Question

find the magnitude and positive direction angle of the vector $\langle - 4 \sqrt { 3 } , - 4 \
angle$.
the magnitude of the vector is
(simplify your answer.)

Explanation:

Step1: Recall the formula for the magnitude of a vector \(\langle x, y

angle\)
The magnitude \(r\) of a vector \(\langle x, y
angle\) is given by the formula \(r=\sqrt{x^{2}+y^{2}}\). For the vector \(\langle - 4\sqrt{3},-4
angle\), we have \(x = - 4\sqrt{3}\) and \(y=-4\).

Step2: Substitute the values of \(x\) and \(y\) into the formula

First, calculate \(x^{2}\) and \(y^{2}\):
\(x^{2}=(-4\sqrt{3})^{2}=(-4)^{2}\times(\sqrt{3})^{2}=16\times3 = 48\)
\(y^{2}=(-4)^{2}=16\)

Then, find \(x^{2}+y^{2}\):
\(x^{2}+y^{2}=48 + 16=64\)

Finally, find the square root of \(x^{2}+y^{2}\):
\(r=\sqrt{64}=8\)

Answer:

\(8\)