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Chapter 10 · Vector Algebra

Exercise 10.3 · Q5: 17(2i^+3j^+6k^)\frac17(2\hat i+3\hat j+6\hat k)⁠, 17(3i^−6j^+2k^)\frac17(3\hat i-6\hat j+2\hat k)⁠, 17(6i^+2j^−3k^)\frac17(6\hat i+2\hat j-3\hat k)⁠: unit and mutually perpendicular

Exercise 10.3Unit vectors and perpendicular vectors · Sections 10.5 and 10.6.1
The question

Show that each of the given three vectors is a unit vector: 17(2i^+3j^+6k^)\frac17(2\hat i + 3\hat j + 6\hat k)⁠, 17(3i^−6j^+2k^)\frac17(3\hat i - 6\hat j + 2\hat k)⁠, 17(6i^+2j^−3k^)\frac17(6\hat i + 2\hat j - 3\hat k)⁠.

Also, show that they are mutually perpendicular to each other.

Exam answer

Let a⃗=17(2i^+3j^+6k^)\vec a = \frac17(2\hat i + 3\hat j + 6\hat k)⁠, b⃗=17(3i^−6j^+2k^)\vec b = \frac17(3\hat i - 6\hat j + 2\hat k) and c⃗=17(6i^+2j^−3k^)\vec c = \frac17(6\hat i + 2\hat j - 3\hat k)⁠.

∣a⃗∣=1722+32+62=1749=1\begin{aligned} |\vec a| &= \frac17\sqrt{2^2 + 3^2 + 6^2} \\ &= \frac17\sqrt{49} = 1 \end{aligned} ∣b⃗∣=1732+(−6)2+22=1749=1\begin{aligned} |\vec b| &= \frac17\sqrt{3^2 + (-6)^2 + 2^2} \\ &= \frac17\sqrt{49} = 1 \end{aligned} ∣c⃗∣=1762+22+(−3)2=1749=1\begin{aligned} |\vec c| &= \frac17\sqrt{6^2 + 2^2 + (-3)^2} \\ &= \frac17\sqrt{49} = 1 \end{aligned}

So a⃗\vec a⁠, b⃗\vec b and c⃗\vec c are unit vectors.

a⃗⋅b⃗=149[2(3)+3(−6)+6(2)]=149(6−18+12)=0\begin{aligned} \vec a\cdot\vec b &= \frac{1}{49}[2(3) + 3(-6) + 6(2)] \\ &= \frac{1}{49}(6 - 18 + 12) = 0 \end{aligned} b⃗⋅c⃗=149[3(6)+(−6)(2)+2(−3)]=149(18−12−6)=0\begin{aligned} \vec b\cdot\vec c &= \frac{1}{49}[3(6) + (-6)(2) + 2(-3)] \\ &= \frac{1}{49}(18 - 12 - 6) = 0 \end{aligned} c⃗⋅a⃗=149[6(2)+2(3)+(−3)(6)]=149(12+6−18)=0\begin{aligned} \vec c\cdot\vec a &= \frac{1}{49}[6(2) + 2(3) + (-3)(6)] \\ &= \frac{1}{49}(12 + 6 - 18) = 0 \end{aligned}

The vectors are nonzero and each scalar product is 0, so each pair is perpendicular.

∴ a⃗\therefore\ \vec a⁠, b⃗\vec b⁠, c⃗\vec c are mutually perpendicular unit vectors. Hence proved.

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

The one idea this problem is built on

Two facts about the scalar product do all the work:

∣r⃗∣=r⃗⋅r⃗,a⃗⋅b⃗=0  ⟺  a⃗⊥b⃗.\begin{aligned} &|\vec r| = \sqrt{\vec r\cdot\vec r}, \\ &\vec a\cdot\vec b = 0 \iff \vec a \perp \vec b. \end{aligned}

So "unit vector" means the squares of the components add up to 1, and "perpendicular" means the products of matching components add up to 0. A fraction 17\frac17 in front can be taken out first: it multiplies a length by 17\frac17 and a scalar product by 149\frac{1}{49}⁠.

Given, to find, and the plan
Given
a⃗=17(2i^+3j^+6k^)\vec a = \frac17(2\hat i + 3\hat j + 6\hat k)⁠, b⃗=17(3i^−6j^+2k^)\vec b = \frac17(3\hat i - 6\hat j + 2\hat k)⁠, c⃗=17(6i^+2j^−3k^)\vec c = \frac17(6\hat i + 2\hat j - 3\hat k)⁠.
To show
∣a⃗∣=∣b⃗∣=∣c⃗∣=1|\vec a| = |\vec b| = |\vec c| = 1⁠, and a⃗⊥b⃗\vec a \perp \vec b⁠, b⃗⊥c⃗\vec b \perp \vec c⁠, c⃗⊥a⃗\vec c \perp \vec a⁠.
Plan
(1) Find the three lengths. (2) Find the three scalar products a⃗⋅b⃗\vec a\cdot\vec b⁠, b⃗⋅c⃗\vec b\cdot\vec c⁠, c⃗⋅a⃗\vec c\cdot\vec a⁠. (3) Conclude, using that the vectors are nonzero.

Basics you need

1Unit vector and length

A unit vector has magnitude 1 (Section 10.3), ∣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) and ∣λa⃗∣=∣λ∣∣a⃗∣|\lambda\vec a| = |\lambda||\vec a| (Section 10.5). Basics: use Pythagoras twice

2Scalar product in components

a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3⁠, and a scalar factor comes out: (λa⃗)⋅(μb⃗)=λμ(a⃗⋅b⃗)(\lambda\vec a)\cdot(\mu\vec b) = \lambda\mu(\vec a\cdot\vec b) (Section 10.6.1, Property 2). Lesson: the dot product

3Perpendicular vectors

For nonzero a⃗\vec a⁠, b⃗\vec b⁠: a⃗⋅b⃗=0  ⟺  a⃗⊥b⃗\vec a\cdot\vec b = 0 \iff \vec a \perp \vec b (Section 10.6.1, Observation 2). Lesson: Examples 13–21

Solution

  1. The length of a⃗\vec a

    ∣a⃗∣=1722+32+62=174+9+36=1749=77=1.\begin{aligned} |\vec a| &= \frac17\sqrt{2^2 + 3^2 + 6^2} \\ &= \frac17\sqrt{4 + 9 + 36} \\ &= \frac17\sqrt{49} = \frac77 = 1. \end{aligned}
    The factor 17\frac17 is positive, so it comes out of the length unchanged: ∣17v⃗∣=17∣v⃗∣\big|\frac17\vec v\big| = \frac17|\vec v|⁠. The numbers 2, 3, 6 are chosen so that 4+9+36=49=724 + 9 + 36 = 49 = 7^2⁠, a perfect square, and the 17\frac17 then cancels the 7 exactly.
    ∣λa⃗∣=∣λ∣∣a⃗∣|\lambda\vec a| = |\lambda||\vec a| (Section 10.5); ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

  2. The lengths of b⃗\vec b and c⃗\vec c

    ∣b⃗∣=1732+(−6)2+22=1749=1,∣c⃗∣=1762+22+(−3)2=1749=1.\begin{aligned} |\vec b| &= \frac17\sqrt{3^2 + (-6)^2 + 2^2} \\ &= \frac17\sqrt{49} = 1, \\ |\vec c| &= \frac17\sqrt{6^2 + 2^2 + (-3)^2} \\ &= \frac17\sqrt{49} = 1. \end{aligned}

    So all three are unit vectors.

    Each vector uses the same three numbers 2, 3, 6, in a different order and with different signs. Squaring removes the signs and the order does not matter in a sum, so each sum of squares is 4+9+36=494 + 9 + 36 = 49 again.
    A unit vector has magnitude 1 (Section 10.3). Lesson: six kinds of vectors

  3. The three scalar products

    a⃗⋅b⃗=149[2(3)+3(−6)+6(2)]=149(6−18+12)=0,b⃗⋅c⃗=149[3(6)+(−6)(2)+2(−3)]=149(18−12−6)=0,c⃗⋅a⃗=149[6(2)+2(3)+(−3)(6)]=149(12+6−18)=0.\begin{aligned} \vec a\cdot\vec b &= \tfrac{1}{49}[2(3) + 3(-6) + 6(2)] \\ &= \tfrac{1}{49}(6 - 18 + 12) = 0, \\ \vec b\cdot\vec c &= \tfrac{1}{49}[3(6) + (-6)(2) + 2(-3)] \\ &= \tfrac{1}{49}(18 - 12 - 6) = 0, \\ \vec c\cdot\vec a &= \tfrac{1}{49}[6(2) + 2(3) + (-3)(6)] \\ &= \tfrac{1}{49}(12 + 6 - 18) = 0. \end{aligned}
    Each vector carries a factor 17\frac17⁠, so a scalar product of two of them carries 17×17=149\frac17 \times \frac17 = \frac{1}{49} in front, and inside the bracket we multiply matching components of the whole-number vectors and add. In every case the positive products are cancelled exactly by the negative ones.
    a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3⁠; (λa⃗)⋅b⃗=λ(a⃗⋅b⃗)(\lambda\vec a)\cdot\vec b = \lambda(\vec a\cdot\vec b) (Section 10.6.1, Property 2). Lesson: the dot product

  4. Conclude: mutually perpendicular

    None of the three vectors is zero (each has length 1), and each pair has scalar product 0. So a⃗⊥b⃗\vec a \perp \vec b⁠, b⃗⊥c⃗\vec b \perp \vec c and c⃗⊥a⃗\vec c \perp \vec a⁠: the three vectors are mutually perpendicular unit vectors.

    X Y Z O a b c |a| = |b| = |c| = 1 a · b = b · c = c · a = 0
    The three vectors drawn from O in space, in a true orthographic view with one scale for all three axes. The view is along a⃗+b⃗+c⃗\vec a + \vec b + \vec c⁠, so the three arrows show at equal angles on the screen, like the corner of a cube; each pair meets at a right angle, marked at O. They are a new set of three perpendicular unit directions, just like i^\hat i⁠, j^\hat j⁠, k^\hat k but turned.
    "Mutually perpendicular" means every pair is perpendicular, so all three pairs must be checked; two of them would not be enough. The rule a⃗⋅b⃗=0  ⟺  a⃗⊥b⃗\vec a\cdot\vec b = 0 \iff \vec a \perp \vec b needs both vectors nonzero, which the unit lengths guarantee.
    Section 10.6.1, Observation 2. Lesson: Examples 13–21
Result

Each of a⃗\vec a⁠, b⃗\vec b⁠, c⃗\vec c has length 1, and a⃗⋅b⃗=b⃗⋅c⃗=c⃗⋅a⃗=0\vec a\cdot\vec b = \vec b\cdot\vec c = \vec c\cdot\vec a = 0⁠. So they are unit vectors and are mutually perpendicular. Hence proved.

(Textbook answer: the answer key has no entry for this "show that" question.)

Check it with numbers

In decimals, a⃗≈0.285714 i^+0.428571 j^\vec a \approx 0.285714\,\hat i + 0.428571\,\hat j +0.857143 k^+ 0.857143\,\hat k⁠. The squares of the exact components are 449≈0.081633\frac{4}{49} \approx 0.081633⁠, 949≈0.183673\frac{9}{49} \approx 0.183673 and 3649≈0.734694\frac{36}{49} \approx 0.734694⁠, and 0.081633+0.1836730.081633 + 0.183673 +0.734694=1.000000+ 0.734694 = 1.000000⁠.

The angles all come out as 90∘90^\circ⁠: cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣=01×1=0\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|} = \frac{0}{1 \times 1} = 0 for every pair.

Another way to see it: put the whole-number vectors 2i^+3j^+6k^2\hat i + 3\hat j + 6\hat k⁠, 3i^−6j^+2k^3\hat i - 6\hat j + 2\hat k⁠, 6i^+2j^−3k^6\hat i + 2\hat j - 3\hat k as the rows of a table. Each row has squares adding to 49, and any two rows have products adding to 0, so dividing by 7 makes every length 1 and keeps every right angle.

Common mistakes
  1. Forgetting the 17\frac17⁠. Without it, ∣2i^+3j^+6k^∣=7|2\hat i + 3\hat j + 6\hat k| = 7⁠, not 1; the factor 17\frac17 is what makes each vector a unit vector.
  2. Checking only two pairs. Mutually perpendicular needs a⃗⋅b⃗\vec a\cdot\vec b⁠, b⃗⋅c⃗\vec b\cdot\vec c and c⃗⋅a⃗\vec c\cdot\vec a all zero.
  3. Losing a sign. In b⃗⋅c⃗\vec b\cdot\vec c⁠, the product of −6-6 and 2 is −12-12⁠, and of 2 and −3-3 is −6-6⁠; one slip gives a nonzero answer.
What to take away
  1. Unit vector: the squares of the components add to 1.
  2. Perpendicular (nonzero vectors): the products of matching components add to 0.
  3. Take a common scalar factor out first: it scales a length by ∣λ∣|\lambda| and a scalar product by λμ\lambda\mu⁠.

Every fact used, and where it comes from

FactUsed inSource
A unit vector has magnitude 1Step 2Section 10.3 (Lesson: six kinds of vectors)
∣λa⃗∣=∣λ∣∣a⃗∣|\lambda\vec a| = |\lambda||\vec a|Steps 1, 2Section 10.5
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}Steps 1, 2Section 10.5.1 (Basics: use Pythagoras twice)
a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3⁠; scalars come outStep 3Section 10.6.1, Property 2 (Lesson: the dot product)
a⃗⋅b⃗=0  ⟺  a⃗⊥b⃗\vec a\cdot\vec b = 0 \iff \vec a \perp \vec b (nonzero vectors)Step 4Section 10.6.1, Observation 2
The questionResultExercise 10.3 Q5; no answer-key entry (Lesson: the dot product)
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