Vectors Lines and Planes

Introduction to Vectors Line and Planes

Vector Lines

1) r  =  a + tb  ,  t  `in`  R

Cartesian Form : Let OXYZ be a right handed rectangular Cartesian coordinate framework . Let i , j , k be unit vectors along the positive X , Y , Z axes respectively .

We wish to find in Cartesian form the equation of the straight line L passing through a given point A (a1 , a2 , a3 ) and having direction cosines (b1 ,  b2 , b3) . We denote that L is the straight line passing through the point A with the position vector  a = a1i + a2j + a3k and parallel to the vector  b = b1i + b2j + b3k .

Hence P(x , y , z) is a point on L  `hArr`  `EE`   t  `in`  R such that  xi + yj - zk  =  OP   =  a + tb .

`hArr`  `EE`   t  `in`  R such that  xi + yj + zk   =  (a1 + tb1) i +  (a2 + tb2) j  +  (a3 + tb3) k

`hArr`  `EE`   t  `in`   R  such that   xi + yj - zk   =   a1 + tb1  ,  y  =  a2 + tb2  ,  z  =  a3 + tb3

`hArr`  x - a1  :  y - a2  :  z - a3  : : b1 : b2 : b3          ---------------->(1)

(1) is known as the Cartesian equation of a straight line in point - Direction cosines form .   It is usually expressed as  `(x-a_1)/b_1=(y-a_2)/b_2=(z-a_3)/b_3=t`    ,    `tin`

Since  kb1 , kb2 , kb3 , k  `!=` 0  are the direction ratios of the line the equation of a straight line in point - Direction ratios are  `(x-a_1)/(kb_1)=(y-a_2)/(kb_2)=(z-a_3)/(kb_3)`

2) r  =  ( 1 - t ) a + tb  ,  t  `in`  R

Cartesian Form:

Let OXYZ  be a right handed rectangular Cartesian coordinate framework . Let i , j , k be unit vectors along the positive X , Y, Z axes respectively . Let OP = r = (x , y , z) .

We wish to find the cartesian form of the equations of the straight line passing through two given points A(a1 , a2 , a3) and B(b1 , b2 , b3)

Let a = (a1 , a2 , a3) and  b  =  (b1 , b2 , b3)

P(x , y , z) is a point on L  `hArr` xi + yj + zk  =   OP  =  (1=i)a + tb for some  t  `in`  R .

`hArr`  (x-a1)i + (y-a2)j + (z-a3)k    =   t[ (b1-a1)i + (b2-a2)j + (b3-a3)k ]  ,    t  `in`  R

`hArr`  (x-a1) :  (y-a2) : (z-a3)  : :  (b1-a1) : (b2-a2) : (b3-a3)   --------------->(2)

(2) is known as the Cartesian form of the straight line passing through given points (a1 , a2 , a3) and (b1 , b2 , b3)  . It is usually written in form

`(x-a_1)/(b_1-a_1)=(y-a_2)/(b_2-a_2)=(z-a_3)/(b_3-a_3)`    =  `t`     ,    t  `in`   R

Properties of Vectors Lines and Planes

Properties

The vector equation of the plane passing through a poi A(a) and parallel to two non-collinear vectors b and c  is   r = a + sb + tc  ;   s , t  `in`  R
The vector equation of the plane passing through three non-collinear points A(a) , B(b)( , C(c)  is  r =  (1 - s - t)a + sb + tc  ;  s , t `in`  R
Let A and B  be the distinct points and c be a vector such that AB and C are non-collinear . Then the vector equation of the plane passing through A and B and parallel to c is  r  =  (1 - s)a + sb + tc   ;  s , t  `in`  R
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Solved Problems on Vectors Lines and Planes

1) Write the vector equation of the straight line passing through the points (2i + j + 3k) , (-4i + 3j - k) . Write also the Cartesian form of the equation .

Solution : If the position vectors of A and B with respect to O are OA = a , OB = b then the vector equation of a straight line passing through A and B is

r   =  (1 - t)a + tb  ,  t is real

Hence the vector equation of the straight line passing through the given points is

r   =  (1 - t) (2i+ j + 3k)  +  t(-4i + 3j - k)

Since r = (x , y , z) , a = (2 , 1 , 3) and b = (-4 , 3 , -1) the Cartesian form of the equation is

`(x-2)/(-4-2)=(y-1)/(3-1)=(z-3)/(-1-3)`

`(x-2)/(-6)=(y-1)/(2)=(z-3)/(-4)`

2) Find the vector equation of the plane passing through the points (1 , -2 , 5) , (0 , -5 , -1) and (-3 , 5 , 0)

Solution : If the position vectors of A , B and C with respect to O are OA = a , OB = b , OC = c then the vector equation of a straight line passing through A,B and C  is     (1 - s - t)a + sb + tc

Hence the vector equation of the straight line passing through the given three points is

r   =   (1 - s - t)(i - 2j + 5k)  +  s(-5j - k) + t(-3i + 5j)