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Chapter 10 · Vector Algebra

Exercise 10.2 · Q6: Sum of i^−2j^+k^\hat i-2\hat j+\hat k⁠, −2i^+4j^+5k^-2\hat i+4\hat j+5\hat k and i^−6j^−7k^\hat i-6\hat j-7\hat k

Exercise 10.2Addition of vectors in component form · Sections 10.4 and 10.5.1
The question

Find the sum of the vectors a⃗=i^−2j^+k^\vec a = \hat i - 2\hat j + \hat k⁠, b⃗=−2i^+4j^+5k^\vec b = -2\hat i + 4\hat j + 5\hat k and c⃗=i^−6j^−7k^\vec c = \hat i - 6\hat j - 7\hat k⁠.

Exam answer

Adding the corresponding components,

a⃗+b⃗+c⃗=(1−2+1)i^+(−2+4−6)j^+(1+5−7)k^=0i^−4j^−1k^\begin{aligned} &\vec a + \vec b + \vec c \\ &= (1 - 2 + 1)\hat i \\ &\quad + (-2 + 4 - 6)\hat j \\ &\quad + (1 + 5 - 7)\hat k \\ &= 0\hat i - 4\hat j - 1\hat k \end{aligned}

∴ a⃗+b⃗+c⃗=−4j^−k^\therefore\ \vec a + \vec b + \vec c = -4\hat j - \hat k⁠.

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

The one idea this problem is built on

Vectors in component form are added component by component: the i^\hat i parts together, the j^\hat j parts together and the k^\hat k parts together. Each of these is ordinary addition of signed numbers, so the whole question reduces to three short sums.

Given, to find, and the plan
Given
a⃗=i^−2j^+k^\vec a = \hat i - 2\hat j + \hat k⁠, b⃗=−2i^+4j^+5k^\vec b = -2\hat i + 4\hat j + 5\hat k⁠, c⃗=i^−6j^−7k^\vec c = \hat i - 6\hat j - 7\hat k⁠.
To find
a⃗+b⃗+c⃗\vec a + \vec b + \vec c⁠.
Plan
(1) List the components of each vector with their signs. (2) Add the i^\hat i⁠, j^\hat j and k^\hat k components separately. (3) Write the sum, dropping the zero term. (4) Check by adding two at a time.

Basics you need

1Sum in component form

If a⃗=a1i^+a2j^+a3k^\vec a = a_1\hat i + a_2\hat j + a_3\hat k and b⃗=b1i^+b2j^+b3k^\vec b = b_1\hat i + b_2\hat j + b_3\hat k⁠, then a⃗+b⃗=(a1+b1)i^\vec a + \vec b = (a_1 + b_1)\hat i + (a2+b2)j^+(a3+b3)k^+\, (a_2 + b_2)\hat j + (a_3 + b_3)\hat k (Section 10.5.1, (i)). Lesson: components

2Adding three vectors

Vector addition is associative, (a⃗+b⃗)+c⃗=a⃗+(b⃗+c⃗)(\vec a + \vec b) + \vec c = \vec a + (\vec b + \vec c)⁠, so the sum of three vectors can be written a⃗+b⃗+c⃗\vec a + \vec b + \vec c without brackets (Section 10.4, Property 2 and Remark). Lesson: adding arrows, head to tail

3Reading components with signs

A term with no number has component 1 (i^\hat i means 1i^1\hat i⁠), and a minus sign belongs to the component (−6j^-6\hat j has component −6-6⁠).

Solution

  1. List the components

    Reading each vector with its signs:

    a⃗: (1, −2, 1),b⃗: (−2, 4, 5),c⃗: (1, −6, −7).\begin{aligned} \vec a &: \ (1,\ -2,\ 1), \\ \vec b &: \ (-2,\ 4,\ 5), \\ \vec c &: \ (1,\ -6,\ -7). \end{aligned}
    Writing the three components of each vector in order (i^\hat i⁠, j^\hat j⁠, k^\hat k⁠) before adding avoids the commonest slip, losing a sign or reading i^\hat i as 0 instead of 1.
    Scalar components of xi^+yj^+zk^x\hat i + y\hat j + z\hat k are x,y,zx, y, z (Section 10.5.1). Lesson: components

  2. Add component by component

    i^:1+(−2)+1=0,j^:−2+4+(−6)=−4,k^:1+5+(−7)=−1.\begin{aligned} \hat i:&\quad 1 + (-2) + 1 = 0, \\ \hat j:&\quad -2 + 4 + (-6) = -4, \\ \hat k:&\quad 1 + 5 + (-7) = -1. \end{aligned}
    Applying the rule for two vectors twice, (a⃗+b⃗)+c⃗(\vec a + \vec b) + \vec c⁠, adds the corresponding components of all three. The i^\hat i⁠, j^\hat j⁠, k^\hat k directions are perpendicular, so the parts along one axis never mix with the parts along another.
    a⃗+b⃗=(a1+b1)i^\vec a + \vec b = (a_1 + b_1)\hat i + (a2+b2)j^+(a3+b3)k^+\, (a_2 + b_2)\hat j + (a_3 + b_3)\hat k (Section 10.5.1, (i)), used twice. Lesson: components

  3. Write the sum

    a⃗+b⃗+c⃗=0i^−4j^−1k^=−4j^−k^.\begin{aligned} \vec a + \vec b + \vec c &= 0\hat i - 4\hat j - 1\hat k \\ &= -4\hat j - \hat k. \end{aligned}

    So a⃗+b⃗+c⃗=−4j^−k^\vec a + \vec b + \vec c = -4\hat j - \hat k.

    A zero component is usually left out, and 1k^1\hat k is written k^\hat k⁠. The i^\hat i parts cancel exactly: a⃗\vec a and c⃗\vec c each go 1 unit along i^\hat i⁠, and b⃗\vec b comes 2 units back. The sum has no x component, so it lies parallel to the yz-plane.
    0i^=0⃗0\hat i = \vec 0 and 0⃗\vec 0 is the additive identity: adding it changes nothing (Sections 10.4 and 10.5). Lesson: adding arrows

  4. Check by adding two at a time

    First a⃗+b⃗\vec a + \vec b⁠, then add c⃗\vec c⁠:

    a⃗+b⃗=−i^+2j^+6k^,(a⃗+b⃗)+c⃗=(−1+1)i^+(2−6)j^+(6−7)k^=−4j^−k^.\begin{aligned} \vec a + \vec b &= -\hat i + 2\hat j + 6\hat k, \\ (\vec a + \vec b) + \vec c &= (-1 + 1)\hat i + (2 - 6)\hat j \\ &\quad + (6 - 7)\hat k \\ &= -4\hat j - \hat k. \end{aligned}
    Associativity says the grouping cannot change the answer, so a second route gives an independent check of the arithmetic. Both routes give −4j^−k^-4\hat j - \hat k⁠.
    Associative property of vector addition (Section 10.4, Property 2). Lesson: adding arrows
Result a⃗+b⃗+c⃗=−4j^−k^.\vec a + \vec b + \vec c = -4\hat j - \hat k.

(Textbook answer: −4j^−k^-4\hat j - \hat k⁠.)

Check it with numbers

Another grouping: b⃗+c⃗=−i^−2j^−2k^\vec b + \vec c = -\hat i - 2\hat j - 2\hat k⁠, and adding a⃗=i^−2j^+k^\vec a = \hat i - 2\hat j + \hat k gives 0i^−4j^−k^0\hat i - 4\hat j - \hat k⁠, the same sum. Its magnitude is 0+16+1=17≈4.1231\sqrt{0 + 16 + 1} = \sqrt{17} \approx 4.1231⁠. Adding all nine components gives (0)+(−4)+(−1)=−5(0) + (-4) + (-1) = -5⁠, and indeed (1−2+1)+(−2+4+5)(1 - 2 + 1) + (-2 + 4 + 5) + (1−6−7)=0+7−12=−5+\, (1 - 6 - 7) = 0 + 7 - 12 = -5⁠.

Common mistakes
  1. Dropping a sign. The k^\hat k component of c⃗\vec c is −7-7⁠, so the k^\hat k sum is 1+5−7=−11 + 5 - 7 = -1⁠, not 1313⁠.
  2. Mixing components. Add i^\hat i with i^\hat i only; −2j^-2\hat j and −2i^-2\hat i are different components.
  3. Writing 0i^0\hat i as a missing answer. A zero component is fine: the sum is −4j^−k^-4\hat j - \hat k⁠, a perfectly good vector.
What to take away
  1. Add vectors in component form one component at a time.
  2. List components with signs first; it prevents most errors.
  3. Regrouping (associativity) gives a free check.

Every fact used, and where it comes from

FactUsed inSource
Scalar components of a vectorStep 1Section 10.5.1 (Lesson: components)
a⃗+b⃗=(a1+b1)i^\vec a + \vec b = (a_1 + b_1)\hat i + (a2+b2)j^+(a3+b3)k^+\, (a_2 + b_2)\hat j + (a_3 + b_3)\hat kSteps 2, 4Section 10.5.1, (i)
Associative property; a⃗+b⃗+c⃗\vec a + \vec b + \vec c without bracketsSteps 2, 4Section 10.4, Property 2 and Remark (Lesson: adding arrows)
Zero vector is the additive identityStep 3Section 10.4
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}CheckSection 10.5.1
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