Question

State and prove parallelogram law of vector addition​

Answer 1

Parallelogram law of vector addition :-

When two vectors with a common origin represent two adjacent sides of a parallelogram with magnitude and direction , the resultant vector is represented both in magnitude and direction by the diagonal passing through that point .

$\tt{{\color{olive}{\vec{R} = \vec{A} + \vec{B}}}} \\\\ \tt{{\color{gray}{\vec{|A|} = A~; \vec{|B|} = B }}}$

$\tt{{\color{olive}{\vec{A} = A~\hat{i}}}} \\\\ \tt{{\color{gray}{\vec{B}= B~cos~\theta~\hat{i} + B~sin~\theta~\hat{j}}}}$

$\tt{\color{orange}{\vec{R} = ( A + B ~cos~\theta)\hat{i} + B~sin~\theta~\hat{j}}} \\\\ \tt{{\vec{|R|} = {\sqrt{(A+ B~cos~\theta)^2 + ( B~sin\theta)^2}}}}$

$\tt{{\vec{|R|} = {\sqrt{A^2 + B^2 + 2AB~cos\theta}}}}$

$\tt{\color{navy}{\alpha= Angle~between~{\vec{R}~and~{\vec{A}}}}}$

$\tt{\bold{ \alpha = tan^ {- 1}({\dfrac{B~sin~\theta}{A+B~cos~\theta}})}} \\\\ \tt{\bold{\vec{|R|} = {\sqrt{A^2 + B^2 + 2AB~cos\theta}}}}$

Answer 2

Answer:

Parallelogram law of vector addition states that if two vectors acting simultaneously on a particle are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point, then their resultant is completely represented in magnitude and direction by the diagonal of that parallelogram drawn from that point.

Proof:

SEE THE ATTACHMENT.

Break B in components i.e $B sin \theta$ and $B cos \theta$

Now, we can see a right angled triangle formed in the diagram.

Use Pythagoras theorem to find the length of Hypotenuse (resultant vector)

$|\vec{R} |^2 = (A+B cos \theta)^2 + ( B sin \theta)^2 \\\\|\vec{R} |^2 = A^2 + B^2 cos^2 \theta +2AB cos \theta + B^2 Sin \theta \\\\ |\vec{R} |^2=A^2 + B^2 (Cos^2 \theta + Sin^2 \theta) + 2AB Cos \theta \\\\ R = \sqrt{A^2 + B^2 + 2AB cos \theta } \:\:\:\: \because sin^2 \theta + cos^2 \theta = 1$