The vector equation of a line The position vector of a set of points are given by r = OR = OA + AB 0 A ABR

Cartesian form and vector form The line goes through (2, -1) with gradient ¾ Cartesian form y – y 0 = m(x – x 0 ) 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

Changing Vector form into Cartesian form x = [1] y = [2] 3x = [3] 3  [1] 4y = [4] 4  [2] 3x – 4y = 10 [3]-[4] 3x – 4y – 10 = 0

Changing Cartesian form into Vector form y = 2x - 3 Gradient = 2 and y intercept = (0, - 3) Vector form Basic unit form: xi + yj = (0i – 3j) + (1i + 2j)

Special directions Parallel to the y-axis Find a Vector equation for the line through (2, 1) parallel to y - axis, and deduce its Cartesian equation. Basic unit vectors: (xi + yj) = (x 1 i + y 1 j)+ (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

Special directions Parallel to the x-axis Find a Vector equation for the line through (1, 3) parallel to x - axis, and deduce its Cartesian equation. Basic unit vectors: (xi + yj) = (x 1 i + y 1 j)+ (xi + yj) (xi + yj) = (i + 3j)+ i A vector parallel to the x-axis is i Cartesian form x = 1+ and y = 3 y x y = 3

The intersection of two lines 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. x: 2 + = 6 +  (1) y: =  (2) = 2 and  = - 2

The intersection of two lines 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. x: 2 + 3s= -1(1) y: 1 = 3 – 2t(2) s = -1 and t = 1

Vectors in three dimensions Position vector r: Example: Find down equation for the line through the given point and in the specified direction. Equation: r = (xi + yj + zk) = (x 1 i + y 1 j + z 1 k) + t(xi + yj + zk)

Pair of lines Given a pair of lines there are three possibilities: 1The lines are parallel. 2The lines are not parallel and they intersect. 3The lines are not parallel and they do not intersect then they are skew.

Example Find whether the following pairs of lines are parallel, intersecting or skew. r 1 = i – 3j + 5k + s(3i – j + 2k) and r 2 = 4i + k + t(12i – 4j +8k) Direction vector of r 2 12i – 4j + 8k = 4(3i – j + 2k) The direction vector of r 2 is multiple of the direction vector of r 1. The lines are parallel.

Example Find whether the following pairs of lines are parallel, intersecting or skew. r 1 = i – j + 3k + s(i – j + k) and r 2 = 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 = The lines intersect. Point of intersection is (-2, 2, 0)

Example Find whether the following pairs of lines are parallel, intersecting or skew. r 1 = i + k + s(i + 3j + 4k) and r 2 = 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: ≠ 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) Direction vector = b – a = Equation: r = a + t(b – a)