Question

Two vectors A and B are inclined to each other at an angle theta Q. Using triangle law of vector addition find the magnitude and direction of their resultant.​

Answer 1

Given:

Two vectors A and B .

Angle Between vector A and B =∅

To Find:

Find magnitude and direction of the resultant vector by Triangle law of addition.

Solution:

Given vectors are vector A and vector B.

1. Now Draw a vector A in the plane with a line having an arrow at the head.

2. Draw a vector B at angle ∅ , starting point of vector B is the end point of vector A.

3. Now join the starting point of vector A and the end point of vector B and name it as vector R.

4.Starting point of vector R is the starting point of vector A and end point of vector R is the end point of vector B.

5. Length of the vector R gives the magnitude and the angle made by vector R made with vector A gives the direction of the resultant .

Now find the resultant analytically:

1. Draw a line AB representing vector A.

2. Draw a line BC representing vector B at an angle ∅.

3.Join A and C , AC represents the resultant and ∠BAC represents the direction.

4. Draw a perpendicular CE on AB extends to E.

∠CBE=∅

∠BAC=β

Now in triangle ACE, apply Pythagoras Theorem,

$AC^2=AE^2+CE^2$

       =$(AB+BE)^2 +CE^2$

$R^2=(A+BE)^2+CE^2$ ....................................(1)

In triangle BEC,

cos∅=BE/BC

        =BE/B

BE=Bcos∅.......................................(2)

sin∅=CE/BC

CE=Bsin∅.................................(3)
from eq1 , eq2,eq3 we get

$R^2$=(A+Bcos∅.)² +(Bsin∅)²

     =A²+B²(cos²∅+sin²∅)+2ABcos∅

R²     =A²+B²+2ABcos∅

$R=\sqrt{A^2+B^2+2ABcos∅}$

 tanβ=CE/AE

        =Bsin∅/(A+Bcos∅)

Answer 2

Answer:

Explanation:

Triangle Law of Vector Addition Derivation

Consider two vectors, P and Q, respectively, represented by the sides OA and AB. Let vector R be the resultant of vectors P and Q.

From triangle OCB,

$O B^{2}=O C^{2}+B C^{2}$

$\begin{aligned}O B^{2}=(O A+A C)^{2}+B C^{2} == > \text { eq. 1) }\end{aligned}$

In triangle $A C B with \Theta$ as the angle between $P , Q$

$\begin{aligned}&\cos \Theta=\frac{A C}{A B} \\&A C=A B \cos \Theta=Q \cos \Theta \\&\sin \Theta=\frac{B C}{A B} \\&B C=A B \sin \Theta=Q \sin \Theta\end{aligned}$

Substituting the values of $A C , B C$ in (eqn.1), we get

$R^{2}=(P+Q \cos \Theta)^{2}+(Q \sin \Theta)^{2}$

$R^{2}=P^{2}+2 P Q \cos \Theta+Q^{2} \cos ^{2} \Theta+Q^{2} \sin ^{2} \Theta$

$R^{2}=P^{2}+2 P Q \cos \Theta+Q^{2}$

therefore,

$R=\sqrt{P^{2}+2 P Q \cos \Theta+Q^{2}}$

Above equation is the magnitude of the resultant vector.

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

From triangle OBC,

$\tan \phi=\frac{B C}{O C}\\=\frac{B C}{O A+A C}$

$\tan \phi=\frac{Q \sin \Theta}{P+Q \cos \Theta}$

therefore,

$\phi=\tan ^{-1}\left(\frac{Q \sin \Theta}{P+Q \cos \Theta}\right)$

Above equation is the direction of the resultant vector.