# Find the Magnitude and Direction of a Vector

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

## Magnitude and Direction of a Vector

If vector v is defined by its components as follows v = < a , b >, its magnitude || v || is given by
|| v || = √(a2 + b2)

and its direction defined as the angle θ in standard position of the terminal side through the origin and point with coordinates (a , b).
Angle θ is found by solving the trigonometric equation
tan(θ) = b / a, such that 0 ≤ θ < 2π.
If the magnitude and direction are known, then
a = || v || cos (θ)
b = || v || sin (θ)

## Questions with Detailed Solutions

Question 1: Find the magnitude and direction of vector v given by its components as
v = < 2 , 2>
Solution to Question 1:
Magnitude: || v || = √(22 + 22) = 2 √ 2
Direction θ: tan(θ) = 2 / 2 = 1
The terminal side of v is in quadrant I, hence θ = 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 + 72) = 14
Direction θ: tan(θ) = 7 / (- 7 √3) = - 1 / √3
The terminal side of u is in quadrant II, hence θ = 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 θ: tan(θ) = - 5√3 / - 5 = √3
The terminal side of u is in quadrant III, hence θ = 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 || = √(42 + 12) = √ 17
Magnitude: || 3 u || = √(122 + 32) = 3 √17
Direction of u: θ1: tan(θ1) = 1 / 4
The terminal side of u is in quadrant I, hence θ1 = arctan(1/4) ≅ 14.04°,
Direction of 3 u: θ2: tan(θ2) = 3 / 12 = 1 / 4
The terminal side of 3 u is in quadrant I, hence θ2 = 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 || = √(12 + 12) = √2
Magnitude: || - 6 u || = √((-6)2 + (-6)2) = 6 √2
Direction of u: θ1: tan(θ1) = 1 / 1
The terminal side of u is in quadrant I, hence θ = arctan(1) = 45°,
Direction of - 6 u: θ2: tan(θ2) = - 6 / - 6 = 1
The terminal side of - 6u is in quadrant III, hence θ = 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 || cosθ and b = || u || sinθ
where
|| u || = 5 and θ = 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 + 42) = 4√(10)
Direction θ: tan(θ) = 4 / (-12) = - 1 / 3
The terminal side of w is in quadrant II, hence θ = 180 - arctan(1/3) ≅ 161.57°