Question

State Law of vector addition

Answer 1

Triangle law of vector addition states that when two vectors are represented by two sides of a triangle in magnitude and direction taken in same order then third side of the third side of that triangle represents in magnitude and direction the resultant of the vectors

Answer 2

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

`$\textbf{Triangle Law of Vector Addition}$`

= 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

[Sin²Q + Cos²Q = 1]

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)