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

Find the magnitude and direction in degrees of the vector v=6i+2 seq31​-example-1

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Answer:

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 \:)

User Yoona
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