Question

Find the magnitude and direction in degrees of the vector v=6i+2 seq31​

Answer 1

Answer:

Please check the explanation.

Step-by-step explanation:

Given the vector

v = 6i + 2√3j

The Magnitude of a vector:

$\mathrm{Computing\:the\:Euclidean\:Length\:of\:a\:vector}:\quad \left|\left(x_1\:,\:\:\ldots \:,\:\:x_n\right)\right|=\sqrt{\sum _{i=1}^n\left|x_i\right|^2}$

$=\sqrt{6^2+\left(2\sqrt{3}\right)^2}$

$=\sqrt{36+12}$

$=\sqrt{48}$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$

$=\sqrt{3}\sqrt{2^4}$

$=4\sqrt{3}$

The Direction of a vector:

tan Ф = y/x

y=2√3

x = 6

tan Ф = y/x

          = 2√3 / 6

           = √3 / 3                  

$\theta \:=tan\:^{-1}\left(\frac{\sqrt{3}}{3}\right)$

$\:\theta \:=\frac{\pi \:}{6}=30^{\circ \:}$