Physics Motion in Plane

It resolves vectors into components, adds corresponding x and y values algebraically, then reconstructs the resultant using magnitude and direction. That is, applying vector addition rules.

Vector resolution is important when analysing motion in two or three dimensions. It's used in vector addition and subtraction, projectile motion, and expressing velocity or acceleration using component form. Resolution with unit vectors also helps in applying kinematic equations to multidimensional problems. 

Vectors can be resolved generally into any two non-parallel directions. These are done more as rectangular components using unit vectors. The latter is most common in physics and engineering to represent them into Cartesian coordinates for simplified vector operations.

Vector resolution is important to break complex motion into simpler, one-dimensional components. With it, we can simplify vector addition and analysis in 2D or 3D motion. Generally, we rely on perpendicular directions. That makes vector operations with displacement, velocity, and acceleration more manageable and accurate.

Scalar multiplication requires one vector and one scalar. Dot product requires two vectors. That's the basic difference. 

Vectors are not just numbers like scalars that have only magnitude. Vectors have direction and magnitude, and multiplying two vectors or multiplying a vector by a real number would yield results as per rules of vector algebra. 

The three different ways of vector multiplication are

• Scalar Multiplication - This is multiplying a vector with a scalar quantity. This operation results in another vector. 
• Dot Product - This is an operation that tells us that two vectors gives a scalar as a result. 
• Cross Product - This is an operation, when two vectors are multiplied, yields a vector that is perpendicular to both. 

Horizontal component of velocity of bomb equals to the velocity of plane. So, bomb and plane are in same vertical line always.

$tan\beta =\frac{Bsin60°}{A-Bcos60°}$

$\Rightarrow \beta =ta{n}^{-1}\left(\frac{\sqrt[]{3}B}{2A-B}\right)$

Let's say $\left|\overrightarrow{OA}\right|=\left|\overrightarrow{OB}\right|=\left|\overrightarrow{OC}\right|=\mathcal{l}$ $\stackrel{}{}\stackrel{}{}\stackrel{}{}$

$\overrightarrow{OA}=\mathcal{l}cos30°\widehat{i}+\mathcal{l}sin30°\widehat{j}$

$\overrightarrow{OA}+\overrightarrow{OB}-\overrightarrow{OC}$

= $\left[\mathcal{l}cos30°\widehat{i}+\mathcal{l}sin30°\widehat{j}\right]+\left[\mathcal{l}cos60°\widehat{i}+\mathcal{l}sin60°\left(-\widehat{j}\right)\right]-\left[\mathcal{l}cos45°\left(-\widehat{i}\right)+\mathcal{l}sin45°\left(+\widehat{j}\right)\right]$ $\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}$

$=\mathcal{l}\left(\frac{\sqrt[]{3}}{2}+\frac{1}{2}+\frac{1}{\sqrt[]{2}}\right)\widehat{i}+\mathcal{l}\left[\frac{1}{2}-\frac{\sqrt[]{3}}{2}-\frac{1}{\sqrt[]{2}}\right]\widehat{j}$

$tan\theta =\frac{\mathcal{l}\left[\frac{1}{2}-\frac{\sqrt[]{3}}{2}-\frac{\sqrt[]{2}}{2}\right]}{\mathcal{l}\left[\frac{\sqrt[]{3}}{2}+\frac{1}{2}+\frac{\sqrt[]{2}}{2}\right]}=\frac{1-\sqrt[]{3}-\sqrt[]{2}}{1+\sqrt[]{3}+\sqrt[]{2}}$