First, we need to break down vector [tex]\( u \)[/tex] given by the expression:
[tex]\[ u = 20\left\langle\cos 30^{\circ}, \sin 30^{\circ}\right\rangle \][/tex]
### Step 1: Calculate the components of the vector [tex]\( u \)[/tex]
1. Cosine of 30 degrees:
[tex]\[ \cos 30^{\circ} \approx 0.866 \][/tex]
2. Sine of 30 degrees:
[tex]\[ \sin 30^{\circ} = 0.5 \][/tex]
3. Magnitude of [tex]\( u \)[/tex]:
[tex]\[ u = 20 \left\langle 0.866, 0.5 \right\rangle \][/tex]
The components of [tex]\( u \)[/tex] are:
[tex]\[ u = \left( 20 \cdot 0.866, 20 \cdot 0.5 \right) \approx (17.32, 10.00) \][/tex]
### Step 2: Calculate the vector [tex]\(-3u\)[/tex]
Since [tex]\(-3u\)[/tex] means scaling vector [tex]\( u \)[/tex] by -3, we get:
[tex]\[ -3u = -3 \left(17.32, 10.00\right) = (-51.96, -30.00) \][/tex]
### Step 3: Calculate the magnitude of [tex]\(-3u\)[/tex]
The magnitude of a vector is calculated by scaling the original magnitude:
[tex]\[ \text{Magnitude of } -3u = 3 \times 20 = 60 \][/tex]
### Step 4: Determine the direction (angle) of [tex]\(-3u\)[/tex]
The direction of the vector is given by the angle it makes with the positive x-axis. For the coordinates [tex]\((-51.96, -30.00)\)[/tex]:
1. Since the vector points into the third quadrant (both components are negative), we add 180° to the angle made by [tex]\( u \)[/tex].
Considering the original direction of [tex]\( u \)[/tex] was 30°:
[tex]\[ \text{Direction of } -3u = 210^{\circ} \][/tex]
Thus, the magnitude and direction of [tex]\(-3u\)[/tex] are:
[tex]\[ \boxed{60; 210^{\circ}} \][/tex]
