 Note: vectors are shown in bold. scalars are shown in normal type

The diagram above shows two vectors A and B with angle p between them.

R is the resultant of A and B

R = A + B

This is the resultant in vector

R is the magnitude of vector R

Similarly A and B are the magnitudes of vectors A and B

R = √(A2 + B2 2ABCos p) or [A2 + B2 2ABCos p]1/2

To give the direction of R we find the angle q that R makes with B

Tan q = (A Sin p)/(B + A Cos q)

A vector is completely defined only if both magnitude and direction are given.

### Question

Two forces of 3 N and 4 N are acting at a point such that the angle between them is 60 degrees. Find the resultant force Magnitude R of the resultant force is R = √(32 + 42 + 2 x 3 x 4 Cos 60 deg)

= √(9 + 16 + 12) = √(37 = 6.08 N

Direction of R is given by finding the angle q

tan q = (3 Sin 60 deg)/(4 + 3 Cos 60 deg) = 0.472

q = tan-1 0.472

= 25.3 deg

Thus R is 6.08 N in magnitude and is at an angle of 25.3 deg to the 4 N force.

### Question

A car goes 5 km east 3 km south, 2 km west and 1 km north. Find the resultant displacement.

First we will make the vector diagram O to A 5 km east

A to B 3 km south

B to C 2 km west

c to D 1 km north

Net displacement is OD

Along the horizontal direction: 5 km east - 2 km west = 3 km east

Along the vertical direction: 3 km south - 1 km north = 2 km south

OD = √(32 + 22 + 2 x 3 x Cos 90 deg)

= √(32 + 22)

= 3.6 km

tan p = 2/3

or p = tan-12/3 = 34 deg

Thus resultant displacement is 3.6 km, 34 deg south of east.

## Rectangular components of a vector

To find the component of a vector along a given axis, we drop a perpendicular on the given axis from the vector For example OA is the given vector. We have to find its component along the the horizontal axis. Let us call it x-axis. We drop a perpendicular AB from A onto the x-axis. The length OB is the component of OA along x-axis. If OA makes angle p with the horizontal axis, then in triangle OAB, OB/OA = Cos P or OB = OA Cos P.

Remember that component of a vector is a scalar quantity. If the component is along the negative direction, we put a (-) sign with it.)

Usually we resolve the vector into components along mutually perpendicular components. OB is the x component OB = OA Cos p.

Similarly component along the vertical direction or the y axis is OC

OCAB is a rectangle.

So OC = AB

look at triangle OAB again,

AB/OA = Sin p

=> AB = OA Sin p = OC

Thus y component OC = OA Sin p.

Note that p is the angle with the horizontal axis.

### Question

Find the x and y components of a 25 m displacement at an angle of 210 deg. OA is the displacement vector. The angle with the horizontal axis is 210 deg - 180 deg = 30 deg

x component = OB = -25 Cos 30 deg = -21.7

y component = AB = -25 Sin 30 deg = -12.5 m

Note that each component is pointing along the negative coordinate direction and thus we must take it as negative.

Now we will solve a problem using the component method

### Question

Find the resultant of the following two displacements: 2 m at 30 deg and 4 m at 120 deg. The angles are taken relative to the x axis.  Rx = 2 Cos 30 deg - 4 Cos 60 deg = - 0.268 m

Ry = 2 Sin 30 deg + 4 Sin 60 degg = 4.46 m

R = √(Rx2 + Ry2)

= √(-0.2682 + 4.462) = 4.47 m tan q = Ry/Rx = 4.46/0.268

=> q = 86.6 deg

p = 180 deg - 86.6 deg = 93.4 deg