Let v = 2j and let u be a vector with length 8 that starts at the origin and rotates in the xy-plane. find the maximum and minimum values of the length of the vector u × v.

We want to maximize/minimize the cross-product between two vectors.

We will find that the vectors that maximize the cross product are:

u = 8i = (8, 0, 0)

u = -8i = (-8, 0, 0)

And the ones that minimize it:

u = 8j = (0, 8, 0)

u = -8j = (0, -8, 0)

For two vectors V and W, the cross-product between them is written as:

V×W = |V|*|W|*sin(θ)

Where θ is the angle between V and W.

Here we know that:

v = 2j = (0, 2, 0)

|v| = 2

And we want to find a vector u, such that:

u = (a, b, c)

|u| = 8

We also know that u rotates on the xy-plane, then c = 0.

u = (a, b, 0)

|u| = 8

Then the cross-product between v and u is:

u × v = |u|*|v|*sin(θ)

        = |8|*|2|*sin(θ) = 16*sin(θ)

This is maximized when θ = 90°.

Knowing that the vector v is along the positive y-axis, and that vector u lives in the xy-plane, the only way that the angle between these vectors is 90° is if u is along the positive (or negative) x-axis.

Then we can have:

u = (8, 0, 0)

u = (-8, 0, 0)

These are the two values of u that maximize the cross-product.

To minimize it, we need to have sin(θ) = 0

Then θ = 0°

This means that u should be along the y-axis, then we have:

u = (0, 8, 0)

u = (0, -8, 0)

If you want to learn more, you can read:

link

The magnitude of the cross product is the product of the magnitudes of the vectors (2 and 8) and the sine of the angle between them. If we take length to be a non-negative measure, then the length of u × v will vary from 0 to 16.

The maximum length is 16.

The minimum length is 0.

sinn is the total amount of students now and m were the students before, you subtract them both (n-m) and divide that by m. then, since to get a percentage from a fraction you have to multiply by 100, your answer should look like this: n-m/m*100

step-by-step explanation:

17

step-by-step explanation:

51/3=17

thus 17 is your answer.