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Chapter 10 · Vector Algebra

Exercise 10.2 · Q4: xx⁠, yy making 2i^+3j^2\hat i+3\hat j equal to xi^+yj^x\hat i+y\hat j

Exercise 10.2Components of a vector, equality of vectors · Section 10.5.1
The question

Find the values of xx and yy so that the vectors 2i^+3j^2\hat i + 3\hat j and xi^+yj^x\hat i + y\hat j are equal.

Exam answer

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

2i^+3j^=xi^+yj^2\hat i + 3\hat j = x\hat i + y\hat j

Comparing the coefficients of i^\hat i and j^\hat j on both sides,

x=2x = 2 and y=3y = 3⁠.

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

The one idea this problem is built on

A vector in component form is fixed by its components: a1i^+a2j^+a3k^a_1\hat i + a_2\hat j + a_3\hat k and b1i^+b2j^+b3k^b_1\hat i + b_2\hat j + b_3\hat k are equal if and only if a1=b1a_1 = b_1⁠, a2=b2a_2 = b_2 and a3=b3a_3 = b_3⁠. So one vector equation splits into one ordinary equation per component, and here each equation gives one unknown at once.

Given, to find, and the plan
Given
The vectors 2i^+3j^2\hat i + 3\hat j and xi^+yj^x\hat i + y\hat j⁠.
To find
The values of xx and yy that make them equal.
Plan
(1) Write the condition 2i^+3j^=xi^+yj^2\hat i + 3\hat j = x\hat i + y\hat j⁠. (2) Compare the i^\hat i components and the j^\hat j components (the k^\hat k components are both 0). (3) Check the answer.

Basics you need

1Equality of vectors in components

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, every vector is xi + yj + zk

2Missing terms are zero

2i^+3j^2\hat i + 3\hat j means 2i^+3j^+0k^2\hat i + 3\hat j + 0\hat k⁠: a vector in the xy-plane is a vector in space with k^\hat k component 0. Basics: three numbers fix a point in space

3Equal vectors

Equal vectors have the same magnitude and the same direction (Section 10.3). Lesson: six kinds of vectors

Solution

  1. Write the condition

    We need

    2i^+3j^=xi^+yj^.2\hat i + 3\hat j = x\hat i + y\hat j.

    Writing in the zero k^\hat k components, this is 2i^+3j^+0k^=xi^+yj^+0k^2\hat i + 3\hat j + 0\hat k = x\hat i + y\hat j + 0\hat k⁠.

    The question asks for the vectors to be equal, so the equation is the whole condition. Writing the k^\hat k terms makes the comparison in the next step complete: the third components agree automatically (0 = 0), so they give no condition.
    Component form of a vector (Section 10.5.1). Lesson: components

  2. Compare the components

    Two vectors are equal if and only if their corresponding components are equal. Comparing the coefficients of i^\hat i and of j^\hat j⁠:

    x=2,y=3.x = 2, \qquad y = 3.
    Why is comparing allowed? Move everything to one side: (x−2)i^+(y−3)j^=0⃗(x - 2)\hat i + (y - 3)\hat j = \vec 0⁠. The magnitude of this vector is (x−2)2+(y−3)2\sqrt{(x-2)^2 + (y-3)^2}⁠, and the zero vector has magnitude 0, so the sum of the two squares is 0. Squares are never negative, so each square must be 0, which forces x=2x = 2 and y=3y = 3⁠. The values are unique.
    Section 10.5.1, (iii): 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; the book's Example 4 uses the same comparison. Lesson: Examples 4–10

  3. Check the answer

    With x=2x = 2 and y=3y = 3⁠, the second vector is 2i^+3j^2\hat i + 3\hat j⁠, the same as the first. Both are the arrow from O to the point (2,3)(2, 3)⁠.

    2î 3ĵ (2, 3) O x y x = 2 steps along î y = 3 steps along ĵ
    The vector 2i^+3j^2\hat i + 3\hat j on a unit grid (equal scales): 2 steps along i^\hat i⁠, then 3 steps along j^\hat j⁠. For xi^+yj^x\hat i + y\hat j to be this same arrow, it must take the same steps, so x=2x = 2 and y=3y = 3⁠.
    Equal vectors have the same magnitude and direction. The component steps fix both: different steps would end at a different point, giving a different length or direction.
    Equal vectors (Section 10.3). Lesson: six kinds of vectors
Result x=2,y=3.x = 2, \qquad y = 3.

(Textbook answer: x=2x = 2⁠, y=3y = 3⁠.)

Check it with numbers

Magnitude: ∣2i^+3j^∣=4+9=13≈3.6056|2\hat i + 3\hat j| = \sqrt{4 + 9} = \sqrt{13} \approx 3.6056⁠, and with x=2x = 2⁠, y=3y = 3 also ∣xi^+yj^∣=13|x\hat i + y\hat j| = \sqrt{13}⁠. Direction: both make tan⁡−132≈56.31∘\tan^{-1}\frac32 \approx 56.31^\circ with the positive x-axis. A wrong pair such as x=3x = 3⁠, y=2y = 2 has the same magnitude 13\sqrt{13} but the angle tan⁡−123≈33.69∘\tan^{-1}\frac23 \approx 33.69^\circ⁠, so it is a different vector.

Common mistakes
  1. Matching magnitudes only. x2+y2=13x^2 + y^2 = 13 has infinitely many solutions (for example x=3,y=2x = 3, y = 2⁠); equal vectors need equal components, not just equal lengths.
  2. Mixing up components. The coefficient of i^\hat i gives xx⁠, the coefficient of j^\hat j gives yy⁠.
What to take away
  1. A vector equation a⃗=b⃗\vec a = \vec b in components is three ordinary equations: a1=b1a_1 = b_1⁠, a2=b2a_2 = b_2⁠, a3=b3a_3 = b_3⁠.
  2. A missing i^\hat i⁠, j^\hat j or k^\hat k term means that component is 0.

Every fact used, and where it comes from

FactUsed inSource
Component form xi^+yj^+zk^x\hat i + y\hat j + z\hat kStep 1Section 10.5.1 (Lesson: components)
a⃗=b⃗\vec a = \vec b if and only if the corresponding components are equalStep 2Section 10.5.1, (iii); Example 4 (Lesson: Examples 4–10)
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}⁠; the zero vector has magnitude 0Step 2 (why); CheckSections 10.5.1 and 10.3
Equal vectors: same magnitude and directionStep 3Section 10.3 (Lesson: six kinds of vectors)
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