Question

show that the resultant of two vectors a and b inclined at an angle θ is r=√(a2+b2+2abcosθ)

Answer 1

Answer:

or , AC = AB cosθ

or , AC = OD cosθ

= Q cosθ [ AB = OD = Q ]

or , BC = AB sinθ

or , BC = OD sinθ

= Q sinθ [ AB = OD = Q }

R = P2+2PQcosθ+Q2.

Let ϕ be the angle made by resultant R with P . Then,

Answer 2

Given:

Two vectors $\vec a$ and $\vec b$ makes angle $\theta$

To Find:

The magnitude of resultant of the given two vectors are$\bold{r = \sqrt{a^2+b^2+2ab\cos \theta}}$

Solution:

Using the given vectors $\vec a$ and $\vec b$ let us draw* a diagram ABCD, where angle between $\vec a$ and $\vec b$ is $\theta$ and $\vec r$ be the resultant vector.

Now in the diagram extend $\Vec {AC}$ to point E(let)

∴From the diagram

$CE = a\cos \theta$

and $ED = a \sin \theta$

Here in the triangle $\Delta AED$, we get,

$|\vec {AE}| = (b + a\cos \theta)$

∴Magnitude of the resultant vector of $\vec a$ and $\vec b$ is,

$|\vec r|^2 = (b + a\cos \theta)^2+(a\sin \theta)^2$

$=> |\vec r|^2 = b^2 + a^2\cos^2 \theta+2ab\cos \theta+a^2\sin^2 \theta$

$=> r^2 = a^2+ b^2+2ab\cos \theta$

$=> r= \sqrt{a^2+ b^2+2ab\cos \theta}$

∴ The resultant of two vectors $\vec a$ and $\vec b$ inclined at an angle θ is $\bold{ r= \sqrt{a^2+ b^2+2ab\cos \theta}}$

(*Diagram is attached below)