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Find the magnitude and direction of the vector <3,9>. Round angles to the nearest degree and other values to the nearest tenth.

User Kamyl
by
7.9k points

2 Answers

2 votes

Answer:

9.5; 72°

Explanation:

User Alex JM
by
7.4k points
4 votes

Answer:

Magnitude: 3√10

Direction angle: 71.56°

Explanation:

The magnitude ||v|| of a vector <a,b>, would be:


||v||=\sqrt{a^(2)+b^(2)


||v||=\sqrt{3^(2)+9^(2)


||v||=√(9+81)


||v||=√(90)


||v||=3√(10)

The vector's reference angle would be:


tan\alpha =|(b)/(a)|


tan\alpha =|(9)/(3)|


tan\alpha =|3|


tan\alpha =3


\alpha = 71.56

If
\alpha is your reference angle, then your direction angle ∅ depends on what quadrant the vector is located in. Since <3,9> is located in Quadrant I, then ∅=
\alpha, which means your direction angle is also 71.56°

User Cordialgerm
by
8.5k points

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