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Chapter 10 · Vector Algebra

Example 4: x,y,zx, y, z making xi^+2j^+zk^x\hat i+2\hat j+z\hat k equal to 2i^+yj^+k^2\hat i+y\hat j+\hat k

ExamplesComponents of a vector, equal vectors · Section 10.5.1
The question

Find the values of xx⁠, yy and zz so that the vectors a⃗=xi^+2j^+zk^\vec a = x\hat i + 2\hat j + z\hat k and b⃗=2i^+yj^+k^\vec b = 2\hat i + y\hat j + \hat k are equal.

Exam answer

Two vectors are equal if and only if their corresponding components are equal.

a⃗=b⃗  ⟺   xi^+2j^+zk^=2i^+yj^+k^\begin{aligned} \vec a = \vec b \iff\ &x\hat i + 2\hat j + z\hat k \\ &= 2\hat i + y\hat j + \hat k \end{aligned}

Comparing the coefficients of i^\hat i⁠, j^\hat j and k^\hat k⁠: x=2x = 2⁠, 2=y2 = y and z=1z = 1⁠.

∴ x=2, y=2, z=1\therefore\ x = 2,\ y = 2,\ z = 1⁠.

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

The one idea this problem is built on

Every vector can be written in exactly one way as a1i^+a2j^+a3k^a_1\hat i + a_2\hat j + a_3\hat k⁠. So two vectors are equal if and only if their corresponding components are equal: the i^\hat i parts agree, the j^\hat j parts agree and the k^\hat k parts agree. One vector equation becomes three ordinary equations, one for each unknown.

Given, to find, and the plan
Given
a⃗=xi^+2j^+zk^\vec a = x\hat i + 2\hat j + z\hat k and b⃗=2i^+yj^+k^\vec b = 2\hat i + y\hat j + \hat k⁠.
To find
xx⁠, yy⁠, zz with a⃗=b⃗\vec a = \vec b⁠.
Plan
(1) Write both vectors with all three components visible (k^=1k^\hat k = 1\hat k⁠). (2) Equate the i^\hat i⁠, j^\hat j and k^\hat k components. (3) Read off x,y,zx, y, z⁠. (4) Substitute back to check.

Basics you need

1Component form

Every vector can be written as xi^+yj^+zk^x\hat i + y\hat j + z\hat k⁠; the numbers x,y,zx, y, z are its scalar components along the axes (Section 10.5.1). Lesson: components

2Equality by components

For 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: a⃗=b⃗\vec a = \vec b if and only if a1=b1a_1 = b_1⁠, a2=b2a_2 = b_2 and a3=b3a_3 = b_3 (Section 10.5.1 (iii)). Lesson: components

3A point is fixed by three numbers

The components of a position vector are the coordinates of its tip, and two points coincide only when all three coordinates agree. Basics: three numbers fix a point

Solution

  1. Line up the components

    Write k^\hat k as 1k^1\hat k⁠, so that both vectors show all three components:

    a⃗=xi^+2j^+zk^,b⃗=2i^+yj^+1k^.\begin{aligned} \vec a &= x\hat i + 2\hat j + z\hat k, \\ \vec b &= 2\hat i + y\hat j + 1\hat k. \end{aligned}
    The rule compares coefficients one by one, so each coefficient must be visible. In b⃗\vec b the coefficient of k^\hat k is 1, and in a⃗\vec a the coefficient of j^\hat j is the number 2; the unknowns sit in different places in the two vectors.
    In xi^+yj^+zk^x\hat i + y\hat j + z\hat k⁠, the numbers x,y,zx, y, z are the scalar components (Section 10.5.1). Lesson: components

  2. Equate corresponding components

    a⃗=b⃗\vec a = \vec b holds if and only if

    x=2⏟i^,2=y⏟j^,z=1⏟k^.\underbrace{x = 2}_{\hat i}, \qquad \underbrace{2 = y}_{\hat j}, \qquad \underbrace{z = 1}_{\hat k}.
    "If and only if" works both ways. If the components agree, the vectors are built from the same multiples of i^,j^,k^\hat i, \hat j, \hat k⁠, so they are equal. If the vectors are equal, then (x−2)i^+(2−y)j^+(z−1)k^=0⃗(x - 2)\hat i + (2 - y)\hat j + (z - 1)\hat k = \vec 0⁠, and the zero vector has all components 0, because i^,j^,k^\hat i, \hat j, \hat k point along three different axes and no combination of two of them can cancel the third.
    Equality of vectors in component form (Section 10.5.1 (iii)). Lesson: components

  3. Read off the values

    Each equation already gives one unknown:

    x=2x = 2⁠, y=2y = 2⁠, z=1z = 1.

    The three equations are independent of each other (each involves one unknown only), so the values are unique: no other choice of x,y,zx, y, z makes the vectors equal.
    Corresponding components must be equal, one axis at a time. Basics: three numbers fix a point

  4. Substitute back

    With these values,

    a⃗=2i^+2j^+1k^,b⃗=2i^+2j^+1k^,\begin{aligned} \vec a &= 2\hat i + 2\hat j + 1\hat k, \\ \vec b &= 2\hat i + 2\hat j + 1\hat k, \end{aligned}

    the same vector, so a⃗=b⃗\vec a = \vec b⁠.

    Substituting back confirms that the values make the two vectors identical in every component, so nothing was lost or misread when the coefficients were compared.
    Two vectors in component form are equal when all their components agree (Section 10.5.1 (iii)). Lesson: Examples 4–10
Result

x=2x = 2⁠, y=2y = 2⁠, z=1z = 1⁠; then a⃗=b⃗=2i^+2j^+k^\vec a = \vec b = 2\hat i + 2\hat j + \hat k⁠.

(Textbook answer: x=2x = 2⁠, y=2y = 2⁠, z=1z = 1⁠.)

Check it with numbers

With x=2x = 2⁠, y=2y = 2⁠, z=1z = 1⁠: a⃗−b⃗=(2−2)i^+(2−2)j^+(1−1)k^=0⃗\vec a - \vec b = (2 - 2)\hat i + (2 - 2)\hat j + (1 - 1)\hat k = \vec 0⁠. Both vectors have magnitude 22+22+12=9=3\sqrt{2^2 + 2^2 + 1^2} = \sqrt9 = 3⁠.

A wrong choice fails at once: with x=1x = 1⁠, a⃗−b⃗=−i^≠0⃗\vec a - \vec b = -\hat i \ne \vec 0⁠, even if yy and zz are right.

Common mistakes
  1. Matching by position in the line instead of by unit vector. Compare the coefficient of i^\hat i with the coefficient of i^\hat i⁠, and so on.
  2. Forgetting the hidden 1. k^\hat k means 1k^1\hat k⁠, so z=1z = 1⁠, not z=0z = 0⁠.
  3. Thinking equal magnitudes are enough. Equal vectors need every component equal; equal lengths alone do not make vectors equal (see Example 5).
What to take away
  1. a⃗=b⃗\vec a = \vec b if and only if a1=b1a_1 = b_1⁠, a2=b2a_2 = b_2⁠, a3=b3a_3 = b_3⁠.
  2. One vector equation in space gives three scalar equations.
  3. Write every coefficient explicitly (k^=1k^\hat k = 1\hat k⁠) before comparing.

Every fact used, and where it comes from

FactUsed inSource
Component form xi^+yj^+zk^x\hat i + y\hat j + z\hat k⁠; scalar componentsStep 1Section 10.5.1 (Lesson: components)
a⃗=b⃗\vec a = \vec b if and only if a1=b1a_1 = b_1⁠, a2=b2a_2 = b_2⁠, a3=b3a_3 = b_3Steps 2, 4Section 10.5.1 (iii)
Difference of vectors by components; the zero vectorStep 2Section 10.5.1 (ii); Section 10.3
Magnitude x2+y2+z2\sqrt{x^2 + y^2 + z^2}CheckSection 10.5.1
The book's working and answerResultExample 4, Section 10.5.1 (Lesson: Examples 4–10)
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