Find the unit vector in the direction of u=(-3,2). Write your answer as a linear combination of the standard unit vectors i and j.

a. u=-3[sqrt(13)/13]i+2[sqrt(13)/13]j
b. u=-3[sqrt(5)/5]i+2[sqrt(5)/5]j
c. u=3[sqrt(5)/5]i-2[sqrt(5)/5]j
d. u=3[sqrt(13)/13]i+2[sqrt(13)/13]j

I do not understand this. Can someone please explain it to me?

|u| = √(9+4) = √13

So, divide u by its length to get a unit vector (length=1)

so, (a)

If you've advanced to the point where you're studying vectors, your teacher should stop insisting on rational denominators. I much prefer 1/√13 to the cumbersome √13/13

just sayin' ...

Hey is there any way that we could talk one on one? Im pretty sure were im the same math class as you and want some help pls!! :))

Sure, I'd be happy to explain it to you!

To find the unit vector in the direction of u=(-3,2), we need to divide each component of u by the magnitude of u.

The magnitude of u is the square root of ( (-3)^2 + 2^2 ), which is √(9 + 4) = √13.

So, dividing each component of u by √13 gives us:

-3/√13 for the i-component, and
2/√13 for the j-component.

The unit vector in the direction of u can be written as a linear combination of the standard unit vectors i and j, with each component multiplied by its respective unit vector.

Therefore, the correct answer is:
u = -3/√13 * i + 2/√13 * j.

Out of the given options, the answer that matches this is:
d. u = 3[sqrt(13)/13]i + 2[sqrt(13)/13]j.

Hope that helps! Let me know if you have any other questions.

To find the unit vector in the direction of u=(-3,2), we first need to find the magnitude (or length) of vector u. The magnitude of a vector is calculated using the formula sqrt(x^2 + y^2), where x and y are the components of the vector.

In this case, the magnitude of u is sqrt((-3)^2 + 2^2) = sqrt(9 + 4) = sqrt(13).

Next, we divide each component of the vector u by its magnitude to obtain the unit vector.

So, the unit vector in the direction of u is:
u_unit = (-3/sqrt(13), 2/sqrt(13))

Now, we can write this unit vector as a linear combination of the standard unit vectors i and j. The i vector represents the x-component of a vector and the j vector represents the y-component.

Therefore, the unit vector can be expressed as:
u_unit = (-3/sqrt(13))i + (2/sqrt(13))j

Hence, the correct answer is: a. u = -3[sqrt(13)/13]i + 2[sqrt(13)/13]j.

To find the unit vector in the direction of u=(-3,2), we need to normalize the vector u by dividing each component by its magnitude.

1. Find the magnitude of vector u using the formula: ||v|| = sqrt(x^2 + y^2), where x and y are the components of vector u.
In this case, the magnitude of u is ||u|| = sqrt((-3)^2 + (2)^2) = sqrt(9 + 4) = sqrt(13).

2. Divide each component of vector u by its magnitude:
u_normalized = (-3/sqrt(13), 2/sqrt(13)).

Now, we can write the answer as a linear combination of the standard unit vectors i and j.

The standard unit vector i represents the direction along the x-axis (horizontal), and j represents the direction along the y-axis (vertical).

From step 2, we have u_normalized = (-3/sqrt(13), 2/sqrt(13)).

Therefore, the correct answer is d. u=3[sqrt(13)/13]i+2[sqrt(13)/13]j.