I need help with my vectors unit. I'm really confused about how to set them up. Please help.

a.) Given u=2i-5j and v=-i+3j, determine -6u-3v?

b.) Find the component form of v if |v|=4 and the angle it makes with the x-axis is 60 degree.

c.) Given u=5i+3j and v+3i+2j, find (u)(v).

d.) Find the angle between the vectors u and v if u=-4i+2j and v=3i+2j.

UPDATE: I figured it out.

-6u-3v = -6(2i-5j) - 3(-i+3j)

= -12i+30j+3i-9j
= -9i+21j
|-6u-3v| = √(81+441) = √522

x = |v| cosθ and y = |v| sinθ
Since cos60° = 1/2 and sin60° = √3/2
v = (4(1/2),4(√3/2)) = (2,2√3)

I assume that (u)(v) means u•v
If so, then
u•v = (5i+3j)•(3i+2j) = 5*3 + 3*2 = 21

Since u•v = |u|*|v| cosθ, we have
-12 + 4 = √20 * √13 cosθ
cosθ = -8/√260
θ = 119.74°

Sure, I can help you with your vectors questions! Let's go through each question step by step:

a.) Given u=2i-5j and v=-i+3j, we want to find -6u-3v.
To find -6u, we simply multiply each component of u by -6:
-6u = (-6)(2i) + (-6)(-5j) = -12i + 30j.

Similarly, to find -3v, we multiply each component of v by -3:
-3v = (-3)(-i) + (-3)(3j) = 3i - 9j.

Finally, we add -6u and -3v together:
-6u - 3v = (-12i + 30j) + (3i - 9j) = -9i + 21j.

So, -6u-3v = -9i + 21j.

b.) We're given that |v| = 4, which means the magnitude or the length of vector v is 4. We also know that the angle it makes with the x-axis is 60 degrees.

We can represent v as v = xi + yj, where x and y are the unknown components of v.

Since the magnitude of v is 4, we can use the Pythagorean theorem to find y:
|v| = sqrt(x^2 + y^2)
4 = sqrt(x^2 + y^2)
Squaring both sides:
16 = x^2 + y^2

Next, we can use the angle it makes with the x-axis to find x and y. The x-component of v can be found by multiplying the magnitude of v (4) by the cosine of the angle (60 degrees):
x = 4 * cos(60) = 4 * (1/2) = 2.

Similarly, the y-component of v can be found by multiplying the magnitude of v (4) by the sine of the angle (60 degrees):
y = 4 * sin(60) = 4 * (√3/2) = 2√3.

So, the component form of v is v = 2i + 2√3j.

c.) Given u = 5i + 3j and v = 3i + 2j, we need to find the dot product of u and v, denoted as (u)(v).

The dot product of two vectors is calculated by multiplying corresponding components and then summing them up. In other words:
(u)(v) = (5 * 3) + (3 * 2) = 15 + 6 = 21.

Therefore, (u)(v) = 21.

d.) We're given u = -4i + 2j and v = 3i + 2j. We need to find the angle between these two vectors.

To find the angle between two vectors, we can use the dot product formula:
u · v = |u| * |v| * cos(θ), where θ is the angle between them.

First, we find the dot product of u and v:
u · v = (-4 * 3) + (2 * 2) = -12 + 4 = -8.

Next, we calculate the magnitudes of u and v:
|u| = √((-4)^2 + 2^2) = √(16 + 4) = √20 = 2√5.
|v| = √(3^2 + 2^2) = √(9 + 4) = √13.

Now we can rearrange the dot product formula to solve for cos(θ):
cos(θ) = (u · v) / (|u| * |v|)
cos(θ) = -8 / (2√5 * √13)
cos(θ) = -8 / (2 * √(5 * 13))
cos(θ) = -8 / (2 * √(65))
cos(θ) = -4 / √65

Finally, we can use an inverse cosine function to find the angle θ:
θ = arccos(-4 / √65)
θ ≈ 134.08 degrees.

Therefore, the angle between u and v is approximately 134.08 degrees.