Find the magnitude and direction of vectors; questions with solutions.

and its direction defined as the angle ? in standard position of the terminal side through the origin and point with coordinates (a , b).

Angle ? is found by solving the trigonometric equation

tan(?) = b / a, such that 0 ? ? < 2?.

If the magnitude and direction are known, then

a = || v || cos (?)

b = || v || sin (?)

Question 1: Find the magnitude and direction of vector v given by its components as

v = < 2 , 2>

__Solution to Question 1:__

Magnitude: || v || = √(2^{2} + 2^{2}) = 2 √ 2

Direction ?: tan(?) = 2 / 2 = 1

The two components of vector v are positive and therefore the terminal side of v is in quadrant I, hence ? = arctan(1) = 45° ,

Question 2: Calculate the magnitude and direction of vector u given by its components as

u = < - 7 √3 , 7>

__Solution to Question 2:__

Magnitude: || u || = √((- 7 √3)^{2} + 7^{2}) = 14

Direction ?: tan(?) = 7 / (- 7 √3) = - 1 / √3

The terminal side of u is in quadrant II, hence ? = 180° - arctan(1 / √3) = 180 - 30 = 150° ,

Question 3: Calculate the magnitude and direction of vector v given by its components as

v = < - 5 , - 5√3 >

__Solution to Question 3:__

Magnitude: || v || = √((- 5)^{2} + (- 5√3)^{2}) = 10

Direction ?: tan(?) = - 5√3 / - 5 = √3

The terminal side of u is in quadrant III, hence ? = 180 + arctan(√3) = 180 + 60 = 240° ,

Question 4: Calculate and compare the magnitude and direction of the vector u and 3 u with u given by

u = < 4 , 1 >

__Solution to Question 4:__

3 u is calculated by applying the scalar multiplication rule

3 u = < 3 × 4 , 3 × 1 > = < 12 , 3 >

Magnitude: || u || = √(4^{2} + 1^{2}) = √ 17

Magnitude: || 3 u || = √(12^{2} + 3^{2}) = 3 √17

Direction of u: ?_{1}: tan(?_{1}) = 1 / 4

The terminal side of u is in quadrant I, hence ?_{1} = arctan(1/4) ? 14.04°,

Direction of 3 u: ?_{2}: tan(?_{2}) = 3 / 12 = 1 / 4

The terminal side of 3 u is in quadrant I, hence ?_{2} = arctan(1/4) ? 14.04°,

Because of the multiplication of u by 3 the magnitude of 3 u is 3 times the magnitude of u but the direction does not change.

Question 5: Calculate and compare the magnitude and direction of the vector u and - 6 u with u given by

u = < 1 , 1 >

__Solution to Question 5:__

Apply the scalar multiplication rule to find - 6 u.

- 6 u = < - 6 × 1 , - 6 × 1 > = < - 6 , - 6>

Magnitude: || u || = √(1^{2} + 1^{2}) = √2

Magnitude: || - 6 u || = √((-6)^{2} + (-6)^{2}) = 6 √2

Direction of u: ?_{1}: tan(?_{1}) = 1 / 1

The terminal side of u is in quadrant I, hence ? = arctan(1) = 45°,

Direction of - 6 u: ?_{2}: tan(?_{2}) = - 6 / - 6 = 1

The terminal side of - 6u is in quadrant III, hence ? = 180 + arctan(1) = 225 °,

Because of the multiplication of u by - 6 the magnitude of - 6 u is 6 times the magnitude of u but the direction has changed by 180° because the terminal side of - 6 u is opposite to the terminal side of u; this is due to the minus sign of - 6.

Question 6: Calculate the components of vector u whose magnitude is 5 and direction given by the angle in standard position and equal to 270°.

__Solution to Question 6:__

Let u = < a , b >. According to the formulas given above,

a = || u || cos? and b = || u || sin?

where

|| u || = 5 and ? = 270°

Hence

a = 5 cos(270°) = 0

b = 5 sin(270°) = - 5

Question 7: Two vectors u and v have magnitudes equal to 2 and 4 and direction, given by the angle in standard position, equal to 90° and 180° respectively. Find the magnitude and direction of the vector 2 u + 3 v

__Solution to Question 7:__

Let us first use the formula given above to find the components of u and v.

u = < 2 cos(90°) , 2 sin(90°) > = < 0 , 2 >

v = < 4 cos(180°) , 4 sin(180°) > = < - 4 , 0 >

Let w = 2 u + 3 v and find the components of w.

w = 2 < 0 , 2 > + 3 < - 4 , 0 > = < - 12 , 4 >

Magnitude: || w || = √((- 12)^{2} + 4^{2}) = 4√(10)

Direction ?: tan(?) = 4 / (-12) = - 1 / 3

The terminal side of w is in quadrant II, hence ? = 180 - arctan(1/3) ? 161.57°

Vector Calculators.