Ooo Vectors

I had to learn basic things about vectors. So thought why not write a blog post :P? 

Also special thanks to papa 💫

 So here we go..

PHYSICAL QUANTITY:-

A quantity that can be measured and expressed in units. We have two types of Physical Quantity

• Scalar:- A physical quantity that has only magnitude. Example: Mass, speed.
• Vectors:- A physical quantity that has both magnitude and direction. Example: Displacement, velocity. 

Now, what are vectors?

VECTORS:-

• A physical quantity that has both magnitude and direction.
• In geometry, a vector is a directed line segment, whose length is the MAGNITUDE of the vector and the arrow indicating the DIRECTION.
• The MAGNITUDE of a vector $(\vec{A})$ is the absolute value of a vector and is indicated as $\boxed{|\vec{A}|}.$

REPRESENTATION OF VECTORS:-

• A vector pointing from $A$ to $B$ is written as $\vec{AB}.$
• We usually define a single point $O$ as the ORIGIN. Then we associate every point $P$ with the vector $\vec{OP}$ or $\vec{P}$

MAGNITUDE OF VECTORS:-

• It's just the distance between the initial point $A$ and endpoint $B.$
• We consider the cartesian plane, let $A$ has coordinate $(x_0,y_0)$ and $B$ has coordinate $(x_1,y_1).$ 
• Then  $\vec{AB}= ~~\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}.$
• Letting $A$ as the origin and end point $B=~~(x,y),$ we get $\vec{AB}=~~\sqrt{x^2+y^2}$ 

EQUALITY OF VECTORS:-

• Two vectors are called equal if their magnitudes and direction are the same. 
• So a vector from $(0,0)$ to $(0,1)$ is same as $(0,2)$ to $(0,3).$

ADDITION OF VECTORS:- 

• We use the Parallelogram Law to add vectors.
• The vector pointing to $(x,y)$ from $(0,0)$ is denoted as $\langle x,y \rangle.$
• If we have two vectors $\langle x_1,y_1 \rangle,~~\langle x_2,y_2 \rangle ,$ then adding we get $\langle x_1+x_2,~~y_1+y_2 \rangle.$
• We also have another way. Let $\vec{AB}=\vec{a},~~ \vec{AC}= \vec{b}.$ Let $D$ be a point such that $ABDC$ is a Parallelogram. We say  $\vec{AB}+\vec{AC}=\vec{AD}.$
• Now, what about the magnitude of $\vec{AD}?$
• Let's drop the perpendicular from $D$ to $AB$ and call the feet $E.$ 
• Suppose $\angle CAB=\theta$ and $\angle DAB=\alpha .$
• So we get $$AD^2=~~(AB+BE)^2+(DE)^2$$ $$= (|\vec{a}|+\vec{b}\cos \theta)^2+~~(|\vec{b}|\sin \theta)^2$$ $$= |\vec{a}|^2+|\vec{a}||\vec{b}|\cos \theta+|\vec{b}|^2.$$
• Hence, the magnitude of $$\boxed{ \vec{a}+ \vec{b}=~~\sqrt{ |\vec{a}|^2+|\vec{a}||\vec{b}|\cos \theta+|\vec{b}|^2}}$$

FUN FACT:- The sum of the coordinate of the points of a parallelogram are always equal.

Note:- We also have triangle law of addition, but that gives the same result ( here we basically join one head of the vector and one vector's tail and the resultant vector form the third side of the triangles, whose two sides are the two given vectors. )

More like, we have $\vec{BD}=\vec{b}$ and we are applying the triangle law. I think parallelogram is more famous (?)

MULTIPLICATION OF A VECTOR BY A SCALAR:- 

• Suppose $\vec{a}$ is a vector and $k$ be any real number ( A scalar quantity).
• We define $\vec{b}=k\vec{a}.$ And the magnitude would be $k|\vec{a}|.$
• The vector $\vec{b}$ has same direction if $k$ is positive, opposite direction if $k$ is negetive.

SUBTRACTION OF VECTORS:- 

• Let $\vec{a}$ and $\vec{b}$ be two vectors. we define $\vec{a}-\vec{b}$ as the sum of vector $\vec{a}$ and $\vec{-b}$
• Note that $\vec{-b}$ is just the vector $\vec{b}$ but with invited direction.

RESOLUTION OF VECTORS:-

• In the above diagram, we have $\vec{a}=\vec{OA}=\vec{OB}+\vec{OC}.$
• But note that $OB=|\vec{a}|\cos \alpha$ and $OC=|\vec{a}|\cos \beta.$

Now, we define $\hat{i},\hat{j},\hat {k}$ as the unit vector along $OX,OY,OX$ (It's like a unit which is introduced to simplify algebric manupilations, we will understand that later)

• So $\vec{a}=|\vec{a}|\cos\alpha\hat{i}+~~|\vec{a}|\cos \beta\hat {j}.$

This we talked in $X-Y$ plane, right? Now, what about $X-Y-Z$ plane? Well.. it's the same!

• We have $$\boxed{\vec{a}=|\vec{a}|\cos\alpha\hat{i}+ ~~| \vec{a}|\cos \beta\hat {j}+~~|\vec{a}|\cos\gamma\hat{k} }$$ where $\vec{a}$ makes angle of $\gamma$ on $Z$ axis. 

By this we can see that any vector in $\Bbb {R_3}$ can be expressed as a linear combination of $3$ unit vectors i.e $\hat{i},\hat{j},\hat {k}.$

P.S. I actually love the next two parts and the above part :P . It's math u see :O

DOT PRODUCT or SCALAR PRODUCT:-

• We define, $\vec{a}\cdot \vec{b}=|\vec{a}|\cdot |\vec{b}|\cos\theta$ where $\theta $ is the angle between vectors $\vec{a}$ and $ \vec{b}.$

Here are few examples which will help you :

$$ \hat{i}\cdot \hat{i}=1$$

$$\hat{j}\cdot \hat{j}=1 $$

$$ \hat{k}\cdot \hat{k}=1$$

$$\hat{i}\cdot \hat{j}=0$$

$$ \hat{j}\cdot \hat{k}=0$$

$$ \hat{i}\cdot \hat{k}=0$$

Well.. this makes sense since $\sin 90=0$ and $\sin 0=1.$

Also, a fun thing to note is that the dot product is commutative and distributive too. Why am I saying this? We will understand when I say about cross-product...

Anyways here's a fun example to solve!

Example:- Find $$(\hat{i}+\hat{2j})\cdot(3\hat{i}-5{j})$$

Solution:- Note that $$(\hat{i}+\hat{2j})\cdot(3\hat{i}-5{j})$$ $$=3\hat{i}\cdot\hat{i}-5\hat{j}{i}+2\hat{j}\cdot 3\hat{i}-10\hat{j}\cdot\hat{j}= 3-0+0-10=-7.$$

CROSS PRODUCT or VECTOR PRODUCT:-

• The cross product or the vector product of the two vectors is $$\boxed{|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\sin\theta}$$ where $\theta$ is the angle between vector $\vec{a}$ and $\vec{b}$
• The direction of vector $\vec{a}\times \vec{b}$ is perpendicular to both $\vec{a}$ and $\vec{b}.$

Here are few examples which will help you :

$$ \hat{i}\times \hat{i}=0$$

$$\hat{j}\times \hat{j}=0 $$

$$ \hat{k}\times \hat{k}=0$$

$$\hat{i}\times \hat{j}=\hat{k}$$

$$ \hat{j}\times\hat{k}=\hat{i}$$

$$ \hat{k}\times\hat{i}=\hat{j}$$

$$\hat{j}\times \hat{i}=-\hat{k}$$

$$ \hat{k}\times\hat{j}=-\hat{i}$$

$$ \hat{i}\times \hat{k}=-\hat{j}$$

Well, by now you must have understood that Cross product isn't commutative. But it does follow distributive law.

Anyways here's another fun example to solve!

Example:- Find $(2\hat{i}+3\hat{j})\times(2\hat{i}+3\hat{j})$

Solution:- Won't write :P , but from this we can conclude that for any vector $\vec{u}$ we have $\vec{u}\times \vec{u}=0.$

AREA, DETERMINANTS and CROSS PRODDUCTS:-

• Note that $|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\sin\theta$
• But we have $$Area=\frac 12 ab\sin\theta=\frac{|\vec{a}||\vec{b}|\sin\theta}{2}=\frac{\vec{a}\times \vec{b}|}{2} $$.

Note that $|\vec{a}\times \vec{b}|$ is scalar

Now, let's do a problem to understand this "Area" topic better.

Examples:- Find the area of triangle with vertices $(-1,-3),~~(4,1),~~(-2,3).$

Solution:- We know that transition doesn't affect the area so..

$$(-1,-3)\rightarrow (0,0)$$

$$(4,1)\rightarrow (5,4)$$

$$(-2,3)\rightarrow (-1,6)$$

Now, we locate these points in the coordinate plane.

Clearly we get $[ABC]=[DBFE]-[ABD]-[ACE]-[BCF]=36-3-10-3=17.$

Now, plotting these points take a bit of time right? So we present two more solutions.

Second solution:-  Remember cross-products? We are going to use that!

We have $\langle 5,4\rangle = 5\hat{i}+4\hat{j}.$ Similarly, we have $\langle 1,6\rangle =-\hat{i}+6\hat{j}.$

So taking cross product, we get $|(5\hat{i}+4\hat{j})\times (\hat{i}+6\hat{j})|=|34\hat{k}|=34.$

So area is $\frac{34}{2}=17.$

Third Solution:- Now we will use matrix :)

Another problem!

Find the area of the triangle (-1,-2,-3),(1,2,3),(3,2,1)$

Solution:- Again transitioning doesn't change the area, so..

$$(-1,-2,-3)\rightarrow (0,0,0)$$

$$(1,2,3)\rightarrow (2,4,6)$$

$$(3,2,1)\rightarrow (4,4,4)$$

Now, we use heron's formula.

Calculating the distances between each point, we get the distance $\sqrt{56},\sqrt{8},\sqrt{48}.$

So we get $s= \sqrt{14}+\sqrt{2}+2\sqrt{3}.$

And so we get $$A=\sqrt{\left(\sqrt{14}+\sqrt{2}+2\sqrt{3}\right)\left(\sqrt{14}+\sqrt{2}+2\sqrt{3}-\sqrt{56}\right)\left(\sqrt{14}+\sqrt{2}+2\sqrt{3}-\sqrt{8}\right)\left(\sqrt{14}+\sqrt{2}+2\sqrt{3}-\sqrt{48}\right)}$$ $$=4\sqrt{6}.$$

Second Solution:- 

We use matrix which is way faster.

Now taking magnitude, using the fact that cross factor is two times the area i.e

Area=$\frac12(\sqrt{64+16^2+64})= 4\sqrt{6}.$

Anyways hope this post helps someone :) If you liked this post and want more posts like these then do subscribe to Sunaina Thinks Absurd.

Sunaina💜