
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  = √(a^{2} + b^{2})
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  = √(2^{2} + 2^{2}) = 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} + 7^{2}) = 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  = √(4^{2} + 1^{2}) = √ 17
Magnitude:  3 u  = √(12^{2} + 3^{2}) = 3 √17
Direction of u: θ_{1}: tan(θ_{1}) = 1 / 4
The terminal side of u is in quadrant I, hence θ_{1} = arctan(1/4) ≅ 14.04°,
Direction of 3 u: θ_{2}: tan(θ_{2}) = 3 / 12 = 1 / 4
The terminal side of 3 u is in quadrant I, hence θ_{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: θ_{1}: tan(θ_{1}) = 1 / 1
The terminal side of u is in quadrant I, hence θ = arctan(1) = 45°,
Direction of  6 u: θ_{2}: tan(θ_{2}) =  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} + 4^{2}) = 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°
More References and Links
Magnitude and Direction Calculator
vectors.
Vector Calculators.
