Question

State triangle law of vector addition. Using the law find the magnitude and direction of resultant of two vectors inclined at an angle theta.​

Answer 1

$\huge\fcolorbox{black}{pink}{Answer}$

$\textsf{\red{Triangle \:Law\:Of\:Vector\:Addition}}$

→It states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction,then the third side of the triangle represents the magnitude and direction of the resultant vector.

$\Huge{\red{\underline{\textsf{Proof}}}}$

Let us consider two vectors P and Q that are represented in the order of magnitude and direction by the sides OA and AB respectively of the triangle OAB. Let R be the resultant of vectors P and Q.

Now,

Derivation of triangle law of vector addition

R=P+Q

From triangle OCB,

OB2=OC2+BC2 OB2=(OA+AC)2+BC2 (eq.1)

In triangle ACB with ϴ as the angle between P and Q

cosΘ=ACAB AC=ABcosΘ=QcosΘ sinΘ=BCAB BC=ABsinΘ=QsinΘ R2=(P+QcosΘ)2+(QsinΘ)2 (after substituting AC and BC in eq.1)

R2=P2+2PQcosΘ+Q2cos2Θ+Q2sin2Θ R2=P2+2PQcosΘ+Q2

therefore, R=P2+2PQcosΘ+Q2−−−−−−−−−−−−−−−−−√

Above equation is the magnitude of the resultant vector.

To determine the direction of the resultant vector, let ɸ be the angle between the resultant vector R and P.

From triangle OBC,

tanϕ=BCOC=BCOA+AC tanϕ=QsinΘP+QcosΘ

therefore, ϕ=tan−1(QsinΘP+QcosΘ)

Above equation is the direction of the resultant vector.

Answer 2

Answer:

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