The vector equation of a line

1

1

x xy yr

The position vector of a set of points are given by

r = OR = OA + AB

0

A

AB R

Cartesian form and vector form

r A p

The line goes through (2, -1) with gradient ¾ Cartesian form

1

1

x xy yr

y – y0 = m(x – x0)

y – -1 = ¾ (x – 2)

y + 1 = ¾ (x – 2)

4y + 4 = 3(x – 2)

4y + 4 = 3x - 6

3x – 4y – 10 = 0

Vector form

2 41 3r

Changing Vector form into Cartesian form

2 41 3r

2 41 3

xy

x = 2 + 4 [1]

y = -1 + 3 [2]

3x = 6 + 12 [3]3 [1]

4y = -4 + 12 [4]4 [2]

3x – 4y = 10[3]-[4]

3x – 4y – 10 = 0

Changing Cartesian form into Vector form

0 13 2r

03xyA

y = 2x - 3

Gradient = 2 and y intercept = (0, - 3)Vector form 12Direction gradient p

1

1

x xy yr

Basic unit form: xi + yj = (0i – 3j) + (1i + 2j)

Special directions

Parallel to the y-axisFind a Vector equation for the line through (2, 1) parallel to y -axis, and deduce its Cartesian equation.

Basic unit vectors: (xi + yj) = (x1i + y1j)+(xi + yj)

(xi + yj) = (2i + 1j)+j

A vector parallel to the y-axis is j

Cartesian form

x = 2 and y = 1 + y

x

x = 2

x = 2

Special directions

Parallel to the x-axisFind a Vector equation for the line through (1, 3) parallel to x -axis, and deduce its Cartesian equation.

Basic unit vectors: (xi + yj) = (x1i + y1j)+(xi + yj)

(xi + yj) = (i + 3j)+i

A vector parallel to the x-axis is i

Cartesian form

x = 1+ and y = 3y

x

y = 3

y = 3

The intersection of two lines

2 1 6 13 2 1 3r and r

47r

Find the position vector of the point where the following lines intersect:

When the lines intersect, the position vector is the same for each of them.

2 1 6 13 2 1 3

xyr

x: 2 + = 6 + (1)

y: 3 + 2 = 1 - 3 (2) = 2 and = - 2

The intersection of two lines

2 3 1 01 0 3 2r s and r t

11r

Find the position vector of the point where the following lines intersect:

When the lines intersect, the position vector is the same for each of them.

2 3 1 01 0 3 2

xyr s t

x: 2 + 3s= -1 (1)

y: 1 = 3 – 2t (2)s = -1 and t = 1

Vectors in three dimensionsPosition vector r:

1

1

1

x x x

r y y t y

z z z

Example: Find down equation for the line through the given point and in the specified direction.

1

2 4 1 2

3

( , , ),

2 1

4 2

1 3

r t

Equation:

r = (xi + yj + zk) = (x1i + y1j + z1k) + t(xi + yj + zk)

Pair of linesGiven a pair of lines there are three possibilities:

1 The lines are parallel.

2 The lines are not parallel and they intersect.

3 The lines are not parallel and they do not intersect then they are skew.

ExampleFind whether the following pairs of lines are parallel, intersecting or skew.

r1 = i – 3j + 5k + s(3i – j + 2k) and

r2 = 4i + k + t(12i – 4j +8k)

Direction vector of r2

12i – 4j + 8k = 4(3i – j + 2k)The direction vector of r2 is multiple of the direction vector of r1.

The lines are parallel.

ExampleFind whether the following pairs of lines are parallel, intersecting or skew. r1 = i – j + 3k + s(i – j + k) and

r2 = 2i + 4j+ 6k+ t(2i + j +3k)

Direction vectors

2i +j + 3k ≠ (i – j + k)The lines are not parallel.

i – j + 3k + s(i – j + k) = 2i + 4j+ 6k+ t(2i + j +3k) i(1 +s) + j(-1-s) + k(3 + s)= i(2 + 2t) + j(4 + t)+ k(6 + 3t)1 + s = 2 + 2t-1 - s = 4 + t

s = - 3 and t = - 2

k: 3 – 3 = 6 - 6 The lines intersect.Point of intersection is (-2, 2, 0)

ExampleFind whether the following pairs of lines are parallel, intersecting or skew. r1 = i + k + s(i + 3j + 4k) and

r2 = 2i + 3j + t(4i -j + k)

Direction vectors

i + 3j + 4k ≠ (4i – j + k)The lines are not parallel.

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

1 + s = 2 + 4t 3s = 3 - t

s = 1 and t = 0

k: 1 + 5 ≠ 0 The lines do not intersect.

The lines are not parallel and do intersect so they are skew.

Example

Find the equation of line which goes through a = (1, 0, 4) and b = (6, 3, -2)

6 1 5

3 0 3

2 4 6

Direction vector = b – a =

Equation: r = a + t(b – a) 1 5

0 3

4 6

r t