Parametric and Cartesian equations of a line, and moving between the forms
Pure Mathematics Sec 3 Term 1, Unit 4, Lesson 1
GoalWrite the parametric and Cartesian (symmetric) equations of a line, and change between the three forms.
New words
parametric equations · x = x₁ + at, y = y₁ + bt, z = z₁ + ct, one equation for each coordinateCartesian equations · (x − x₁)/a = (y − y₁)/b = (z − z₁)/c, with t eliminatedparameter · the number t that picks one point of the line
Explanation
Write the vector equation r = A + t d one coordinate at a time. With A = (x₁, y₁, z₁) and d = (a, b, c) we get the parametric equations (in Arabic المعادلات البارامترية): x = x₁ + at, y = y₁ + bt, z = z₁ + ct. Solving each for t gives t = (x − x₁)/a = (y − y₁)/b = (z − z₁)/c, so the three fractions are equal; these are the Cartesian (symmetric) equations (in Arabic المعادلة الكارتيزية). This form needs a, b and c all different from 0, because we cannot divide by 0. If a = 0, then x = x₁ does not change, and the line is written x = x₁ together with (y − y₁)/b = (z − z₁)/c. To read a point and a direction from a Cartesian equation, make sure every numerator is x, y or z minus a number, with coefficient 1: divide top and bottom by the coefficient. For example (2x + 1)/4 = (x + ½)/2, and (3 − y)/3 = (y − 3)/(−3). Then the numbers on top give the point and the numbers underneath give the direction. Caution: a single equation ax + by + cz + d = 0 is a plane, not a line in space. A line in space needs two equations, such as the two equalities of the Cartesian form. Through two points A and B, take d = AB and then use either point.
form
the line through (1, −2, 3) with direction (3, 2, −4)
vector
r = (1, −2, 3) + t(3, 2, −4)
parametric
x = 1 + 3t, y = −2 + 2t, z = 3 − 4t
Cartesian
(x − 1)/3 = (y + 2)/2 = (z − 3)/(−4)
Worked examples
Through P(1, −2, 3) with direction (3, 2, −4): parametric x = 1 + 3t, y = −2 + 2t, z = 3 − 4t, and Cartesian (x − 1)/3 = (y + 2)/2 = (z − 3)/(−4), as in the table.
Through A(2, 1, −3) and B(5, −1, 0): d = AB = (3, −2, 3). Parametric x = 2 + 3t, y = 1 − 2t, z = −3 + 3t. Cartesian (x − 2)/3 = (y − 1)/(−2) = (z + 3)/3. Check: t = 1 gives (5, −1, 0) = B.
A zero in the direction. Through (4, 1, −2) with direction (0, 2, 5): x = 4, y = 1 + 2t, z = −2 + 5t. The Cartesian form is x = 4 together with (y − 1)/2 = (z + 2)/5.
From Cartesian to a point and a direction: (2x + 1)/4 = (3 − y)/3 = (z + 5)/6. Rewrite as (x + ½)/2 = (y − 3)/(−3) = (z + 5)/6. The point is (−½, 3, −5) and the direction is (2, −3, 6), of size 7, so the direction cosines are (2/7, −3/7, 6/7).
Class activity
Translate and check. Choose any point and any direction with small whole numbers, and write all three forms of the line. Then choose a value of t, find the point, and check that its coordinates satisfy the Cartesian equation.
Practice
1
Write the parametric equations of the line through (−2, 0, 4) with direction (1, 5, −2).
x = −2 + t, y = 5t, z = 4 − 2t.
2
Write the Cartesian equations of the line through (1, 1, 1) and (3, 2, −1).
d = (2, 1, −2), so (x − 1)/2 = (y − 1)/1 = (z − 1)/(−2).
3
Write x = 3 − 2t, y = 1 + t, z = −4t in Cartesian form.
The point is (3, 1, 0) and d = (−2, 1, −4), so (x − 3)/(−2) = y − 1 = z/(−4).
4
Find a point and a direction vector for (x + 3)/2 = (2y − 4)/6 = (5 − z)/3.
(2y − 4)/6 = (y − 2)/3 and (5 − z)/3 = (z − 5)/(−3). The point is (−3, 2, 5) and d = (2, 3, −3).
5
Show that (7, −1, −2) is on the line (x − 1)/3 = (y − 3)/(−2) = (z − 2)/(−2).
The three fractions are 6/3 = 2, (−4)/(−2) = 2 and (−4)/(−2) = 2. They are equal, so the point is on the line (t = 2).
