Question

State and prove triangle law of vector addition.

Answer 1

It states if two vectors can be represented in magnitude and direction by two sides of triangle taken in order ,then the resultant vector can be represented in magnitude and direction by third side of the triangle taken in opposite order.

Answer 2

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


$\textbf{Triangle Law :}$

= 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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