☰ خطة العام · الرياضيات البحتة (بالإنجليزية)

Lines in space: direction vectors, three forms of equation and the angle between lines

الأسبوع 12 · المستقيم في الفراغ: متجه الاتجاه والمعادلة المتجهة والمعادلات البارامترية والمعادلة الكارتيزية والزاوية بين مستقيمين وتوازي المستقيمات وتعامدها وتقاطعها وتخالفها · 3 دروس
1

The direction vector of a line in space, direction ratios and the vector equation of a line

Pure Mathematics Sec 3 Term 1, Unit 4, Lesson 1
GoalFind a direction vector of a line in space, and write the vector equation of the line.
New words
direction vector · any non-zero vector parallel to the line; for points A and B on the line, d = ABdirection ratios · the components a, b, c of a direction vector; only their ratios mattervector equation · r = A + t d, where A is a known point, d is a direction vector and t is any real number
Explanation

In a plane a line is fixed by its slope and one point. In space, one number is not enough to describe a tilt, so we use a whole vector. A line in space is fixed by one point A on it and one direction. A direction vector d is any non-zero vector parallel to the line: if A and B lie on the line, d = AB. Every non-zero multiple of d is also a direction vector, so the components a, b, c of d, called the direction ratios (in Arabic نسب الاتجاه), matter only through their ratios. The direction cosines of Week 9 are the components of the unit vector d/‖d‖; so ℓ² + m² + n² = 1 and the direction ratios are proportional to ℓ, m, n. To go from ratios to cosines, divide by ‖d‖. The axes have direction vectors (1, 0, 0), (0, 1, 0) and (0, 0, 1). A zero in d tells you something: if the y-component is 0, the y-coordinate never changes along the line, so the line is parallel to the xz-plane. Now the vector equation. Take any point of the line with position vector r = (x, y, z). Starting at A and moving along the line by a multiple of d, we get r = A + t d, where t is any real number, called the parameter. Each value of t gives one point of the line: t = 0 gives A, t = 1 gives A + d, and a negative t gives points on the other side of A. The same line has many vector equations, because we may choose a different point A or a different multiple of d.

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Worked examples
Let A(1, 2, 3) and B(4, 4, 7) lie on a line (the segment AB is the diagonal of the box in the picture). A direction vector is AB = (3, 2, 4) and ‖AB‖ = √29. The vector equation is r = (1, 2, 3) + t(3, 2, 4). For t = 1 we get B, for t = −1 the point (−2, 0, −1), and for t = ½ the midpoint (5/2, 3, 5).
From ratios to cosines. For the ratios (4, −4, 2), ‖d‖ = 6 and the direction cosines are (2/3, −2/3, 1/3). From cosines to ratios: the cosines (2/7, 3/7, 6/7) give the ratios (2, 3, 6) when multiplied by 7.
The line through P(2, 5, −1) with direction (1, −3, 2) is r = (2, 5, −1) + t(1, −3, 2). For t = 1 the point is (3, 2, 1), and for t = −2 it is (0, 11, −5).
Is the point (5, −4, 5) on that line? The x-coordinate 2 + t = 5 gives t = 3. Then y = 5 − 9 = −4 and z = −1 + 6 = 5, both correct, so yes. The point (6, −1, 3) needs t = 4 from x, but then y = 5 − 12 = −7, so it is not on the line.
Class activity

A stretched string. Choose two corners of a table and call one of them the origin; measure each as (x, y, z) with the table edges and the height as axes. Write the direction vector from one to the other and the vector equation of the line along which a taut string would run. Then find the point at t = ½ and check that it is halfway.

Practice
1
Find a direction vector of the line through A(−1, 3, 2) and B(2, −1, 6).
2
Write the vector equation of the line through (0, −2, 5) with direction (4, 1, −3), and find the point for t = 2.
3
Find the direction cosines of the line with direction ratios (2, −3, 6).
4
Is the point (3, 3, −4) on the line r = (1, 5, −2) + t(1, −1, −1)?
5
A line has direction ratios (3, 0, −4). What is special about it, and what are its direction cosines?