Question

Two vectors having magnitude 12 and 13 are inclined at an angle 45 degrees to each other. Find the resultant vector and the angle it makes with x axis.

Answer 1

use appropriate vector relations.

Answer 2

Answer:

The resultant vector and the angle it makes is $\frac{\tan ^{-1} b \sin \theta}{a+b \cos \theta}$ and 23.41⁰  respectively.

Explanation:

Let us assume 2 vectors as a and b

Therefore magnitude of a=12  

Magnitude of b=13

Q is the angle at which it is inclined=45⁰

We have to find the resultant vector and the angle  

To find resultant vector

$R=\sqrt{a^{2}+b^{2}+2 a b \cos \theta}$

Substitute the values of a and b in the above eqn  

We get

$R=\sqrt{12^{2}+13^{2}+31 \times 2 \cos 45^{\circ}}$

By simplification

$R=\sqrt{144+169+31 \times 2\left(\frac{1}{\sqrt{2}}\right)}$

=√(313+220.65)  

=√533.65  

=23.10

To find the angle

We know that the direction of the resultant vector=α=$\frac{\tan ^{-1} b \sin \theta}{a+b \cos \theta}$

$\begin{array}{l}{\alpha=\frac{\tan ^{-1} 13 \sin \theta}{12+13 \cos 45^{\circ}}} \\ {\alpha=\tan ^{-1} 13\left(\frac{1}{\sqrt{2}}\right) / 12+13\left(\frac{1}{\sqrt{2}}\right)} \\ {\alpha=\tan ^{-1}\left[\frac{13}{12 \sqrt{2}+13}\right]} \\ {\alpha=\tan ^{-1} \frac{13}{29.968}} \\ {\alpha=\tan ^{-1}(0.433)}\end{array}$

=23.41⁰