Question

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

Answer 1

9.5; 72°

Explanation:

Answer 2

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°