Vector Subtraction, Resolution and 3D Representation

class - XI, Vector / By

Subtraction of vectors:

Let us consider  and  creates angle θ. To subtract  from  we can write + (- ) [where –  is the opposite vector of ]. The angle between  and –  is (180 – θ) and  is the resultant of  and – .

The magnitudes of  is R2 = P2 + Q2+2P.Q  

Or, R2 = P2 +Q2– 2PQ 

If creates angle  with then tan =

Or, tan = .

Resolution of vectors:

Resolution of a vector into its components is the process to determining a set of vectors, whose resultant is the given vector. Each vector in that set is called a component of the given vector.

Prove: Let us consider (= ) is resolute along the line OA and OB creates angles  and  with OC respectively. From point C, CD and CE are drawn parallel to OB and OA respectively. OD and OE represent  and  respectively.

From ODC, COD = , OCD = , then ODC = 1800– (  + ).

From trigonometry, we can write = =

Or, = =

Or, =  =

So, P =  and Q = .

Resolution of a vector into two rectangular components:

Let us consider OX and OY are two mutually perpendicular axes, where O is the origin. OB represents creates angle θ with +ve x axis. Let BA and BC are the perpendiculars on OX and OY axis respectively.  and  are the components of and represented by OA and OC respectively. I.e.  = + .

The magnitudes of  and  are represented as P = R  [ = ]

and Q = R [ = ]

So that R = and tan = .

Representation of a vector by coordinates:

Let us consider OX, OY and OZ are three perpendicular axes, where O is the origin. Let P is a point with coordinates (x, y, z) and  =  is the position vector of P. , and are the unit vectors along + ve X, Y and Z axis. Let PB is the perpendicular drawn on x-y plane. BA and BC are the perpendicular drawn on X and Y axis respectively. PD is the perpendicular drawn on Z axis.

Therefore  = x,  = = y and  = = z and their magnitudes are OA = x, AB = OC = y and BP = OD = z.

From OAB,  =  +  and from OBP,  =  +  

Or,  = + +

So,  = x + y + z.

Magnitude: From OAB, OB2 = OA2 + AB2 and from OBP, OP2 = OB2 + BP2

Or, OP2 = OA2 + AB2 + BP2

So, R2 = x2 + y2 + z2

Direction: Let us consider ,  and  are the angles of R with +ve X, Y and Z axis respectively. Then, cos = , cos = , cos =  where cos, cos and cos  are the direction cosines of .

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