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Chapter 10 · Vector Algebra

Example 23: Unit vector perpendicular to a⃗+b⃗\vec a+\vec b and a⃗−b⃗\vec a-\vec b⁠, a⃗=i^+j^+k^\vec a=\hat i+\hat j+\hat k⁠, b⃗=i^+2j^+3k^\vec b=\hat i+2\hat j+3\hat k

ExamplesVector product, unit normal · Section 10.6.3
The question

Find a unit vector perpendicular to each of the vectors (a⃗+b⃗)(\vec a + \vec b) and (a⃗−b⃗)(\vec a - \vec b)⁠, where a⃗=i^+j^+k^\vec a = \hat i + \hat j + \hat k⁠, b⃗=i^+2j^+3k^\vec b = \hat i + 2\hat j + 3\hat k⁠.

Exam answer

a⃗+b⃗=2i^+3j^+4k^\vec a + \vec b = 2\hat i + 3\hat j + 4\hat k and a⃗−b⃗=−j^−2k^\vec a - \vec b = -\hat j - 2\hat k⁠.

A vector perpendicular to both a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b is

c⃗=(a⃗+b⃗)×(a⃗−b⃗)=∣i^j^k^2340−1−2∣=i^(−6+4)−j^(−4−0)+k^(−2−0)=−2i^+4j^−2k^\begin{aligned} \vec c &= (\vec a + \vec b)\times(\vec a - \vec b) \\ &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{vmatrix} \\ &= \hat i(-6 + 4) - \hat j(-4 - 0) \\ &\quad + \hat k(-2 - 0) \\ &= -2\hat i + 4\hat j - 2\hat k \end{aligned} ∣c⃗∣=4+16+4=24=26|\vec c| = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt6 Unit vector=±c⃗∣c⃗∣=±−2i^+4j^−2k^26\begin{aligned} \text{Unit vector} &= \pm\frac{\vec c}{|\vec c|} \\ &= \pm\frac{-2\hat i + 4\hat j - 2\hat k}{2\sqrt6} \end{aligned}

∴\therefore the required unit vector is ±16(−i^+2j^−k^)\pm\dfrac{1}{\sqrt6}(-\hat i + 2\hat j - \hat k)⁠.

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

The one idea this problem is built on

The vector product u⃗×v⃗\vec u\times\vec v is, by definition, perpendicular to both u⃗\vec u and v⃗\vec v⁠. So (a⃗+b⃗)×(a⃗−b⃗)(\vec a + \vec b)\times(\vec a - \vec b) points in a direction perpendicular to both given vectors; dividing it by its length gives a unit vector, and its negative is the other one.

Given, to find, and the plan
Given
a⃗=i^+j^+k^\vec a = \hat i + \hat j + \hat k⁠, b⃗=i^+2j^+3k^\vec b = \hat i + 2\hat j + 3\hat k⁠.
To find
A unit vector perpendicular to both a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b⁠.
Plan
(1) Find a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b⁠. (2) Find c⃗=(a⃗+b⃗)×(a⃗−b⃗)\vec c = (\vec a + \vec b)\times(\vec a - \vec b) by a determinant. (3) Check that c⃗\vec c is perpendicular to both. (4) Divide by ∣c⃗∣|\vec c|⁠, giving both signs. (5) Confirm with (a⃗+b⃗)×(a⃗−b⃗)=−2(a⃗×b⃗)(\vec a + \vec b)\times(\vec a - \vec b) = -2(\vec a\times\vec b)⁠.

Basics you need

1The vector product is perpendicular to both factors

u⃗×v⃗=∣u⃗∣∣v⃗∣sin⁡θ n^\vec u\times\vec v = |\vec u||\vec v|\sin\theta\,\hat n⁠, where n^\hat n is a unit vector perpendicular to both u⃗\vec u and v⃗\vec v (Definition 3, Section 10.6.3). Lesson: the cross product

2The determinant form

u⃗×v⃗\vec u\times\vec v is the determinant with i^,j^,k^\hat i, \hat j, \hat k on top and the components of u⃗\vec u⁠, then v⃗\vec v⁠, below (Section 10.6.3). Basics: a determinant measures a signed area

3Unit vector in a direction

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

4Rules of the vector product

a⃗×a⃗=0⃗\vec a\times\vec a = \vec 0⁠, b⃗×a⃗=−a⃗×b⃗\vec b\times\vec a = -\vec a\times\vec b⁠, and ×\times distributes over ++ (Section 10.6.3, Observations 2 and 6, Property 3). Lesson: Examples 22–25

Solution

  1. The two given vectors

    a⃗+b⃗=(1+1)i^+(1+2)j^+(1+3)k^=2i^+3j^+4k^,a⃗−b⃗=(1−1)i^+(1−2)j^+(1−3)k^=−j^−2k^.\begin{aligned} \vec a + \vec b &= (1 + 1)\hat i + (1 + 2)\hat j + (1 + 3)\hat k \\ &= 2\hat i + 3\hat j + 4\hat k, \\ \vec a - \vec b &= (1 - 1)\hat i + (1 - 2)\hat j + (1 - 3)\hat k \\ &= -\hat j - 2\hat k. \end{aligned}
    Add or subtract matching components. The i^\hat i part of a⃗−b⃗\vec a - \vec b is 0, which matters in the determinant: it puts a 0 in the bottom row.
    Sum and difference by components (Section 10.5.1). Lesson: components

  2. The vector product

    c⃗=(a⃗+b⃗)×(a⃗−b⃗)=∣i^j^k^2340−1−2∣=i^(−6+4)−j^(−4−0)+k^(−2−0)=−2i^+4j^−2k^.\begin{aligned} \vec c &= (\vec a + \vec b)\times(\vec a - \vec b) \\ &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{vmatrix} \\ &= \hat i(-6 + 4) - \hat j(-4 - 0) \\ &\quad + \hat k(-2 - 0) \\ &= -2\hat i + 4\hat j - 2\hat k. \end{aligned}
    The three minors are (3)(−2)−(4)(−1)=−6+4(3)(-2) - (4)(-1) = -6 + 4⁠, (2)(−2)−(4)(0)=−4(2)(-2) - (4)(0) = -4 and (2)(−1)−(3)(0)=−2(2)(-1) - (3)(0) = -2⁠; the middle one takes a minus sign, so −(−4)=+4-(-4) = +4⁠. The book names this vector c⃗\vec c⁠.
    Determinant form of the vector product; expansion along the top row. Basics: a determinant measures a signed area

  3. Check: c⃗\vec c is perpendicular to both

    c⃗⋅(a⃗+b⃗)=−4+12−8=0,c⃗⋅(a⃗−b⃗)=0−4+4=0.\begin{aligned} \vec c\cdot(\vec a + \vec b) &= -4 + 12 - 8 = 0, \\ \vec c\cdot(\vec a - \vec b) &= 0 - 4 + 4 = 0. \end{aligned}
    The products are (−2)(2)(-2)(2)⁠, (4)(3)(4)(3)⁠, (−2)(4)(-2)(4) for a⃗+b⃗\vec a + \vec b⁠, and (−2)(0)(-2)(0)⁠, (4)(−1)(4)(-1)⁠, (−2)(−2)(-2)(-2) for a⃗−b⃗\vec a - \vec b⁠. Two nonzero vectors are perpendicular exactly when their scalar product is 0. Both products vanish, so c⃗\vec c really is perpendicular to each given vector; a slip in a minor would show up here.
    u⃗⋅v⃗=0  ⟺  u⃗⊥v⃗\vec u\cdot\vec v = 0 \iff \vec u \perp \vec v (Section 10.6.1, Observation 2). Lesson: the dot product

  4. Make it a unit vector, both ways

    ∣c⃗∣=(−2)2+42+(−2)2=24=26,\begin{aligned} |\vec c| &= \sqrt{(-2)^2 + 4^2 + (-2)^2} \\ &= \sqrt{24} = 2\sqrt6, \end{aligned} c⃗∣c⃗∣=−2i^+4j^−2k^26=−16i^+26j^−16k^.\begin{aligned} \frac{\vec c}{|\vec c|} &= \frac{-2\hat i + 4\hat j - 2\hat k}{2\sqrt6} \\ &= -\frac{1}{\sqrt6}\hat i + \frac{2}{\sqrt6}\hat j - \frac{1}{\sqrt6}\hat k. \end{aligned}

    Its negative, 16i^−26j^+16k^\frac{1}{\sqrt6}\hat i - \frac{2}{\sqrt6}\hat j + \frac{1}{\sqrt6}\hat k⁠, is also a unit vector perpendicular to both. So the answer is ±16(−i^+2j^−k^)\pm\frac{1}{\sqrt6}(-\hat i + 2\hat j - \hat k)⁠.

    O a + b, length √29 a − b, length √5 θ ≈ 155.99° ĉ: out of the page −ĉ: into the page area = |c| = 2√6 ≈ 4.90
    a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b drawn in their own plane, true to scale. Every vector perpendicular to both is perpendicular to this plane: c^\hat c points straight out of the page (towards you) and −c^-\hat c straight into it. The parallelogram on the two vectors has area ∣c⃗∣=26|\vec c| = 2\sqrt6⁠.
    Dividing by the length keeps the direction and makes the length 1: 16+46+16=1\frac16 + \frac46 + \frac16 = 1⁠. A plane has two perpendicular directions, opposite to each other, and that is the book's Note: the second unit vector comes from (a⃗−b⃗)×(a⃗+b⃗)=−c⃗(\vec a - \vec b)\times(\vec a + \vec b) = -\vec c⁠.
    c^=1∣c⃗∣c⃗\hat c = \frac{1}{|\vec c|}\vec c (Section 10.5); the Note after Example 23. Lesson: Examples 22–25

  5. A shortcut that confirms the direction

    (a⃗+b⃗)×(a⃗−b⃗)=a⃗×a⃗−a⃗×b⃗+b⃗×a⃗−b⃗×b⃗=0⃗−a⃗×b⃗−a⃗×b⃗−0⃗=−2(a⃗×b⃗).\begin{aligned} &(\vec a + \vec b)\times(\vec a - \vec b) \\ &= \vec a\times\vec a - \vec a\times\vec b + \vec b\times\vec a - \vec b\times\vec b \\ &= \vec 0 - \vec a\times\vec b - \vec a\times\vec b - \vec 0 = -2(\vec a\times\vec b). \end{aligned}

    Here a⃗×b⃗=i^(3−2)−j^(3−1)+k^(2−1)=i^−2j^+k^\vec a\times\vec b = \hat i(3 - 2) - \hat j(3 - 1) + \hat k(2 - 1) = \hat i - 2\hat j + \hat k⁠, and −2(i^−2j^+k^)=−2i^+4j^−2k^=c⃗-2(\hat i - 2\hat j + \hat k) = -2\hat i + 4\hat j - 2\hat k = \vec c⁠.

    Expanding uses distributivity, a⃗×a⃗=b⃗×b⃗=0⃗\vec a\times\vec a = \vec b\times\vec b = \vec 0 and b⃗×a⃗=−a⃗×b⃗\vec b\times\vec a = -\vec a\times\vec b⁠. It shows that the plane of a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b is the plane of a⃗\vec a and b⃗\vec b themselves, so the same two unit normals serve both pairs.
    Section 10.6.3, Property 3 and Observations 2 and 6. Lesson: the cross product
Result

A unit vector perpendicular to both a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b is −16i^+26j^−16k^-\frac{1}{\sqrt6}\hat i + \frac{2}{\sqrt6}\hat j - \frac{1}{\sqrt6}\hat k⁠; the other one is its negative, 16i^−26j^+16k^\frac{1}{\sqrt6}\hat i - \frac{2}{\sqrt6}\hat j + \frac{1}{\sqrt6}\hat k⁠.

(Textbook answer: c⃗∣c⃗∣=−16i^+26j^−16k^\frac{\vec c}{|\vec c|} = \frac{-1}{\sqrt6}\hat i + \frac{2}{\sqrt6}\hat j - \frac{1}{\sqrt6}\hat k⁠, with c⃗=−2i^+4j^−2k^\vec c = -2\hat i + 4\hat j - 2\hat k and ∣c⃗∣=26|\vec c| = 2\sqrt6⁠; its Note gives the second unit vector 16i^−26j^+16k^\frac{1}{\sqrt6}\hat i - \frac{2}{\sqrt6}\hat j + \frac{1}{\sqrt6}\hat k⁠.)

Check it with numbers

16≈0.408248\frac{1}{\sqrt6} \approx 0.408248 and 26≈0.816497\frac{2}{\sqrt6} \approx 0.816497⁠; the squares of the three components are exactly 16,46,16\frac16, \frac46, \frac16⁠, adding to 1.

∣c⃗∣=26≈4.898979|\vec c| = 2\sqrt6 \approx 4.898979⁠. The angle between a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b⁠: cos⁡θ=−11295=−11145≈−0.913500\cos\theta = \frac{-11}{\sqrt{29}\sqrt5} = \frac{-11}{\sqrt{145}} \approx -0.913500⁠, so θ≈155.99∘\theta \approx 155.99^\circ⁠, and 145sin⁡θ=145−121=24\sqrt{145}\sin\theta = \sqrt{145 - 121} = \sqrt{24}⁠, the length of c⃗\vec c again.

Common mistakes
  1. Stopping at c⃗\vec c⁠. −2i^+4j^−2k^-2\hat i + 4\hat j - 2\hat k is perpendicular to both, but its length is 262\sqrt6⁠, not 1; it must be divided by ∣c⃗∣|\vec c|⁠.
  2. Using the scalar product. (a⃗+b⃗)⋅(a⃗−b⃗)=−11(\vec a + \vec b)\cdot(\vec a - \vec b) = -11 is a number; a vector perpendicular to both comes from the vector product.
  3. Thinking the other sign is wrong. Both ±16(−i^+2j^−k^)\pm\frac{1}{\sqrt6}(-\hat i + 2\hat j - \hat k) are correct; the question asks for one of them.
What to take away
  1. A vector perpendicular to both u⃗\vec u and v⃗\vec v is u⃗×v⃗\vec u\times\vec v⁠; the unit vectors are ±u⃗×v⃗∣u⃗×v⃗∣\pm\frac{\vec u\times\vec v}{|\vec u\times\vec v|}⁠.
  2. Check a vector product by dotting it with each factor.
  3. (a⃗+b⃗)×(a⃗−b⃗)=−2(a⃗×b⃗)(\vec a + \vec b)\times(\vec a - \vec b) = -2(\vec a\times\vec b)⁠: it has the same direction line as a⃗×b⃗\vec a\times\vec b⁠.

Every fact used, and where it comes from

FactUsed inSource
Sum and difference of vectors by componentsStep 1Section 10.5.1 (Lesson: components)
u⃗×v⃗\vec u\times\vec v is perpendicular to both u⃗\vec u and v⃗\vec vIdea, Step 2Section 10.6.3, Definition 3
u⃗×v⃗\vec u\times\vec v as a determinant in componentsSteps 2, 5Section 10.6.3 (Basics: a determinant measures a signed area)
u⃗⋅v⃗=0  ⟺  u⃗⊥v⃗\vec u\cdot\vec v = 0 \iff \vec u \perp \vec vStep 3Section 10.6.1, Observation 2
c^=1∣c⃗∣c⃗\hat c = \frac{1}{|\vec c|}\vec c⁠; magnitude from componentsStep 4Sections 10.5 and 10.5.1
Two opposite unit normals to a planeStep 4The Note after Example 23
a⃗×a⃗=0⃗\vec a\times\vec a = \vec 0⁠; b⃗×a⃗=−a⃗×b⃗\vec b\times\vec a = -\vec a\times\vec b⁠; distributivityStep 5Section 10.6.3, Observations 2 and 6, Property 3
The book's working and answerResultExample 23, Section 10.6.3 (Lesson: Examples 22–25)
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