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Chapter 10 · Vector Algebra

Exercise 10.4 · Q2: Unit vector perpendicular to a⃗+b⃗\vec a+\vec b and a⃗−b⃗\vec a-\vec b⁠, a⃗=3i^+2j^+2k^\vec a=3\hat i+2\hat j+2\hat k⁠, b⃗=i^+2j^−2k^\vec b=\hat i+2\hat j-2\hat k

Exercise 10.4Vector product, unit vector perpendicular to two vectors · Section 10.6.3
The question

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

Exam answer

a⃗+b⃗=4i^+4j^\vec a + \vec b = 4\hat i + 4\hat j and a⃗−b⃗=2i^+4k^\vec a - \vec b = 2\hat i + 4\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^440204∣=i^(16−0)−j^(16−0)+k^(0−8)=16i^−16j^−8k^\begin{aligned} \vec c &= (\vec a + \vec b)\times(\vec a - \vec b) \\ &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} \\ &= \hat i(16 - 0) - \hat j(16 - 0) \\ &\quad + \hat k(0 - 8) \\ &= 16\hat i - 16\hat j - 8\hat k \end{aligned} ∣c⃗∣=256+256+64=576=24\begin{aligned} |\vec c| &= \sqrt{256 + 256 + 64} \\ &= \sqrt{576} = 24 \end{aligned} Unit vector=±c⃗∣c⃗∣=±16i^−16j^−8k^24\begin{aligned} \text{Unit vector} &= \pm\frac{\vec c}{|\vec c|} \\ &= \pm\frac{16\hat i - 16\hat j - 8\hat k}{24} \end{aligned}

∴\therefore the required unit vector is ±(23i^−23j^−13k^)\pm\left(\dfrac23\hat i - \dfrac23\hat j - \dfrac13\hat k\right)⁠.

The answer key prints ±23i^  23j^  13k^\pm\frac23\hat i\ \ \frac23\hat j\ \ \frac13\hat k with the signs between the terms missing; the signs are as above.

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 perpendicular to both u⃗\vec u and v⃗\vec v⁠. So

c⃗=(a⃗+b⃗)×(a⃗−b⃗)\vec c = (\vec a + \vec b)\times(\vec a - \vec b)

is perpendicular to each of a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b⁠. Dividing by its length gives a unit vector, and its negative is the other one.

Given, to find, and the plan
Given
a⃗=3i^+2j^+2k^\vec a = 3\hat i + 2\hat j + 2\hat k and b⃗=i^+2j^−2k^\vec b = \hat i + 2\hat j - 2\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) Take their vector product c⃗\vec c and check it. (3) Divide by ∣c⃗∣|\vec c|⁠, with both signs. (4) Check by a shortcut and see the picture. (5) Compare with the answer key.

Basics you need

1Adding and subtracting in components

Add or subtract the matching components: (a1+b1)i^+(a2+b2)j^+(a3+b3)k^(a_1 + b_1)\hat i + (a_2 + b_2)\hat j + (a_3 + b_3)\hat k (Section 10.5.1). Lesson: components

2The 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 with n^\hat n perpendicular to both u⃗\vec u and v⃗\vec v (Definition 3, Section 10.6.3); in components it is the determinant with i^,j^,k^\hat i, \hat j, \hat k on top. Lesson: the cross product

3Unit vector in a given direction

c^=c⃗∣c⃗∣\hat c = \frac{\vec c}{|\vec c|} has length 1 and the direction of c⃗\vec c (Section 10.5). Basics: use Pythagoras twice

4Two directions are perpendicular to a plane

If c^\hat c is perpendicular to two non-parallel vectors, so is −c^-\hat c⁠. The book makes the same point in Example 23, which has the same pattern.

Solution

  1. The two vectors

    a⃗+b⃗=(3+1)i^+(2+2)j^+(2−2)k^=4i^+4j^,a⃗−b⃗=(3−1)i^+(2−2)j^+(2+2)k^=2i^+4k^.\begin{aligned} \vec a + \vec b &= (3 + 1)\hat i + (2 + 2)\hat j + (2 - 2)\hat k \\ &= 4\hat i + 4\hat j, \\ \vec a - \vec b &= (3 - 1)\hat i + (2 - 2)\hat j + (2 + 2)\hat k \\ &= 2\hat i + 4\hat k. \end{aligned}
    Add (or subtract) matching components. In a⃗−b⃗\vec a - \vec b the last component is 2−(−2)=42 - (-2) = 4⁠. The missing parts are zeros: a⃗+b⃗\vec a + \vec b has no k^\hat k part and a⃗−b⃗\vec a - \vec b has no j^\hat j part, and these zeros go into the determinant.
    Sum and difference in components (Section 10.5.1). Lesson: components

  2. Their vector product

    c⃗=(a⃗+b⃗)×(a⃗−b⃗)=∣i^j^k^440204∣=i^(16−0)−j^(16−0)+k^(0−8)=16i^−16j^−8k^.\begin{aligned} \vec c &= (\vec a + \vec b)\times(\vec a - \vec b) \\ &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} \\ &= \hat i(16 - 0) - \hat j(16 - 0) + \hat k(0 - 8) \\ &= 16\hat i - 16\hat j - 8\hat k. \end{aligned}

    Check: c⃗⋅(a⃗+b⃗)=64−64+0=0\vec c\cdot(\vec a + \vec b) = 64 - 64 + 0 = 0 and c⃗⋅(a⃗−b⃗)=32+0−32=0\vec c\cdot(\vec a - \vec b) = 32 + 0 - 32 = 0⁠.

    The minors are (4)(4)−(0)(0)=16(4)(4) - (0)(0) = 16⁠, (4)(4)−(0)(2)=16(4)(4) - (0)(2) = 16 and (4)(0)−(4)(2)=−8(4)(0) - (4)(2) = -8⁠; the j^\hat j term takes a minus sign. Both scalar products are 0, so c⃗\vec c is perpendicular to each of the two vectors, as the definition promises.
    u⃗×v⃗\vec u\times\vec v is perpendicular to u⃗\vec u and v⃗\vec v (Definition 3, Section 10.6.3). Lesson: the cross product

  3. Divide by the length, with both signs

    ∣c⃗∣=162+(−16)2+(−8)2=256+256+64=576=24,\begin{aligned} |\vec c| &= \sqrt{16^2 + (-16)^2 + (-8)^2} \\ &= \sqrt{256 + 256 + 64} = \sqrt{576} = 24, \end{aligned} ±c⃗∣c⃗∣=±16i^−16j^−8k^24=±(23i^−23j^−13k^).\begin{aligned} \pm\frac{\vec c}{|\vec c|} &= \pm\frac{16\hat i - 16\hat j - 8\hat k}{24} \\ &= \pm\left(\frac23\hat i - \frac23\hat j - \frac13\hat k\right). \end{aligned}
    Dividing a nonzero vector by its own length keeps the direction and makes the length 1. Since 576=242576 = 24^2⁠, the division is exact: 1624=23\frac{16}{24} = \frac23 and 824=13\frac{8}{24} = \frac13⁠. As a check, 49+49+19=1\frac49 + \frac49 + \frac19 = 1⁠.
    Unit vector c⃗∣c⃗∣\frac{\vec c}{|\vec c|} (Section 10.5). Basics: use Pythagoras twice

  4. A shortcut, and the picture

    Expanding the product with the distributive law, 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:

    (a⃗+b⃗)×(a⃗−b⃗)=−2(a⃗×b⃗).(\vec a + \vec b)\times(\vec a - \vec b) = -2(\vec a\times\vec b).

    Here a⃗×b⃗=i^(−4−4)−j^(−6−2)+k^(6−2)\vec a\times\vec b = \hat i(-4 - 4) - \hat j(-6 - 2) + \hat k(6 - 2) =−8i^+8j^+4k^= -8\hat i + 8\hat j + 4\hat k⁠, and −2(−8i^+8j^+4k^)=16i^−16j^−8k^=c⃗-2(-8\hat i + 8\hat j + 4\hat k) = 16\hat i - 16\hat j - 8\hat k = \vec c⁠.

    O a + b a − b θ ≈ 71.57° |a + b| = 4√2 |a − b| = 2√5 area = |c| = 24 ĉ: out of the page −ĉ: into the page
    a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b drawn in their own plane, true to scale, about 71.57∘71.57^\circ apart. Turning a⃗+b⃗\vec a + \vec b towards a⃗−b⃗\vec a - \vec b is anticlockwise as drawn, so c^=c⃗∣c⃗∣\hat c = \frac{\vec c}{|\vec c|} points out of the page and −c^-\hat c into it. The shaded parallelogram has area ∣c⃗∣=24|\vec c| = 24⁠.
    The shortcut computes c⃗\vec c a second, independent way and agrees. It also explains the answer: a vector perpendicular to a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b is perpendicular to the plane of a⃗\vec a and b⃗\vec b⁠, which is the direction of a⃗×b⃗\vec a\times\vec b⁠. A plane has exactly two unit normals, one on each side, hence the ±\pm⁠.
    Distributivity, a⃗×a⃗=0⃗\vec a\times\vec a = \vec 0 and b⃗×a⃗=−a⃗×b⃗\vec b\times\vec a = -\vec a\times\vec b (Section 10.6.3, Observations 2 and 6, Property 3). Lesson: Examples 22–25

  5. The answer key

    The key prints ±23i^  23j^  13k^\pm\frac23\hat i\ \ \frac23\hat j\ \ \frac13\hat k⁠: the two signs between the terms are missing (blank in the printed key). Reading them as plus signs would give a wrong answer, because

    (23i^+23j^+13k^)⋅(4i^+4j^)=163≠0.\left(\tfrac23\hat i + \tfrac23\hat j + \tfrac13\hat k\right)\cdot(4\hat i + 4\hat j) = \tfrac{16}{3} \ne 0.

    With minus signs, as found above, the vector is perpendicular to both.

    The numbers 23,23,13\frac23, \frac23, \frac13 in the key match our components; only the signs are lost. Testing a claimed answer with the scalar product settles which signs are right.
    Perpendicular vectors have scalar product 0 (Section 10.6.1, Observation 2). Lesson: the dot product
Result

The unit vectors perpendicular to both a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b are ±(23i^−23j^−13k^)\pm\left(\frac23\hat i - \frac23\hat j - \frac13\hat k\right)⁠.

(Textbook answer: printed as ±23i^  23j^  13k^\pm\frac23\hat i\ \ \frac23\hat j\ \ \frac13\hat k⁠, with the two signs between the terms missing; the correct answer is ±(23i^−23j^−13k^)\pm\left(\frac23\hat i - \frac23\hat j - \frac13\hat k\right)⁠.)

Check it with numbers

23≈0.666667\frac23 \approx 0.666667 and 13≈0.333333\frac13 \approx 0.333333⁠; exactly, (23)2+(23)2+(13)2=4+4+19=1\left(\frac23\right)^2 + \left(\frac23\right)^2 + \left(\frac13\right)^2 = \frac{4 + 4 + 1}{9} = 1⁠.

∣a⃗+b⃗∣=32|\vec a + \vec b| = \sqrt{32}⁠, ∣a⃗−b⃗∣=20|\vec a - \vec b| = \sqrt{20}⁠, (a⃗+b⃗)⋅(a⃗−b⃗)=8(\vec a + \vec b)\cdot(\vec a - \vec b) = 8⁠, so cos⁡θ=8640≈0.316228\cos\theta = \frac{8}{\sqrt{640}} \approx 0.316228⁠, θ≈71.57∘\theta \approx 71.57^\circ and sin⁡θ=24640≈0.948683\sin\theta = \frac{24}{\sqrt{640}} \approx 0.948683⁠.

Then ∣a⃗+b⃗∣∣a⃗−b⃗∣sin⁡θ=640⋅24640=24=∣c⃗∣|\vec a + \vec b||\vec a - \vec b|\sin\theta = \sqrt{640}\cdot\frac{24}{\sqrt{640}} = 24 = |\vec c|⁠, since 640−82=576=242640 - 8^2 = 576 = 24^2⁠.

Common mistakes
  1. Giving c⃗\vec c instead of a unit vector. 16i^−16j^−8k^16\hat i - 16\hat j - 8\hat k is perpendicular to both but has length 24; the question asks for length 1.
  2. Losing the zeros. a⃗+b⃗=4i^+4j^+0k^\vec a + \vec b = 4\hat i + 4\hat j + 0\hat k and a⃗−b⃗=2i^+0j^+4k^\vec a - \vec b = 2\hat i + 0\hat j + 4\hat k⁠; a missing 0 shifts the entries of the determinant.
  3. Copying the key's signs as plus signs. 23i^+23j^+13k^\frac23\hat i + \frac23\hat j + \frac13\hat k is not perpendicular to a⃗+b⃗\vec a + \vec b⁠; the minus signs are part of the answer.
What to take away
  1. A unit vector perpendicular to u⃗\vec u and v⃗\vec v is ±u⃗×v⃗∣u⃗×v⃗∣\pm\frac{\vec u\times\vec v}{|\vec u\times\vec v|}⁠.
  2. (a⃗+b⃗)×(a⃗−b⃗)=−2(a⃗×b⃗)(\vec a + \vec b)\times(\vec a - \vec b) = -2(\vec a\times\vec b)⁠: the same direction line as a⃗×b⃗\vec a\times\vec b⁠.
  3. Check a unit vector twice: its components squared add to 1, and its scalar product with each given vector is 0.

Every fact used, and where it comes from

FactUsed inSource
Sum and difference of vectors in componentsStep 1Section 10.5.1 (Lesson: components)
u⃗×v⃗\vec u\times\vec v is perpendicular to u⃗\vec u and v⃗\vec v⁠; determinant formStep 2Section 10.6.3, Definition 3 (Lesson: the cross product)
Expansion of a determinant along a rowSteps 2, 4Determinants (Chapter 4) (Basics: a determinant measures a signed area)
Unit vector c⃗∣c⃗∣\frac{\vec c}{|\vec c|}⁠; magnitude from componentsStep 3Sections 10.5 and 10.5.1
Distributive law; a⃗×a⃗=0⃗\vec a\times\vec a = \vec 0⁠; b⃗×a⃗=−a⃗×b⃗\vec b\times\vec a = -\vec a\times\vec bStep 4Section 10.6.3, Property 3, Observations 2 and 6
Scalar product zero for perpendicular vectorsSteps 2, 5Section 10.6.1, Observation 2
The question and the answerStep 5, ResultExercise 10.4 Q2; NCERT answer key (Lesson: the cross product, exercises)
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