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Grade 12 · Lesson 2.3

Operations on vectors in coordinates

Name: Date: Time spent: min

Goals

Remember

Exercises

Exercise 1

Let u⃗=(2,−1,3)\vec{u}=(2, -1, 3) and v⃗=(−1,4,0)\vec{v}=(-1, 4, 0). Find u⃗+v⃗\vec{u}+\vec{v}, 3u⃗−2v⃗3\vec{u}-2\vec{v} and u⃗⋅v⃗\vec{u}\cdot\vec{v}.

Exercise 2

Let a⃗=(2,−3,6)\vec{a}=(2, -3, 6). Find ∣a⃗∣|\vec{a}| and the vector e⃗\vec{e} with the same direction as a⃗\vec{a} and length 11.

Exercise 3

Find the angle between the vectors a⃗=(1,1,0)\vec{a}=(1, 1, 0) and b⃗=(0,1,1)\vec{b}=(0, 1, 1).

Exercise 4

Among the three vectors u⃗=(1,2,−1)\vec{u}=(1, 2, -1), v⃗=(3,−1,1)\vec{v}=(3, -1, 1), w⃗=(1,1,3)\vec{w}=(1, 1, 3), which pairs are perpendicular?

Exercise 5

Find xx, yy such that the vectors a⃗=(1,x,3)\vec{a}=(1, x, 3) and b⃗=(−2,4,y)\vec{b}=(-2, 4, y) are collinear.

Exercise 6

Let A(3,2,−1)A(3, 2, -1), B(1,1,1)B(1, 1, 1), C(2,3,3)C(2, 3, 3). Prove that triangle ABCABC has a right angle at BB and find its area.

Exercise 7

Let a⃗=(1,−2,2)\vec{a}=(1, -2, 2) and b⃗=(2,2,1)\vec{b}=(2, 2, 1). Is each claim true or false?

a) ∣a⃗∣=3|\vec{a}|=3.

b) a⃗⊥b⃗\vec{a}\perp\vec{b}.

c) a⃗−b⃗=(−1,−4,3)\vec{a}-\vec{b}=(-1, -4, 3).

d) ∣a⃗+b⃗∣=6|\vec{a}+\vec{b}|=6.

Exercise 8

A person pulls a sledge with the force F⃗=(10,5,−2)\vec{F}=(10, 5, -2) (unit: N). The sledge moves along the displacement d⃗=(3,4,0)\vec{d}=(3, 4, 0) (unit: m). Find the angle between the pulling force and the direction of motion (in degrees, rounded to the nearest whole number).

Exercise 9

(Revisiting Vectors and their operations in space.) In the previous lesson, by decomposing vectors, we found that the angle between AC→\overrightarrow{AC} and A′D→\overrightarrow{A'D} in the cube ABCD.A′B′C′D′ABCD.A'B'C'D' is 60∘60^\circ. Find it again with coordinates, for the cube with edge 11, AA at the origin, B(1,0,0)B(1, 0, 0), D(0,1,0)D(0, 1, 0), A′(0,0,1)A'(0, 0, 1).

Exercise 10 Challenge

Let A(2,1,1)A(2, 1, 1) and B(4,−1,1)B(4, -1, 1). Find the points MM on the axis OxOx such that triangle MABMAB has a right angle at MM.

Mini test

15 minutes, without looking back at the lesson. Part I: circle one letter; Part II: write T or F for each statement; Part III: write the answer in the box.

Part I. Multiple choice

Question 1. In the system OxyzOxyz, let a⃗=(2,−1,3)\vec{a}=(2, -1, 3) and b⃗=(1,0,−2)\vec{b}=(1, 0, -2). The coordinates of a⃗−2b⃗\vec{a}-2\vec{b} are
A. (0,−1,−1)(0, -1, -1).B. (0,−1,7)(0, -1, 7).C. (4,−1,−1)(4, -1, -1).D. (1,−1,5)(1, -1, 5).
Question 2. Let u⃗=(3,−2,1)\vec{u}=(3, -2, 1) and v⃗=(1,4,−2)\vec{v}=(1, 4, -2). The scalar product u⃗⋅v⃗\vec{u}\cdot\vec{v} is
A. 77.B. (3,−8,−2)(3, -8, -2).C. −7-7.D. −3-3.
Question 3. Which vector is perpendicular to a⃗=(1,2,−2)\vec{a}=(1, 2, -2)?
A. (1,2,2)(1, 2, 2).B. (2,−1,2)(2, -1, 2).C. (−2,−4,4)(-2, -4, 4).D. (2,1,2)(2, 1, 2).
Question 4. The angle between the vectors a⃗=(1,0,1)\vec{a}=(1, 0, 1) and b⃗=(−1,1,0)\vec{b}=(-1, 1, 0) is
A. 120∘120^\circ.B. 60∘60^\circ.C. 45∘45^\circ.D. 150∘150^\circ.

Part II. True or false

Question 1. In the system OxyzOxyz, let A(1,0,2)A(1, 0, 2), B(3,1,0)B(3, 1, 0), C(0,2,2)C(0, 2, 2).
a) AB→=(2,1,−2)\overrightarrow{AB}=(2, 1, -2).T / F
b) Triangle ABCABC has a right angle at AA.T / F
c) The area of triangle ABCABC is 353\sqrt5.T / F
d) The angle ∠ABC\angle ABC is greater than 30∘30^\circ. T / F

Part III. Short answer

Question 1. Let A(1,2,−1)A(1, 2, -1), B(3,0,1)B(3, 0, 1) and C(x,y,5)C(x, y, 5). Given that AA, BB, CC are collinear, find x+yx+y.
Question 2. A force F⃗=(4,−2,3)\vec{F}=(4, -2, 3) (unit: N) moves an object from M(1,0,0)M(1, 0, 0) to N(5,3,2)N(5, 3, 2) (unit: m). Find the angle between the force F⃗\vec{F} and the direction of motion MN→\overrightarrow{MN}, in degrees, to 1 decimal place.

Score: ........ / 10

Hints (only when stuck)

Exercise 1.

What you need: add, subtract and multiply by a number coordinate by coordinate; u⃗⋅v⃗=x1x2+y1y2+z1z2\vec{u}\cdot\vec{v}=x_1x_2+y_1y_2+z_1z_2.

Steps:

  1. Compute 3u⃗3\vec{u} and 2v⃗2\vec{v} first.
  2. Multiply coordinates in the same position and add.
Exercise 2.

What you need: ∣a⃗∣=x2+y2+z2|\vec{a}|=\sqrt{x^2+y^2+z^2}; 1∣a⃗∣a⃗\dfrac{1}{|\vec{a}|}\vec{a} has the same direction as a⃗\vec{a} and length 11.

Steps:

  1. Compute ∣a⃗∣|\vec{a}|.
  2. Divide each coordinate by ∣a⃗∣|\vec{a}|.
Exercise 3.

What you need: cos⁡(a⃗,b⃗)=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos(\vec{a}, \vec{b})=\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}.

Steps:

  1. Compute a⃗⋅b⃗\vec{a}\cdot\vec{b}, ∣a⃗∣|\vec{a}|, ∣b⃗∣|\vec{b}|.
  2. Find the angle with that cosine.
Exercise 4.

What you need: a⃗⊥b⃗  ⟺  a⃗⋅b⃗=0\vec{a}\perp\vec{b}\iff\vec{a}\cdot\vec{b}=0.

Steps:

  1. Compute the three scalar products u⃗⋅v⃗\vec{u}\cdot\vec{v}, u⃗⋅w⃗\vec{u}\cdot\vec{w}, v⃗⋅w⃗\vec{v}\cdot\vec{w}.
  2. Pick those equal to 00.
Exercise 5.

What you need: b⃗\vec{b} is collinear with a⃗\vec{a}   ⟺  \iff b⃗=ka⃗\vec{b}=k\vec{a}.

Steps:

  1. Find kk from the x-coordinates: −2=k⋅1-2=k\cdot1.
  2. Use kk for the other two coordinates.
Exercise 6.

What you need: right angle at BB   ⟺  \iff BA→⋅BC→=0\overrightarrow{BA}\cdot\overrightarrow{BC}=0; the area of a right-angled triangle is half the product of the two shorter sides.

Steps:

  1. Find BA→\overrightarrow{BA}, BC→\overrightarrow{BC} and their scalar product.
  2. Find BABA, BCBC.
Exercise 7.

What you need: the formulas for length, scalar product and subtraction in coordinates.

Steps:

  1. Compute ∣a⃗∣|\vec{a}| and a⃗⋅b⃗\vec{a}\cdot\vec{b}.
  2. Subtract coordinate by coordinate.
  3. Compute a⃗+b⃗\vec{a}+\vec{b} and its length.
Exercise 8.

What you need: cos⁡(F⃗,d⃗)=F⃗⋅d⃗∣F⃗∣∣d⃗∣\cos(\vec{F}, \vec{d})=\dfrac{\vec{F}\cdot\vec{d}}{|\vec{F}||\vec{d}|}; in physics, F⃗⋅d⃗\vec{F}\cdot\vec{d} is the work done by the force (unit: J).

Steps:

  1. Compute F⃗⋅d⃗\vec{F}\cdot\vec{d}.
  2. Compute ∣F⃗∣|\vec{F}|, ∣d⃗∣|\vec{d}| and the cosine of the angle.
Exercise 9.

What you need: the coordinates of the vertices of the cube; the cosine formula for the angle.

Steps:

  1. Write C(1,1,0)C(1, 1, 0) and find AC→\overrightarrow{AC}, A′D→\overrightarrow{A'D}.
  2. Compute the cosine.
Exercise 10.

What you need: a point on OxOx has the form M(t,0,0)M(t, 0, 0); right angle at MM   ⟺  \iff MA→⋅MB→=0\overrightarrow{MA}\cdot\overrightarrow{MB}=0.

Steps:

  1. Write MA→\overrightarrow{MA}, MB→\overrightarrow{MB} in terms of tt.
  2. Set the scalar product equal to 00 and solve.

Self-check

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