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Chapter 10 · Vector Algebra

Example 8: Unit vector along the sum of 2i^+2j^−5k^2\hat i+2\hat j-5\hat k and 2i^+j^+3k^2\hat i+\hat j+3\hat k

ExamplesSum of vectors in components, unit vector · Sections 10.4, 10.5 and 10.5.1
The question

Find the unit vector in the direction of the sum of the vectors, a⃗=2i^+2j^−5k^\vec a = 2\hat i + 2\hat j - 5\hat k and b⃗=2i^+j^+3k^\vec b = 2\hat i + \hat j + 3\hat k⁠.

Exam answer

Let c⃗=a⃗+b⃗\vec c = \vec a + \vec b⁠. Adding corresponding components,

c⃗=(2+2)i^+(2+1)j^+(−5+3)k^=4i^+3j^−2k^\begin{aligned} \vec c &= (2 + 2)\hat i + (2 + 1)\hat j \\ &\quad + (-5 + 3)\hat k \\ &= 4\hat i + 3\hat j - 2\hat k \end{aligned} ∣c⃗∣=42+32+(−2)2=29\begin{aligned} |\vec c| &= \sqrt{4^2 + 3^2 + (-2)^2} \\ &= \sqrt{29} \end{aligned} c^=1∣c⃗∣c⃗=129(4i^+3j^−2k^)\begin{aligned} \hat c &= \frac{1}{|\vec c|}\vec c \\ &= \frac{1}{\sqrt{29}}(4\hat i + 3\hat j - 2\hat k) \end{aligned}

∴ c^=429i^+329j^−229k^\therefore\ \hat c = \dfrac{4}{\sqrt{29}}\hat i + \dfrac{3}{\sqrt{29}}\hat j - \dfrac{2}{\sqrt{29}}\hat k⁠.

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

The one idea this problem is built on

Two short jobs in a row. First add the vectors: in component form, add the i^\hat i parts, the j^\hat j parts and the k^\hat k parts. Then turn the sum c⃗\vec c into a unit vector by dividing it by its own length: c^=1∣c⃗∣c⃗\hat c = \frac{1}{|\vec c|}\vec c⁠. The unit vector must be built from the sum, not from a⃗\vec a and b⃗\vec b separately.

Given, to find, and the plan
Given
a⃗=2i^+2j^−5k^\vec a = 2\hat i + 2\hat j - 5\hat k⁠, b⃗=2i^+j^+3k^\vec b = 2\hat i + \hat j + 3\hat k⁠.
To find
The unit vector in the direction of a⃗+b⃗\vec a + \vec b⁠.
Plan
(1) Add: c⃗=a⃗+b⃗\vec c = \vec a + \vec b⁠. (2) Compute ∣c⃗∣|\vec c|⁠. (3) Form c^=1∣c⃗∣c⃗\hat c = \frac{1}{|\vec c|}\vec c⁠. (4) Check ∣c^∣=1|\hat c| = 1⁠.

Basics you need

1Adding vectors

Geometrically, by the triangle law: put b⃗\vec b at the head of a⃗\vec a⁠; the sum runs from the tail of a⃗\vec a to the head of b⃗\vec b (Section 10.4). In components, a⃗+b⃗=(a1+b1)i^+(a2+b2)j^+(a3+b3)k^\vec a + \vec b = (a_1 + b_1)\hat i + (a_2 + b_2)\hat j + (a_3 + b_3)\hat k (Section 10.5.1 (i)). Lesson: adding arrows, head to tail

2Magnitude from components

∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

3The unit vector

For c⃗≠0⃗\vec c \ne \vec 0⁠, c^=1∣c⃗∣c⃗\hat c = \frac{1}{|\vec c|}\vec c has length 1 and the direction of c⃗\vec c (Section 10.5). Lesson: components

Solution

  1. Add the two vectors

    Let c⃗=a⃗+b⃗\vec c = \vec a + \vec b⁠. Adding corresponding components,

    c⃗=(2+2)i^+(2+1)j^+(−5+3)k^=4i^+3j^−2k^.\begin{aligned} \vec c &= (2 + 2)\hat i + (2 + 1)\hat j + (-5 + 3)\hat k \\ &= 4\hat i + 3\hat j - 2\hat k. \end{aligned}
    O A C a, length √33 b, length √14 a + b, length √29 OA + AC = OC
    The triangle law for this sum, with a⃗=OA→\vec a = \overrightarrow{OA}⁠, b⃗=AC→\vec b = \overrightarrow{AC} and a⃗+b⃗=OC→\vec a + \vec b = \overrightarrow{OC}⁠. The triangle is drawn in its own plane, true to scale: its sides are ∣a⃗∣=33|\vec a| = \sqrt{33}⁠, ∣b⃗∣=14|\vec b| = \sqrt{14} and ∣a⃗+b⃗∣=29|\vec a + \vec b| = \sqrt{29}⁠.
    Placing b⃗\vec b at the head of a⃗\vec a moves us 2 then 2 more along x, 2 then 1 more along y, and −5-5 then +3+3 along z. The total movement in each direction is the sum of the two movements, so the components simply add. Watch the signs in the k^\hat k part: −5+3=−2-5 + 3 = -2⁠.
    Triangle law (Section 10.4); sum in component form (Section 10.5.1 (i)). Lesson: adding arrows, head to tail

  2. Find the length of the sum

    ∣c⃗∣=42+32+(−2)2=16+9+4=29.\begin{aligned} |\vec c| &= \sqrt{4^2 + 3^2 + (-2)^2} \\ &= \sqrt{16 + 9 + 4} = \sqrt{29}. \end{aligned}
    The unit vector is c⃗\vec c divided by its own length, so we need ∣c⃗∣|\vec c| and not ∣a⃗∣+∣b⃗∣|\vec a| + |\vec b|⁠. Since 29≠0\sqrt{29} \ne 0⁠, c⃗\vec c is not the zero vector and has a definite direction.
    ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

  3. Divide by the length

    c^=1∣c⃗∣c⃗=129(4i^+3j^−2k^)=429i^+329j^−229k^.\begin{aligned} \hat c &= \frac{1}{|\vec c|}\vec c = \frac{1}{\sqrt{29}}(4\hat i + 3\hat j - 2\hat k) \\ &= \frac{4}{\sqrt{29}}\hat i + \frac{3}{\sqrt{29}}\hat j - \frac{2}{\sqrt{29}}\hat k. \end{aligned}

    This is the required unit vector.

    Multiplying by the positive scalar 129\frac{1}{\sqrt{29}} keeps the direction of c⃗=a⃗+b⃗\vec c = \vec a + \vec b and makes the length 129⋅29=1\frac{1}{\sqrt{29}}\cdot\sqrt{29} = 1⁠. Each component is divided by 29\sqrt{29}⁠, and the minus sign of the k^\hat k part stays.
    c^=1∣c⃗∣c⃗\hat c = \frac{1}{|\vec c|}\vec c (Section 10.5); λc⃗\lambda\vec c by components (Section 10.5.1 (iv)). Lesson: components

  4. Check that it is a unit vector

    ∣c^∣2=1629+929+429=2929=1.|\hat c|^2 = \frac{16}{29} + \frac{9}{29} + \frac{4}{29} = \frac{29}{29} = 1.
    The squares of the components of a unit vector must add to 1, so this confirms Steps 2 and 3. It cannot catch a wrong sum in Step 1 (any nonzero vector divided by its own length passes this test), so we also re-add the components: 2+2=42 + 2 = 4⁠, 2+1=32 + 1 = 3⁠, −5+3=−2-5 + 3 = -2⁠.
    A unit vector has magnitude 1 (Section 10.3). Lesson: Examples 4–10
Result

a⃗+b⃗=4i^+3j^−2k^\vec a + \vec b = 4\hat i + 3\hat j - 2\hat k⁠, ∣a⃗+b⃗∣=29|\vec a + \vec b| = \sqrt{29}⁠, and the required unit vector is c^=429i^+329j^−229k^\hat c = \frac{4}{\sqrt{29}}\hat i + \frac{3}{\sqrt{29}}\hat j - \frac{2}{\sqrt{29}}\hat k⁠.

(Textbook answer: c^=129(4i^+3j^−2k^)=429i^+329j^−229k^\hat c = \frac{1}{\sqrt{29}}(4\hat i + 3\hat j - 2\hat k) = \frac{4}{\sqrt{29}}\hat i + \frac{3}{\sqrt{29}}\hat j - \frac{2}{\sqrt{29}}\hat k⁠.)

Check it with numbers

29≈5.385165\sqrt{29} \approx 5.385165⁠, so c^≈0.742781i^+0.557086j^−0.371391k^\hat c \approx 0.742781\hat i + 0.557086\hat j - 0.371391\hat k⁠, and 0.7427812+0.5570862+0.3713912≈0.551724+0.310345+0.137931=1.0000000.742781^2 + 0.557086^2 + 0.371391^2 \approx 0.551724 + 0.310345 + 0.137931 = 1.000000⁠.

The lengths in the triangle: ∣a⃗∣=33≈5.744563|\vec a| = \sqrt{33} \approx 5.744563 and ∣b⃗∣=14≈3.741657|\vec b| = \sqrt{14} \approx 3.741657⁠, whose sum 9.4862209.486220 is much more than ∣a⃗+b⃗∣≈5.385165|\vec a + \vec b| \approx 5.385165⁠. So dividing by ∣a⃗∣+∣b⃗∣|\vec a| + |\vec b| would give a vector of length about 0.5676830.567683⁠, not a unit vector.

Common mistakes
  1. Dividing by ∣a⃗∣+∣b⃗∣|\vec a| + |\vec b|⁠. The length of a sum is not the sum of the lengths; divide by ∣a⃗+b⃗∣|\vec a + \vec b|⁠.
  2. Adding the unit vectors a^+b^\hat a + \hat b⁠. That points along a different direction in general and does not have length 1.
  3. Sign slip in the k^\hat k part. −5+3=−2-5 + 3 = -2⁠, so the answer has −229k^-\frac{2}{\sqrt{29}}\hat k⁠.
What to take away
  1. Add vectors component by component; that is the triangle law in numbers.
  2. Unit vector of a sum: add first, then divide by the length of the sum.
  3. ∣a⃗+b⃗∣|\vec a + \vec b| is usually less than ∣a⃗∣+∣b⃗∣|\vec a| + |\vec b| (the third side of a triangle).

Every fact used, and where it comes from

FactUsed inSource
Triangle law of vector additionStep 1Section 10.4 (Lesson: adding arrows)
a⃗+b⃗=(a1+b1)i^+(a2+b2)j^+(a3+b3)k^\vec a + \vec b = (a_1 + b_1)\hat i + (a_2 + b_2)\hat j + (a_3 + b_3)\hat kStep 1Section 10.5.1 (i)
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}Steps 2, 4Section 10.5.1 (Basics: use Pythagoras twice)
c^=1∣c⃗∣c⃗\hat c = \frac{1}{|\vec c|}\vec c⁠; λc⃗\lambda\vec c by componentsStep 3Section 10.5; Section 10.5.1 (iv)
Unit vector: magnitude 1Step 4Section 10.3
The book's working and answerResultExample 8, Section 10.5.1 (Lesson: Examples 4–10)
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