Question

find the magnitude and direction angle of the following vector. write your angle in degrees rounded to four decimal places.

u = 10i - 3j

Explanation:

Step1: Calculate the magnitude of the vector

For a vector \(\mathbf{u}=a\mathbf{i}+b\mathbf{j}\), the magnitude \(|\mathbf{u}|\) is given by the formula \(|\mathbf{u}|=\sqrt{a^{2}+b^{2}}\).
Here, \(a = 10\) and \(b=- 3\).
So, \(|\mathbf{u}|=\sqrt{10^{2}+(-3)^{2}}=\sqrt{100 + 9}=\sqrt{109}\approx10.4403\)

Step2: Calculate the direction angle \(\theta\)

The formula for the direction angle \(\theta\) of a vector \(\mathbf{u}=a\mathbf{i}+b\mathbf{j}\) is \(\tan\theta=\frac{b}{a}\).
Here, \(a = 10\) and \(b=-3\), so \(\tan\theta=\frac{-3}{10}=-0.3\).
Since \(a>0\) and \(b<0\), the vector lies in the fourth - quadrant.
\(\theta=\arctan(-0.3)+360^{\circ}\)
Using a calculator, \(\arctan(-0.3)\approx - 16.6992^{\circ}\), and \(\theta\approx360^{\circ}-16.6992^{\circ}=343.3008^{\circ}\)

Answer:

The magnitude of the vector \(\mathbf{u}\) is approximately \(10.4403\) and the direction angle is approximately \(343.3008^{\circ}\)