Vectors

Vectors

Introducing 3-dimensional vectors

(a)

A 3-dimensiaonal vector is given with reference to 3 mutually perpendicular axes - O_x , O_y O_z . That is, between each axes there is 90 degrees. (between x and y & between y and z)

(b)

A general vector is ai + bj + ck =  $\left(_{b_c}^a\right)$ 
i = unit vector in the direction of the +ve x - axis.

j = unit vector in the direction of the +ve y - axis.

k = unit vector in the direction of the +ve z - axis. 

Examples of general vectors 
 $3i\ -5j\ +7k=\left(_{-5_7}^3\right)$  
 $2i+j-3k=\left(_{1_{-3}}^2\right)$  
 $4i+11k=\left(_{0_{11}}^4\right)$  

Note : 
A unit vector has a magnitude of one unit. 
A vector can be written in cartesian form or vector form. 
The "k" , "j" and "i" are underlined.

(c)

The magnitude of a vector OP = $ai+bj+ck\ is\ \left|OP\right|=\ \sqrt{a^2+b^2+c^2}$  . 

Ex 
 $OB\ =6i-10j-15k$  
 $\therefore\left|OB\right|=\sqrt{\left(6\right)^2+\left(-10\right)^2+\left(-15\right)^2}$  
 $=\sqrt{361}$  
 $=19$  

The unit vector in the direction of a vector  $P=\frac{P}{\left|P\right|}$   $=\frac{vector\ }{magnitude\ }$  

Note : the vector is in the cartesian form. 
Ex 
 $a\ =\left(_{5_{14}}^2\right)$  
 $vector\ in\ cartesian\ form\ =\ 2i+5j+14k$  
 $\left|a\right|=\sqrt{\left(2\right)^2+\left(5\right)^2+\left(14\right)^2}$  
 $=\sqrt{225}$  
 $=15$  

 $U.V\ =\ \frac{2i+5j+14k}{15}$  
 $=\frac{1}{15}\left(2i+5j+14k\right)\ or\ \frac{2}{15}i+\frac{1}{3}j+\frac{14}{15}k$  

(d)

The position vector of a point P is  $\overrightarrow{OP}$  , the vector directed from the origin (O) to the point P. 

A general vector can be written in terms of the position vectors of its endpoints.

 $\overrightarrow{AB}\ =\overrightarrow{AO}\ +\overrightarrow{OB}$  
 $=-\overrightarrow{OA}\ +\overrightarrow{OB}$  
 $=\overrightarrow{OB}\ -\overrightarrow{OA}$  

 $\therefore\overrightarrow{AB}\ =\overrightarrow{OB\ }-\overrightarrow{OA}$  

Ex 
 $\overrightarrow{OP\ }\ =\left(_{-5_1}^3\right)$      $\overrightarrow{OQ\ }\ =\left(_{3_2}^{-4}\right)$  
 $\overrightarrow{PQ}\ =\overrightarrow{OQ}\ -\overrightarrow{OP}$  
 $=\left(3_2^{-4}\right)-\left(_{-5_1}^3\right)$  
 $=\left(_{8_1}^{-7}\right)$  

(e)

Two vectors are parallel if one is a scalar multiple of the other or if the ratios of their corresponding components are equal.

Ex

 $\overrightarrow{AB}\ =4i-2j+8k\ \ \ \ \ \overrightarrow{CD}\ =6i-3j+12k$ 
 The 2 vectors are parallel since  $\overrightarrow{AB}\ =\frac{2}{3}\overrightarrow{CD}\ \ \ \ or\ \overrightarrow{CD}\ =\frac{3}{2}\overrightarrow{AB}$  
OR 
 $\frac{4}{6}=\frac{-2}{-3}=\frac{8}{12}\left(=\frac{2}{3}\right)$  
The corresponding ratios all equal to the same fraction 2/3 .

Note : Equal vectors have the same magnitude and direction.

(f)

The scalar product (dot product) of 2 vectors a and b can be defined by  $a\ \cdot b\$  .

Given  $a=a1\ i\ +a2\ j\ +\ a3\ c\ \ \&\ \ b=b1i\ +b2\ j\ +b3\ k$  
 $a\cdot b=\left(a1\ \times b1\right)+\left(a2\times b2\right)+\left(a3\times b3\right)$  
Ex
 $a=3i\ +5j+7k\ \ \ \&\ b\ =-6i+3j+4k$  
 $a\cdot b=\left(3\times-6\right)+\left(5\times3\right)+\left(7\times4\right)$  
 $=-18+15+28$  
 $=25$  

Angle between 2 vectors (scalar product)

The formula is denoted by   $a\cdot b=\left|a\right|\times\left|b\right|\times\cos\ \theta$  , where theta is the angle between the directions of the 2 vectors. 

Ex
 $\overrightarrow{OA}\ =-4i+6j+10k\ and\ \overrightarrow{OB}=-13i+4j+7k$  
 $\overrightarrow{OA}\ \cdot\ \overrightarrow{OB}\ =\left|\overrightarrow{OA}\right|\times\left|\overrightarrow{OB}\right|\ \cos\ <AOB$  
 $\left(_{6_{10}}^{-4}\right)\cdot\left(4_7^{-13}\right)=\sqrt{\left(-4\right)^2+\left(6\right)^2+\left(10\right)^2\ }\times\sqrt{\left(-13\right)^2+\left(4\right)+\left(7\right)^2}<AOB$  
 $-4\left(-13\right)+6\left(4\right)+10\left(7\right)=\sqrt{16+36+100}\times\sqrt{169+16+49}\ \cos<AOB$   $146=\sqrt{152}\times\sqrt{234}\ \cos\ <AOB$  
 $146=\sqrt{35568}\cos<AOB$  
 $\cos\ AOB=\frac{146}{\sqrt{35568}}$  
 $AOB=\cos^{-1}\left(\frac{146}{\sqrt{35568}}\right)$  
 $\therefore AOB=\ 39.3\degree$  

Note

If the 2 vectors a and b are perpendicular to each other ( theta = 90 ) then  $a\cdot b=0$  .

Example:
 $a=20i-16j+5k\ \ \&\ b=-5i-5j+4k$  
 $a\cdot b=20\left(-5\right)+\left(-16\right)\left(-5\right)+5\left(4\right)$  
 $=-100+80+20$  
 $=0$