Addition Vectors, Components. Units. Illustration Excercise. Analytic Method

ANALYTIC METHOD, VECTORS ADDITION.

1.- COMPONENTS
The graphical sum often has not enough exactitude and is not useful when the vectors are in three dimensions. As every vector can be represented as the sum of two other vectors, these vectors are called the components of the original vector. Usually the components are chosen along two mutually perpendicular directions. For example, assume the vector V below in the figure. It can be split in the component V_x parallel to the x axis and the component V_y parallel to the y axis.

We use coordinate axis x-y with origin at the tail of vector V. Notice that V = V_x + V_y according to the parallelogram rule.

The magnitudes of V_x and V_y are denoted V_x and V_y, and are numbers, positive or negatives as they point at the positive or negative side of the x-y axis.

Notice also that V_x = Vcos and V_y = Vsen.

2.- UNIT VECTORS
Vector quantities can often be expressed in terms of unit vectors. A unit vector is a vector whose magnitude is equal to one and dimensionless. They are used to specify a determinated direction. The symbols i, j y k represent unit vectors pointing in the directions x, y and z positives, respectively.

Now V can be written V = V_xi + V_yj.

If we need to add the vector A = A_x i + A_y j with
the vector B = B_x i + B_y j we write
R = A + B = A_x i + A_y j + B_x i + B_y j = (A_x + B_x)i + (A_y + B_y)j.

The components of R are R_x = A_x + B_x and R_y = A_y + B_y

Exercise, Example: Use of components and unit vectors.

A boyscout walks 22 km in North direction, and then he walks in direction 60º Southeast during 47.0 km. Find the components of the resulting vector displacement from the starting point, its magnitude and angle with the x axis.

Solution: The two displacements are shown in the figure, where we choose the positive x axis pointing to East and the positive y axis pointing to North.

The resultant displacement D is the sum of D₁ and D₂.

Using unit vectors:

D₁ = 22 j
D₂ = 47cos60º i - 47sen60º j
Then D = D₁+D₂ = 22 j + 47cos60º i - 47sen60º j = 23.5 i - 18.7 j
and the resultant vector is completely specified with an x component D_x = 23.5 km and a y component D_y = -18.7 km. (Note D_x and D_y are scalars).

The same resultant vector can be specified giving its magnitude and angle:
D² = D_x² + D_y² = (23.5 km)² + (-18.7 km)² finding D = 30 km.
tan = D_y/D_x = -18.7/23.5 = -0.796 finding = -38.5º (under the x axis) or 38.5 Southeast.

More on This Theme:
· Vectors, Scalars: Initial Page
· Vectors, Addition Tools, Problems