Vectors : Planes and Lines

Planes and Lines

The equation of a 3-dimensional Line

The line has a unique position in space if

(a) It has a known direction and passes through a known fixed point or

(b) It passes through 2 known fixed points.

Points to note:

* The line does not have a gradient.

*The direction of a line can be found using 2 points on the line or a vector that is parallel to the line is known.

*2 lines in space can be either parallel, skewed, or interesting at one point.

*The equation of a line can be represented in 3 forms; Vector equation, Parametric equation, Cartesian equation.

Vector equation of a line.

r = a + tb

where

t = parameter (any real number/constant)

a = any known point on the line

r = position vector of any point on the line

For each value of the parameter t , this equation gives the position vector of 1 point on the line.

Vector equation of a Line

r = a + tb

where

t = parameter (any real number/constant)

a = any known point on the line

r = position vector of any point on the line

b = the direction vector of the line

For each value of the parameter t , this equation gives the position vector of 1 point on the line.

Parametric equation of a line

Substituting  $\overline{r}=\left(_{y_z}^x\right)\ \$  ,  $\overline{a}\ =\left(_{y_{1_{z_1}}}^{x_1}\right)$  ,  $\overline{b}\ =\left(_{m_n}^l\right)$  gives 

 $\left(_{y_z}^x\right)\ =\left(_{y_{1_{z_1}}}^{x_1}\right)\ +t\left(_{m_n}^l\right)$   $\Longrightarrow\left(_{y_z}^x\right)\ =\left(_{y_{1_{z_1}}}^{x_1}\right)+\left(_{tm_{tn}}^{tl}\right)$   $\Longrightarrow\left(_{y_z}^x\right)\ =\left(_{y_1+tm_{z_1+tn}}^{x_1+tl}\right)$  

Parametric equation of a line
 $x=x_1+tl$  
 $y=y_1+tm$  
 $z=z_1+tn$  

where  $x_1,y_1,\ z_{1\ ,\ l\ ,\ m\ ,\ n\ }$  represent integers.

Cartesian equation of a line

By rearranging this parametric form to make t the subject of the formula :

 $t=\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$  

Therefore the cartesian equation of a line is 
 $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$  

where  $x_1,\ y_1,\ z_1,\ l\ ,\ m\ ,\ n$  are integers.
Note: this is a standard formula therefore if x1,y1 or z1 is -ve then the operation will be + instead of -.

Examples

1.                                          Parametric form               Cartesian form:
Vector form :                      $x=4+6t$                $\frac{x-4}{6}=\frac{y-3}{-5}=\frac{z+2}{2}$  

 $\overline{r}\ =\left(_{3_{-2}}^4\right)\ +t\left(_{-5_2}^6\right)$       $y=3-5t$  
                                            $z=-2+2t$  

2. Vector form :                 Parametric form :     Cartesian equation : 
 $\overline{r}\ =\left(_{-7_{-2}}^5\right)+t\left(_{-3_0}^4\right)$       $x=5+4t$               $\frac{x-5}{4}=\frac{y+7}{-3},\ z=-2$    *z is fixed
                                              $y=-7-3t$    
                                              $z=-2$  

3. Parametric form:         Vector form :                         Cartesian form:
 $x=3-4\lambda$                   $\overline{r}\ =\left(_{6_0}^3\right)\ +\lambda\left(_{0_9}^{-4}\right)$                     $\frac{x-3}{-4}=\frac{z}{9},y=6$  *y is fixed
 $y=6$                     
 $z=9\lambda$  

ex

Find the vector of a line which passes through the points  $A\left(3,-4,7\right)\ and\ B\ \left(5,-6,15\right)$  .

 $\overrightarrow{AB}\ =\overrightarrow{OB}\ -\overrightarrow{OA}\$  
 $=\left(_{-6_{15}}^5\right)-\left(_{-4_7}^3\right)$  
 $=\left(_{-2_8}^2\right)$  
 $\therefore$  The direction vector is  $\left(_{-1_4}^1\right)$  
(  divide by 2)
 $\overline{r}\ =\overline{a}\ +t\overline{b}$  
 $Vector\ equation\ :\ \overline{r}\ =\left(_{-4_7}^3\right)\ +t\left(_{-1_4}^1\right)\ or\ \overline{r}\ =\left(_{-6_{15}}^5\right)+t\left(_{-1_4}^1\ \right)$  

ex2

The line passes through the point  $\left(2,4,5\right)$  in the direction -2i + 3j + 8k. Find the value of p and q so that the point (p,10,q) lies on the line. 

 $\overline{r}\ =\left(_{4_5}^2\right)\ +t\left(_{3_8}^{-2}\right)\$  is the equation of the line 
 $\left(p,10,q\right)\ lies\ on\ the\ line$  
 $\left(_{10_q}^p\right)=\left(_{4+3t_{5+8t}}^{2-2t}\right)$      
 $10=4+3t$             $\therefore p=2-2\left(2\right)$        $\therefore q=5+8\left(2\right)$  
 $6=3t$                           $p=2-4$                  $q=5+16$  
 $t=2$                               $p=-2$                   $q=21$  

Relative position of 2 lines in space

The location of 2 lines in space fits one of the following 3 situations:

(a) They are parallel.

(b) The lines are not parallel and intersect at one point.

(c)The lines are not parallel and do not intersect. Such lines are said to be skewed.

2 lines are parallel if their direction multiples of each other. (These lines don't deal with gradients)

2 lines intersect if they is a point common to both lines.

ex1

 $l_1:\ \overline{r}\ =\left(_{3_{-7}}^2\right)+\lambda\left(_{-2_3}^4\right)$  
 $l_2\ :\ \overline{r}\ =\left(_{1_9}^4\right)+\mu\left(_{4_{-6}}^{-8}\right)\$  
 $\left(_{-2_3}^4\right)\ =\frac{1}{2}\left(_{4_{-6}}^{-8}\right)\ or\ \left(_{4_{-6}}^{-8}\right)\ =-2\left(_{-2_3}^4\right)or\ \frac{-4}{8}=\frac{-2}{4}=\frac{3}{-6}=\left(-\frac{1}{2}\right)$  
The lines are parallel since their direction vectors are multiples of each other. 

ex2

 $l_1:\ \overline{r}\ =\left(_{-1_3}^1\right)\ +\mu\left(\ _{-1_1}^1\right)$  
 $l_2\ :\ \overline{r}\ =\left(_{4_6}^2\right)+\lambda\left(_{1_3}^2\right)\$  
The lines are not parallel since the direction vectors are not multiples of each other.
If they intersect: 
 $\left(_{-1-\mu_{3+\mu}}^{1+\mu}\right)=\left(_{4+\lambda_{6+3\lambda}}^{2+2\lambda}\right)$                               $sub\lambda=-2\ into\ \left(2\right)$  
 $1+\mu=2+2\lambda\ \left(1\right)$                                $-1-\mu=4-2$  
 $-1-\mu=4+\lambda\longrightarrow\mu=-5-\lambda\left(2\right)$       $-1-\mu=2$  
 $3+\mu=6+3\lambda\ \left(3\right)$                                        $-\mu=3\rightarrow\mu=-3$  
sub (2) into (3)                                       
 $3+\left(-5-\lambda\right)=6+3\lambda$             sub  $\lambda=-2\ and\ \mu=-3$  into (1)
 $3-5-\lambda=6+3\lambda$                      $1-3=2+2\left(-2\right)$  
 $-2-\lambda=6+3\lambda$                             $-2=-2$  
 $-8=4\lambda\rightarrow\lambda=-2$                 The values satisfy equ (1) Therefore the lines intersect.

Finding the point of intersection

when  $\mu=-3$              or                 when  $\lambda=-2$ 
 $\overline{r}\ =\left(_{-1_3}^1\ \right)+\left(-3\right)\left(_{-1_1}^1\right)\$         $\overline{r}\ =\left(_{4_6}^2\right)+\left(-2\right)\left(_{1_3}^2\right)\$  

 $=\left(_{-1_3}^1\right)\ +\left(_{3_{-3}}^{-3}\right)\$                             $=\left(_{4_6}^2\right)\ +\left(_{-2_{-6}}^{-4}\right)$  
 $=\left(_{2_0}^{-2}\right)$                                              $=\left(_{2_0}^{-2}\right)$  

ex3

 $l_1:\overline{r}\ =\left(_{0_1}^1\right)\ +\lambda\left(_{3_4}^1\right)$  
 $l_2\ :\ \overline{r}\ =\left(_{3_0}^2\right)\ +\mu\left(_{-1_1}^4\right)\$  

These 2 lines are not parallel because their direction vectors are not multiples of each other. The lines either intersect or skewed. If they intersect :
 $\left(_{3\lambda_{1+4\lambda}}^{1+\lambda}\right)=\left(_{3-\mu_{\mu}}^{2+4\mu}\right)$  
 $1+\mu=2+4\mu\left(1\right)$  
 $3\lambda=3-\mu\left(2\right)$  
 $1+4\lambda=\mu\left(3\right)$                     $1+4\left(\frac{2}{7}\right)=\mu$                  sub  $\lambda=\frac{2}{7}and\ \mu=\frac{15}{7}$  into (1)
sub (3) into (2)                         $1+\frac{8}{7}=\mu$                         $1+\frac{2}{7}=2+4\left(\frac{15}{7}\right)$  
 $3\lambda=3-\left(1+4\lambda\right)$               $\mu=\frac{15}{7}$                                      $\frac{9}{7}\ne\frac{74}{7}$  
 $3\lambda=3-1-4\lambda$                        These 2 values does not satisfy the equation 1. The lines do not intersect     
 $7\lambda=2\rightarrow\lambda=\frac{2}{7}$                        and are skewed. 
 $sub\lambda=\frac{2}{7}into\ \left(3\right)$  

Cross Product

If the vectors  $\overline{a\ }\ and\ \overline{b}\$  are both parallel to a given plane, then their cross product  $\overline{a}\ \times\overline{b}\$  is a vector which is perpendicular to the plane. 

*Cross product uses the Determinant idea of vectors.  $\left|_b^a\ _d^c\right|\ =ad-bc$  
* The standard final form is xi - yj + zk , where x ,y and z are integers. 
Consider  $\overline{a}\ =\left(_{a_{2_{a3}}}^{a_1}\right)\ and\ \overline{b}\ =\left(_{b_{2_{b_3}}}^{b_1}\ \right)$  
 $\overline{a}\ \times\overline{b}\ =\left(_{a_{2_{a_3}}}^{a_1}\right)\ \times\left(_{b_{2_{b_3}}}^{b_1}\right)$  
 $=\left|_{a_3}^{a_2}\ \ \ _{b_3}^{b_2}\right|i\ -\ \left|_{a_3}^{a_1}\ _{b_3}^{b_1}\right|j\ +\left|_{a_2}^{a_1}\ _{b_2}^{b_1}\right|k$  
 $=\left(a_2\ b_3-a_3b_2\right)i-\left(a_1b_3-a_3b_1\right)j+\left(a_1b_2-a_2b_1\right)k$  

Exs

 $\left(_{7_4}^5\right)\times\left(_{3_6}^1\right)$  
 $\left|_4^7\ _6^3\right|\ i-\left|_4^5\ _6^1\right|j\ +\left|_7^5\ _3^1\right|k$  
 $\left(42-12\right)i-\left(30-4\right)j+\left(15-7\right)k$  
 $30i-24j+8k$  

 $\left(_{-2_8}^3\right)\times\left(_{7_{-2}}^6\right)$  
 $\left|_8^{-2}\ _{-2}^7\right|i-\left|_8^3\ _{-2}^6\right|j+\left|_{-2}^3\ _7^6\right|k\$  
 $\left(4-56\right)i-\left(-6-48\right)j+\left(21-\left(-12\right)\right)k$  
 $-52i+54j+33k$  

The equation of a plane

Vector equation of a plane 

 $\overline{r}\ \cdot\ \overline{n}\ =d\$  
where  $d=\overline{a}\ \cdot\overline{n}\ \ \ \ \ and\ \overline{a}\$  is a point on the plane. (dot product using the 2 vectors gives the constant d.)

Points to note:
*The equation of a plane needs a point on the plane and the normal vector to the plane in order to determine the equation. 
*The cross product of 2 vectors parallel to a given plane gives the normal vector to the plane. 
*The equation of a plane can be in vector form or cartesian form. 

If  $\overline{n}\ =\ \left(_{b_c}^a\right)\ and\ vector\ \overline{r}\ =\left(_{y_z}^x\right)$  , then we have 

 $\overline{r}\ \cdot\overline{n}\ =d\ \rightarrow\left(_{y_z}^x\right)\ \cdot\left(_{b_c}^a\right)\ =\ d$  
 $ax+by+cz=d$  

Cartesian equation of a plane 
 $ax+by+cz=d$  
where  a, b, c are intergers and d is a constant.

exs

1.

vector equation of a plane :  $\overline{r}\ \cdot\left(_{4_7}^5\right)=10$  

Cartesian equation of the plane :  $5x+4y+7z=10$  

2.The point (3, 4 ,7) lies on a plane whose normal vector is 2i - j + 5k. Find the vector equation and Cartesian equation of the plane.
 $\overline{r}\ \cdot\overline{n}\ =\overline{a}\ \cdot\overline{n}\$  
 $\overline{r}\ \cdot\left(_{-1_5}^2\right)\ =\left(_{4_7}^3\right)\ \cdot\left(_{-1_5}^2\right)$  
 $\overline{r}\ \cdot\left(_{-1_5}^2\right)=6-4+35$  
 $\overline{r}\cdot\left(_{-1_5}^2\right)=37$  (vector equation) 
 $2x-y+5z=37$  (cartesian equation)

3. The plane  $\pi$  contains the 3 points  $A\left(1,3,2\right)\ ,\ B\left(2,4,1\right)\ and\ C\left(6,-2,3\right).$  Find the equation of the plane.

 $\overrightarrow{AC}\ =\ \overrightarrow{OC}\ -\overrightarrow{OA}$  
 $=\left(_{-2_3}^6\right)\ -\left(_{3_2}^1\right)$  
 $=\left(_{-5_1}^5\right)$  

 $\overrightarrow{AB}\ =\overrightarrow{OB}-\overrightarrow{OA}$  
 $=\left(_{4_1}^2\right)\ -\left(_{3_2}^1\right)$  
 $\left(_{1_{-1}}^1\right)$  

The normal vector is perpendicular to both  $\overrightarrow{AB}\ and\ \overrightarrow{AC}$   . Therefore, it is parallel to  $\overrightarrow{AB}\ \times\overrightarrow{AC}$  .                                                 vector equation of plane :      

 $\overrightarrow{AB}\ \times\overrightarrow{AC}\ =\left(_{1_{-1}}^1\right)\ \times\left(_{-5_1}^5\right)$                          $r\cdot\left(_{3_5}^2\right)\ =\left(_{3_2}^1\right)\ \cdot\left(_{3_5}^2\right)$  
 $=\left|_{-1}^1\ _1^{-5}\right|i\ -\left|_{-1}^1\ _1^5\right|j\ +\ \left|_1^1\ _{-5}^5\right|k$                      $\overline{r}\ \cdot\ \overline{n}\ =\overline{a}\ \cdot\ \overline{n}$  
 $=\left(1-5\right)i-\left(1+5\right)j+\left(-5-5\right)k$  $\ \ \ \ \ \ \ \ \ \ \ \overline{r}\ \cdot\left(_{3_5}^2\right)=2+9+10\$                                                                                                               $\overline{r}\ \cdot\left(_{3_5}^2\right)=21$  
 $=-4i-6j-10k$                                        Cartesian equation of plane : 
Normal vector to plane  $\left(_{3_5}^2\right)$                               $2x+3y+5z=21$  
(divide by -2)

Relationship between a line and a plane

A line can be :
*parallel (and external) to a plane
*Contained in a plane
* intersecting a plane at one point.

Consider the line L and the plane  $\pi$  whose equations ae respectively  $\overline{r}\ =\overline{a}\ +\ t\ \overline{b}\ and\ \overline{r}\ \cdot\ \overline{n}\ =d$  . If L is parallel to  $\pi$  or is contained in  $\pi$  , then the vector b and n are perpendicular to each other, that is  $\overline{b}\ \cdot\overline{n}\ =0$  .

If the known point on L is also known point on  $\pi$   then the line is contained in the plane . 

 $If\ \overline{b}\ \cdot\ \overline{n}\ \ne0,$  the line  is not parallel to the plane but instead, intersects the plane at one point. 

Egs

1.  $l:\ \overline{r}\ =\left(_{-1_4}^6\right)\ +\mu\left(_{-1_4}^1\ \right)\ and\ \pi\ :\ x+5y+z=5$  
Direction vector of line =  $\left(_{-1_4}^1\right)\$  
Normal vector of plane =  $\left(_{5_1}^1\right)\$  
 $\overline{b}\ \cdot\ \overline{n}\ =\left(_{-1_4}^1\right)\ \cdot\left(_{5_1}^1\right)$  
 $=1-5+4$  
 $=0$  
The line is either parallel to the plane or is contained in the plane. 
(6,-1,4) is a point on the line. Sub point into equation of plane.
6 + 5(-1) + 4 = 5 
1 + 4 = 5 
5 = 5 
The point also lies in the plane therefore the line L is contained in the plane pi.
 

2.  $L:\overline{r}\ =\left(_{1_5}^3\right)\ +\ \lambda\left(_{-1_4}^1\right)\ and\ \pi:\ x+5y+z=5$  

Direction vector of the line =  $\left(_{-1_4}^1\right)$  
Normal vector of plane=  $\left(_{5_1}^1\right)$  
 $\overline{b}\ \cdot\overline{n}\ =\left(_{-1_4}^1\right)\cdot\left(_{5_1}^1\right)$  
 $=1-5+4$  
 $=0$  
The lie is either parallel or is contained in the plane. 
(3,1,5) is a point on  the line . sub point into plane.
3 + 5(1) + 5 = 5
13   $\ne$  5 
Therefore the point does not lie on the plane and the line is parallel to the plane.

3.  $l:\overline{r}\ =\left(_{0_1}^1\right)+t\left(_{1_{-3}}^2\right)\ and\ \pi:\ 3x+2y+4z=11\$  

Direction vector of the line =  $\left(_{1_{-3}}^2\right)$  
Normal vector of plane = $\left(_{2_4}^3\right)$  
 $\overline{b}\ \cdot\ \overline{n}\ =\left(_{1_{-3}}^2\right)\cdot\left(_{2_4}^3\right)$  
 $=6+2-12$  
 $=-4$  
The line is not parallel to the plane. It intersects at one point. Substituting the parametric form of the line into the equation of the plane gives the value of t .
 $x=1+2t,\ y=t,\ z=1-3t$  (sub into pi)                    finding  point of intersection                                                                                                                                           $\overline{r}\ =\left(_{0_1}^1\right)+\left(-1\right)\left(_{1_{-3}}^2\right)$  
 $3\left(1+2t\right)+2t+4\left(1-3t\right)=11$                                            $=\left(_{0_1}^1\right)+\left(_{-1_3}^{-2}\right)$  
 $3+6t+2t+4-1+2t=11$                                                $=\left(_{-1_4}^{-1}\right)$  
 $7-4t=11$                                                 The point of intersection is (-1,-1,4).
 $4t=-4$  
t = -1