Mathematical
Vector
Addition
Mathematical Addition of Vectors
The process of adding vectors can be accurately done
using basic trigonometry. If you follow each step carefully,
you will break down each vector into it's x and y componets
and determine the magnitude and direction of the resultant
vector. We have provided you with a "Vector Worksheet" to
help you organize your work.
Let's say that we are adding three vectors
A, B, and C.
STEP #1 - Deconstruct each vector into
it's X & Y components.
A) To do this, you must first find the Theta angle
to the x axis for each vector.
00
30 Km/hr @ 450
60 Km/hr @ 3150
900
2700
60 Km/hr @ 1350
1800
B. Now calculate the X & Y components of each vector treating the
magnitude of each vector as the hypotenuse of a right triangle
00
X component = Magnitude x cos
-x = (60)(cos45) = -42.4
+y
2700
900
-x
y component = Magnitude x sin
y = (60)(sin45) = +42.4
1800
60 Km/hr @ 3150
60 Km/hr @ 1350
30 Km/hr @ 450
= 45o
= 45o
= 450
-42.4
00
+y
2700
90
-x
0
+42.4
1800
B. Now calculate the X & Y components of each vector treating the
magnitude of each vector as the hypotenuse of a right triangle
00
X component = Magnitude x cos
-x = (60)(cos45) = -42.4
+x = (30)(cos45) = +21.2
+y
+y
2700
+x
-x
900
y component = Magnitude x sin
y = (60)(sin45) = +42.4
+y = (30)(sin45) = +21.2
1800
60 Km/hr @ 3150
60 Km/hr @ 1350
30 Km/hr @ 450
= 45o
= 45o
= 450
00
+ 21.2
+x = (30)(cos45)
+y = +21.2
+y
2700
+x
-x
900
+42.4
+21.2
1800
-42.4
B. Now calculate the X & Y components of each vector treating the
magnitude of each vector as the hypotenuse of a right triangle
00
X component = Magnitude x cos
-x = (60)(cos45) = -42.4
+x = (30)(cos45) = +21.2
+x = (60)(cos45) = +42.4
+y
+y
270
+x
0
900
+x
-x
y component = Magnitude x sin
-Y
y = (60)(sin45) = +42.4
+y = (30)(sin45) = +21.2
-Y = (60)(sin45) = -42.4
1800
60 Km/hr @ 3150
60 Km/hr @ 1350
30 Km/hr @ 450
= 45o
= 45o
= 450
STEP #2 - List and add all x components and y
components. Including all signs. These sums are
the components of the resultant vector.
00
+ 21.2
+42.4
+63.6
-42.4
-42.4
+x = (30)(cos45)+y= +21.2
+y
2700
+x
+21.2
900
+x
-x
-Y
+42.4
+21.2
-42.4
0
180
+63.6
-42.4
+21.2
STEP #3 - Convert the resultant components into
navigational vector notation.
To do this, first use the Pythagorean Theorem to determine the hypotenuse
00
+ 21.2
+42.4
+63.6
-42.4
-42.4
+21.2
+x = (30)(cos45) = +21.2
+y
+y
2700
+x
-x
+21.2
900
+x
+21.2
-Y
+42.4
+21.2
-42.4
0
180
+63.6
-42.4
+21.2
Resultant = √21.2 +21.2
2
2
The magnitude of the resultant is 29.9 Km/hr.
Now, find the Theta θ of the resultant using the inverse
tangent formula. tan-1= l y/x l = θR
00
+ 21.2
+42.4
+63.6
-42.4
-42.4
+x = (30)(cos45) = +21.2
+y
+y
2700
+x
+21.2
900
+x
-x
-Y
+42.4
+21.2
-42.4
0
180
+63.6
-42.4
+21.2
Since the θR = 45 , the navigational direction
o
of the resultant vector is NAV = 90 - θR = 45 .
o
00
+ 21.2
+42.4
+63.6
NAVR = 45o
-42.4
-42.4
θR=45o
+y = (30)(cos45) = +21.2
+x
+y
+x
270
+x
-x
+21.2
900
0
-Y
+42.4
+21.2
180
-42.4
0
+63.6
-42.4
29.9 Km/hr
+21.2
45o
Use this guide to get the navigational angle in other
quadrants.
Quadrant #1: NAV = 90 - θR
Quadrant #2: NAV = 270 + θR
Quadrant #3: NAV = 270 - θR
Quadrant #4: NAV = 90 + θR
Congratulations! You have successfully
calculated a vector addition!
The resultant vector = 29.9 Km/hr @ 45
o