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Chapter 10 · Vector Algebra

Exercise 10.2 · Q5: Scalar and vector components from (2,1)(2,1) to (−5,7)(-5,7)

Exercise 10.2Vector joining two points, components · Sections 10.5.1 and 10.5.2
The question

Find the scalar and vector components of the vector with initial point (2,1)(2, 1) and terminal point (−5,7)(-5, 7)⁠.

Exam answer

Let A(2, 1) be the initial point and B(−5, 7) the terminal point. Then

AB→=(−5−2)i^+(7−1)j^=−7i^+6j^\begin{aligned} \overrightarrow{AB} &= (-5 - 2)\hat i + (7 - 1)\hat j \\ &= -7\hat i + 6\hat j \end{aligned}

Scalar components: −7-7 and 66⁠.

Vector components: −7i^-7\hat i and 6j^6\hat j⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The vector from a point A(x1,y1)A(x_1, y_1) to a point B(x2,y2)B(x_2, y_2) is "terminal minus initial": AB→=(x2−x1)i^+(y2−y1)j^\overrightarrow{AB} = (x_2 - x_1)\hat i + (y_2 - y_1)\hat j⁠. Once it is in the form xi^+yj^x\hat i + y\hat j⁠, the numbers x,yx, y are its scalar components and the vectors xi^,yj^x\hat i, y\hat j are its vector components.

Given, to find, and the plan
Given
Initial point A(2,1)A(2, 1) and terminal point B(−5,7)B(-5, 7)⁠.
To find
The scalar components and the vector components of AB→\overrightarrow{AB}⁠.
Plan
(1) Compute AB→\overrightarrow{AB} as terminal minus initial. (2) Read off the scalar components (the numbers) and the vector components (the numbers times i^\hat i⁠, j^\hat j⁠). (3) Check on a grid that the steps lead from A to B.

Basics you need

1Vector joining two points

For P1(x1,y1,z1)P_1(x_1, y_1, z_1) and P2(x2,y2,z2)P_2(x_2, y_2, z_2)⁠, P1P2→=(x2−x1)i^\overrightarrow{P_1P_2} = (x_2 - x_1)\hat i + (y2−y1)j^+(z2−z1)k^+\, (y_2 - y_1)\hat j + (z_2 - z_1)\hat k (Section 10.5.2). In the plane the k^\hat k part is 0. Lesson: components

2Scalar and vector components

For r⃗=xi^+yj^+zk^\vec r = x\hat i + y\hat j + z\hat k⁠, the numbers x,y,zx, y, z are the scalar components and xi^,yj^,zk^x\hat i, y\hat j, z\hat k are the vector components of r⃗\vec r along the axes (Section 10.5.1). Lesson: components

3Points in the plane

(x,y)(x, y) means xx units along the x-axis and yy units along the y-axis from O; a negative xx is to the left. Basics: numbers fix a point

Solution

  1. Name the points and form the vector

    Let A(2,1)A(2, 1) be the initial point and B(−5,7)B(-5, 7) the terminal point. Terminal minus initial, coordinate by coordinate:

    AB→=(−5−2)i^+(7−1)j^=−7i^+6j^.\begin{aligned} \overrightarrow{AB} &= (-5 - 2)\hat i + (7 - 1)\hat j \\ &= -7\hat i + 6\hat j. \end{aligned}
    By the triangle law, OA→+AB→=OB→\overrightarrow{OA} + \overrightarrow{AB} = \overrightarrow{OB}⁠, so AB→=OB→−OA→=(−5i^+7j^)−(2i^+j^)\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (-5\hat i + 7\hat j) - (2\hat i + \hat j)⁠. The order matters: the vector starts at the initial point, so the initial point's coordinates are subtracted. Reversing the order would give BA→=7i^−6j^\overrightarrow{BA} = 7\hat i - 6\hat j⁠, the opposite vector.
    P1P2→=OP2→−OP1→\overrightarrow{P_1P_2} = \overrightarrow{OP_2} - \overrightarrow{OP_1} (Section 10.5.2, Fig 10.15); the same calculation as Example 10. Lesson: Examples 4–10

  2. Read off the components

    In the form AB→=xi^+yj^\overrightarrow{AB} = x\hat i + y\hat j we have x=−7x = -7 and y=6y = 6⁠. So

    scalar components: −7-7 and 66; vector components: −7i^-7\hat i and 6j^6\hat j.

    Scalar components are numbers, and they carry the sign: −7-7 says "7 units in the negative x direction". Vector components are the two arrows −7i^-7\hat i and 6j^6\hat j whose sum is AB→\overrightarrow{AB}⁠. Since the points lie in the plane, there is no k^\hat k component (it is 0).
    "Here, xx⁠, yy and zz are called as the scalar components of r⃗\vec r⁠, and xi^x\hat i⁠, yj^y\hat j and zk^z\hat k are called the vector components" (Section 10.5.1). Lesson: components

  3. See it on the grid

    A(2, 1) B(−5, 7) −7î (7 units left) 6ĵ O x y
    The vector AB→\overrightarrow{AB} (blue) on a unit grid with equal scales. Its vector components (gold) take you from A first 7 units to the left (−7i^-7\hat i⁠) and then 6 units up (6j^6\hat j⁠), arriving at B.

    Starting at A(2,1)A(2, 1)⁠: going 7 units left reaches (−5,1)(-5, 1)⁠, and then 6 units up reaches (−5,7)=B(-5, 7) = B⁠. So −7i^+6j^-7\hat i + 6\hat j really does carry A to B.

    The two vector components, placed head to tail, add up to AB→\overrightarrow{AB} by the triangle law. The picture also shows why the first scalar component is negative: B is to the left of A.
    Triangle law of vector addition (Section 10.4). Lesson: adding arrows, head to tail
Result

AB→=−7i^+6j^\overrightarrow{AB} = -7\hat i + 6\hat j⁠. Scalar components: −7-7 and 66⁠. Vector components: −7i^-7\hat i and 6j^6\hat j⁠.

(Textbook answer: −7-7 and 66⁠; −7i^-7\hat i and 6j^6\hat j⁠.)

Check it with numbers

Initial point plus the components gives the terminal point: (2+(−7), 1+6)=(−5,7)(2 + (-7),\ 1 + 6) = (-5, 7)⁠. The length of AB→\overrightarrow{AB} is (−7)2+62=85≈9.2195\sqrt{(-7)^2 + 6^2} = \sqrt{85} \approx 9.2195⁠, the same as the distance from A to B by the distance formula, (−5−2)2+(7−1)2=85\sqrt{(-5-2)^2 + (7-1)^2} = \sqrt{85}⁠. Its direction makes about 139.40∘139.40^\circ with the positive x-axis (pointing up and to the left), as in the figure.

Common mistakes
  1. Subtracting the wrong way round. Initial minus terminal gives 7i^−6j^7\hat i - 6\hat j⁠, which is BA→\overrightarrow{BA}⁠, the opposite vector.
  2. Dropping the sign. The scalar component is −7-7⁠, not 7; the sign records the direction along the axis.
  3. Swapping the two kinds of component. Scalar components are the numbers −7,6-7, 6⁠; vector components are the vectors −7i^,6j^-7\hat i, 6\hat j⁠.
What to take away
  1. AB→\overrightarrow{AB} = (terminal) −- (initial), coordinate by coordinate.
  2. In xi^+yj^+zk^x\hat i + y\hat j + z\hat k⁠: scalar components x,y,zx, y, z⁠; vector components xi^,yj^,zk^x\hat i, y\hat j, z\hat k⁠.
  3. Check by adding the components to the initial point: you must land on the terminal point.

Every fact used, and where it comes from

FactUsed inSource
P1P2→=(x2−x1)i^\overrightarrow{P_1P_2} = (x_2 - x_1)\hat i + (y2−y1)j^+(z2−z1)k^+\, (y_2 - y_1)\hat j + (z_2 - z_1)\hat kStep 1Section 10.5.2 and Fig 10.15; Example 10 (Lesson: Examples 4–10)
Scalar components and vector componentsStep 2Section 10.5.1 (Lesson: components)
Triangle law OA→+AB→=OB→\overrightarrow{OA} + \overrightarrow{AB} = \overrightarrow{OB}Steps 1, 3Section 10.4 (Lesson: adding arrows)
∣xi^+yj^∣=x2+y2|x\hat i + y\hat j| = \sqrt{x^2 + y^2}⁠; distance formulaCheckSections 10.5.1, 10.5.2
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