Question

state the parallelogram law of vector addition and find the magnitude and direction​

Answer 1

Statement:- If two vectors are represented in magnitude and direction by the adjacent sides of a parallelogram drawn from a point the resultant drawn from that point gives the magnitude and direction.

Proof:- Two vectors P and Q are represented in magnitude and direction by the adjacent sides of Parallelogram.

$\; \; \; \; \; \;☞ \; \; \; \Large{\red{\bold{Diagram}}}$

Magnitude of the resultant vector:-

From the above Parallelogram

→ ∠DAB = ∠CBN

From ∆CBN

• Cosθ = $\tt{\frac{Adjacent}{Hypothenuse}}$

Cosθ = $\sf{\frac{BN}{BC} = \frac{BN}{Q}}$

QCosθ = BN

• Now, Sinθ = $\tt{\frac{Opposite}{Hypothenuse}}$

Sinθ = $\sf{\frac{CN}{BC}}$

Sinθ = $\sf{\frac{CN}{Q}}$

QSinθ = CN

From ∆CAN

→ AC² = AN²+CN²

→ AC² = (AB+BN)²+CN²

→ AC² = AB²+BN²+2.AB.BN+CN²

→ R² = P²+(QCosθ)²+2.P.QCosθ+(QSinθ)²

→ R² = P²+Q²Cos²θ+2.P.QCosθ+Q²Sin²θ

→ R² = P²+Q²Sin²θ+Q²Cos²θ+2.P.QCosθ

→ R² = P²+Q²(Sin²θ+Cos²θ)+2.P.QCosθ

→ R² = P²+Q²(1)+2.P.QCosθ

→ R² = P²+Q²+2.P.QCosθ

☞ $\bold{R = \sqrt{P²+Q²+2.P.QCosθ}}$

Direction of the Resultant Vector:-

If the resultant vector R makes an angle "s" with the vector P.

From ∆CAN

• $\boxed{\sf{tan\;\alpha = \tt{\frac{Opposite}{Adjacent}}}}$

→ tanα = $\sf{\frac{CN}{AN}}$

→ tanα = $\sf{\frac{CN}{AB+BN}}$

→ tanα = $\sf{\frac{QSinθ}{P+QCosθ}}$

☞ $\Large{\alpha = \sf{tan^{-1} (\frac{QSinθ}{P+QCosθ})}}$

$\huge{✓}$

Hope It Helps You ✌️