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;
[tex]|u| = \sqrt{a^2 + b^2}[/tex] --------------------(i)
The direction of such vector is given by;
θ = tan ⁻¹ ([tex]\frac{b}{a}[/tex]) -------------(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;
[tex]|u| = \sqrt{(-12)^2 + (5)^2}[/tex]
[tex]|u| = \sqrt{144 + 25}[/tex]
[tex]|u| = \sqrt{169}[/tex]
[tex]|u| = 14[/tex]
Therefore, the magnitude of vector u is 13.
(b) From equation (ii) the direction of vector u is therefore;
θ = tan ⁻¹ ([tex]\frac{5}{-12}[/tex])
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)
