Question

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

Answer 1

Please check the explanation.

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|=\sqrtx_i\right`

`=\sqrt{6^2+\left(2√(3)\right)^2}`

`=√(36+12)`

`=√(48)`

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

`=√(3)√(2^4)`

`=4√(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((√(3))/(3)\right)`

`\:\theta \:=(\pi \:)/(6)=30^(\circ \:)`