Question

State and prove triangle law of vector addition.

Answer 1

$Here \: is \: the \: answer \: of \: your \: question$


`$\textbf{Triangle Law}$`

= It states that if two vectors acting simultaneously at a point are represented in magnitude and direction by the two sides of a triangle taken in same order. And their resultant is represented in magnitude and direction by the third side of the triangle taken in opposite order.

→ [Diagram is in attachment]


`$\textbf{Prove}$`

Consider two vectors A vector and B vector represented by OP and PQ. Let the angle between A vector and B vector is Q (theta) by the two sides of a triangle. Resultant to be OD vector by third side of triangle taken in opposite order. Draw DN perpendicular to OP produced.

`$\textbf{Magnitude of R vector}$`

In ∆ OND (By Pythagoras)

(R)² = (ON)² + (ND)²

(R)² = (OP + PN)² + (ND)²

(R)² = (A + PN)² + (NQ)² ..............(S)

In ∆ PDN

PN ÷ PD = Cos Q

PN ÷ B = Cos Q

PN = B Cos Q ..........(1)

ND ÷ PQ = Sin Q

ND ÷ B = Sin Q

ND = B Sin Q .............(2)

Put value of (1) and (2) in (S)

(R)² = (A + B Cos Q)² + (B Sin Q)²

(R)² = A² + B² Cos²Q + 2AB Cos Q + B² Sin² Q

R = √A² + B² (Sin²Q + Cos²Q) + 2AB CosQ

R = √A² + B² + 2AB Cos Q

`$\textbf{Direction of R vector}$`

Let R vector make an angle Π with A vector.

tan Π = DN ÷ ON

= B Sin Q ÷ OP + PN

= B Sin Q ÷ A + B Cos Q

Answer 2

Triangle Law of Vector Addition

Statement of Triangle Law

If 2 vectors acting simultaneously on a body are represented both in magnitude and direction by 2 sides of a triangle taken in an order then the resultant(both magnitude and direction) of these vectors is given by 3rd side of that triangle taken in opposite order.

Derivation of the law

Consider two vectors P and Q acting on a body and represented both in magnitude and direction by sides OA and AB respectively of a triangle OAB. Let θ be the angle between P and Q. Let R be the resultant of vectors P and Q. Then, according to triangle law of vector addition, side OB represents the resultant of P and Q.

For example:

, we have

                  R = P + Q

Now, expand A to C and draw BC perpendicular to  OC.

From triangle OCB,

             

In triangle ACB,

              

Also,

              

Magnitude of resultant:

Substituting value of AC and BC in (i), we get

              

which is the magnitude of resultant.

Direction of resultant: Let ø be the angle made by resultant R with P. Then,

From triangle OBC,

              

which is the direction of resultant