State and prove triangle law of vector addition

Asked by admin @ 29/08/2021 in Physics viewed by 326 People

1.State the triangle law of vector addition and prove it.

Answered by admin @ 29/08/2021

$\huge\mathfrak{A{\textbf{\red{nswer :}}}}$

$\textbf{Triangle Law :}$

= 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

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