Lessons 17 · 1 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 11, Parts 1 and 2, 1st edition. ZAMIN NASHR, Tashkent, 2018
17
Vectors in space and operations on them
Textbook, Part 1: pp. 122–132
GoalFind the coordinates and length of vectors in space, perform operations on them (addition, multiplication by a number, dot product), and apply the conditions for collinearity and perpendicularity.
New words
coordinates of a vector · vektor koordinatalaricollinear vectors · kollinear vektorlardot (scalar) product · skalar ko‘paytmaunit coordinate vector · ort
Explanation
The vector AB with initial point A(x₁, y₁, z₁) and endpoint B(x₂, y₂, z₂) has coordinates (x₂ − x₁, y₂ − y₁, z₂ − z₁); equal vectors have equal corresponding coordinates. The length of a(a₁, a₂, a₃) is |a| = √(a₁² + a₂² + a₃²). Addition and multiplication by a number work coordinatewise: a + b = (a₁ + b₁, a₂ + b₂, a₃ + b₃), λa = (λa₁, λa₂, λa₃); geometrically the triangle, parallelogram and parallelepiped rules give AB + BC = AC, AB + AD = AC, AB + AD + AA₁ = AC₁. Non-zero a and b are collinear if a = λb, i.e. their coordinates are proportional. With the unit vectors i(1, 0, 0), j(0, 1, 0), k(0, 0, 1) we write a = a₁i + a₂j + a₃k. The dot product is a · b = |a||b| cos φ = a₁b₁ + a₂b₂ + a₃b₃ (a number!). Consequences: cos φ = (a · b)/(|a||b|); a ⊥ b ⇔ a · b = 0; a · a = |a|².
Worked examples
A(1, 2, −1), B(4, 6, 11): AB = (3, 4, 12), |AB| = √(9 + 16 + 144) = 13. For a(1, −2, 2) and b(2, 2, −1): a · b = 2 − 4 − 2 = −4, |a| = 3, |b| = 3, cos φ = −4/9, so the angle is obtuse (cos < 0).
If a(2, m, 1) and b(3, −1, −4) are perpendicular: 6 − m − 4 = 0 ⇒ m = 2. For a(2, −4, 6) and b(−1, 2, −3): a = −2b, so they are collinear and oppositely directed. If a = (1, 2, 3) then a = i + 2j + 3k.
Class activity
“Vector journey”: on paper, record three consecutive “steps” (vectors a, b, c — for example 2 m east, 1 m north, 1.5 m up) in coordinates starting from a corner. Compute the coordinates of a + b + c and its length, and compare with the straight-line distance.
Practice
1
Find the coordinates of AB for A(3, −1, 2) and B(−2, 4, 6).
(−5, 5, 4)
2
Find the length of the vector a(2, 3, 6).
7
3
Compute the dot product of a(1, 2, 3) and b(4, −1, 0).
2
4
Why does a · b = 0 not mean that a = 0 or b = 0?
a · b = |a||b| cos φ is also zero when cos φ = 0, i.e. φ = 90° (perpendicular vectors). For example (1, 0, 0) · (0, 1, 0) = 0.