + 10sqrt(3)j.
Now let's move on to the second question.
To find |2v - u|, we need to calculate the vector 2v - u, then find its magnitude.
Given v = 5i - 3j and u = 10i + 10sqrt(3)j, we can calculate 2v - u:
2v - u = 2(5i - 3j) - (10i + 10sqrt(3)j)
2v - u = 10i - 6j - 10i - 10sqrt(3)j
2v - u = -6j - 10sqrt(3)j
Combining like terms:
2v - u = (-6 - 10sqrt(3))j
To find the magnitude of a vector, we use the equation:
|a| = sqrt(a_x^2 + a_y^2)
In this case, our vector is -6 - 10sqrt(3)j, so we can find its magnitude:
|2v - u| = sqrt((-6)^2 + (-10sqrt(3))^2)
|2v - u| = sqrt(36 + 300)
|2v - u| = sqrt(336)
Therefore, |2v - u| is equal to sqrt(336) or approximately 18.33.