Question

State and prove parallelogram law of vector addition and determine magnitude and direction of resultant vector.

Answer 1

i. Parallelogram law of vector add addition:

If two vectors of same type starting from the same point (tails cit the same point), are represented in magnitude and direction by the two adjacent sides of a parallelogram then, their resultant vector is given in magnitude and direction, by the diagonal of the parallelogram starting from the same point.

ii. Proof:

a. Consider two vectors \(\overset\rightarrow{P}\) and \(\overset\rightarrow{Q}\) of the same type, with their tails at the point O’ and θ’ is the angle between \(\overset\rightarrow{P}\) and \(\overset\rightarrow{Q}\) as shown in the figure below.

b. Join BC and AC to complete the parallelogram OACB, with \(\overline{OA}\) = \(\overset\rightarrow{P}\) and \(\overline{AC}\) = \(\overset\rightarrow{Q}\) as the adjacent sides. 

We have to prove that diagonal \(\overline{OC}\) = \(\overset\rightarrow{R}\), the resultant of sum of the two given vectors.

c. By the triangle law of vector addition, we have,

\(\overset\longrightarrow{OA}\) + \(\overset\longrightarrow{AC}\) = \(\overset\longrightarrow{OC}\) … (1)

As \(\overset\longrightarrow{AC}\) is parallel to \(\overset\longrightarrow{OB}\),

\(\overset\longrightarrow{AC}\) = \(\overset\longrightarrow{OB}\) = \(\overset\rightarrow{Q}\)

Substituting \(\overset\longrightarrow{OA}\) and \(\overset\longrightarrow{OC}\) in equation (1) we have,

\(\overset\rightarrow{P}\) + \(\overset\rightarrow{Q}\) = \(\overset\rightarrow{R}\)

Hence proved.

iii. Magnitude of resultant vector:

a. To find the magnitude of resultant vector \(\overset\rightarrow{R}\) = \(\overset\longrightarrow{OC}\), draw a perpendicular from C to meet OA extended at S.

b. In right angle triangle ASC,

c. Using Pythagoras theorem in right angled triangle, OSC

(OC)² = (OS)² + (SC)²

= (OA + AS)² + (SC)²

∴ (OC)² = (OA)² + 2(OA).(AS) + (AS²) + (SC)² . . . .(4)

d. From right angle triangle ASC,

(AS)² + (SC)² = (AC)² …. (5)

e. From equation (4) and (5), we get

(OC)² = (OA)² + 2(OA) (AS) + (AC)² …(6)

f. Using (2) and (6), we get

(OC)² = (OA)² + (AC)² + 2(OA)(AC) cos θ

∴ R² = P² + Q² + 2 PQ cos θ

Equation (7) gives the magnitude of resultant vector \(\overset\rightarrow{R}\).

v. Direction of resultant vector:

\(\overset\rightarrow{R}\) make an angle α with \(\overset\rightarrow{P}\).

From equations (2) and (3), we get

Equation (9) represents direction of resultant vector.

[Note: If β is the angle between \(\overset\rightarrow{R}\) and \(\overset\rightarrow{Q}\), it can be similarly derived that