Question

Find the magnitude and the positive direction angle for u. u = (-12, 5) theta = degree (Round to the nearest tenth as needed.) For the vector r = (-9, -10), find -9r.

Answer 1

Question:

1. Find the magnitude and the positive direction angle for u.

u = (-12, 5)

For theta degree, round to the nearest tenth as needed.

2. For the vector r = (-9, -10), find -9r.

Answer:

(1)

The magnitude of vector u is 13.

The positive direction of vector u is 157.4° (to the nearest tenth)

(2)

-9r = (81, 90)

Step-by-step explanation:

(1) For a given vector u = (a, b), the magnitude of u is given by;

$|u| = \sqrt{a^2 + b^2}$            --------------------(i)

The direction of such vector is given by;

θ = tan ⁻¹ ($\frac{b}{a}$)                    -------------(ii)

Where;

a and b are the x and y components of the vector.

Now,

From the question,

u = (-12, 5)

(a) From equation (i), the magnitude of u is therefore;

$|u| = \sqrt{(-12)^2 + (5)^2}$

$|u| = \sqrt{144 + 25}$

$|u| = \sqrt{169}$

$|u| = 14$

Therefore, the magnitude of vector u is 13.

(b) From equation (ii) the direction of vector u is therefore;

θ = tan ⁻¹ ($\frac{5}{-12}$)

θ = tan ⁻¹ (-0.41667)

θ = -22.62°

The direction here is negative, but we have been told to find the positive direction.

From the given x and y components of the vector, it can be deduced that the the vector lies in the negative x direction and the positive y direction.

Therefore, the vector lies in the second quadrant. This is shown in the diagram attached to this response.

To get the positive direction, m, we need to add 180° to the result of θ = -22.62° i.e

m = 180° + (-22.62)°

m = 157.38°

Therefore, the positive direction of vector u is 157.4° (to the nearest tenth)

(2) For the vector r = (-9, -10), we are asked to calculate -9r.

Since;

r = (-9, -10)

-9r = -9(-9, -10)    [Multiply by -9]

-9r = (81, 90)

Therefore, -9r = (81, 90)