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Chapter 10 · Vector Algebra

Example 5: i^+2j^\hat i+2\hat j and 2i^+j^2\hat i+\hat j⁠: equal magnitudes, unequal vectors

ExamplesComponents of a vector, magnitude and equality · Section 10.5.1
The question

Let a⃗=i^+2j^\vec a = \hat i + 2\hat j and b⃗=2i^+j^\vec b = 2\hat i + \hat j⁠. Is ∣a⃗∣=∣b⃗∣|\vec a| = |\vec b|⁠? Are the vectors a⃗\vec a and b⃗\vec b equal?

Exam answer ∣a⃗∣=12+22=5∣b⃗∣=22+12=5\begin{aligned} |\vec a| &= \sqrt{1^2 + 2^2} = \sqrt5 \\ |\vec b| &= \sqrt{2^2 + 1^2} = \sqrt5 \end{aligned}

So ∣a⃗∣=∣b⃗∣|\vec a| = |\vec b|⁠.

Two vectors are equal only if their corresponding components are equal. Here the coefficient of i^\hat i is 1 in a⃗\vec a and 2 in b⃗\vec b⁠, and 1≠21 \ne 2⁠.

∴ ∣a⃗∣=∣b⃗∣=5\therefore\ |\vec a| = |\vec b| = \sqrt5⁠, but a⃗\vec a and b⃗\vec b are not equal.

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

The one idea this problem is built on

A vector carries two pieces of information: its magnitude (how long) and its direction (which way). Equal vectors must agree in both. Here the two vectors have the same length, 5\sqrt5⁠, but point in different directions, so they are not equal. In components: equal vectors need every component equal, and a⃗\vec a⁠, b⃗\vec b have their components swapped.

Given, to find, and the plan
Given
a⃗=i^+2j^\vec a = \hat i + 2\hat j⁠, b⃗=2i^+j^\vec b = 2\hat i + \hat j⁠.
To find
Whether ∣a⃗∣=∣b⃗∣|\vec a| = |\vec b|⁠, and whether a⃗=b⃗\vec a = \vec b⁠.
Plan
(1) Compute ∣a⃗∣|\vec a| and ∣b⃗∣|\vec b| with ∣xi^+yj^∣=x2+y2|x\hat i + y\hat j| = \sqrt{x^2 + y^2}⁠. (2) Compare them. (3) Compare the components to decide equality. (4) See the two directions in a figure.

Basics you need

1Magnitude from components

∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}⁠, found by Pythagoras (Section 10.5.1). In the plane the zz term is 0. Basics: use Pythagoras twice

2Equal vectors

Equal vectors have the same magnitude and the same direction (Section 10.3); in components, a⃗=b⃗\vec a = \vec b if and only if the corresponding components are equal (Section 10.5.1 (iii)). Lesson: six kinds of vectors worth naming

3Components as steps on a grid

i^+2j^\hat i + 2\hat j means 1 unit along the x-axis and 2 units along the y-axis: the arrow from O to the point (1, 2). Basics: three numbers fix a point

Solution

  1. Magnitude of a⃗\vec a

    a⃗=i^+2j^\vec a = \hat i + 2\hat j has components 1 and 2, so

    ∣a⃗∣=12+22=1+4=5.|\vec a| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt5.
    The arrow from O to (1, 2) is the hypotenuse of a right triangle with legs 1 (along x) and 2 (along y), so its length is 12+22\sqrt{1^2 + 2^2} by Pythagoras.
    ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2} for r⃗=xi^+yj^+zk^\vec r = x\hat i + y\hat j + z\hat k (Section 10.5.1). Basics: use Pythagoras twice

  2. Magnitude of b⃗\vec b

    b⃗=2i^+j^\vec b = 2\hat i + \hat j has components 2 and 1, so

    ∣b⃗∣=22+12=4+1=5.|\vec b| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt5.
    The same two numbers appear, in the other order, and the sum of squares does not depend on the order. So the two lengths must agree.
    Magnitude of a vector in component form (Section 10.5.1). Lesson: components

  3. Compare the magnitudes

    Both are 5\sqrt5⁠, so yes, ∣a⃗∣=∣b⃗∣|\vec a| = |\vec b|.

    The magnitude is a single non-negative number, so comparing magnitudes is comparing two numbers: 5=5\sqrt5 = \sqrt5⁠.
    The magnitude (length) of a vector is a scalar (Section 10.2). Lesson: a vector is a length with a direction

  4. Decide whether the vectors are equal

    Compare corresponding components:

    i^:  1 in a⃗,  2 in b⃗;j^:  2 in a⃗,  1 in b⃗.\begin{aligned} \hat i:&\ \ 1 \text{ in } \vec a,\ \ 2 \text{ in } \vec b; \\ \hat j:&\ \ 2 \text{ in } \vec a,\ \ 1 \text{ in } \vec b. \end{aligned}

    The components are different (1≠21 \ne 2⁠), so a⃗\vec a and b⃗\vec b are not equal.

    x y A(1, 2) B(2, 1) O OA = î + 2ĵ, length √5 OB = 2î + ĵ, length √5 same length, different directions dashed: radius √5
    a⃗=OA→\vec a = \overrightarrow{OA} to A(1, 2) and b⃗=OB→\vec b = \overrightarrow{OB} to B(2, 1), drawn with equal scales on both axes. Both tips lie on the circle of radius 5\sqrt5 about O (equal magnitudes), but the arrows point in different directions (unequal vectors).
    Vectors are equal if and only if all their corresponding components are equal; one mismatch is enough to make them unequal. In the picture, a⃗\vec a is steep (rising 2 for every 1 across) and b⃗\vec b is shallow (rising 1 for every 2 across): same length, different directions, and equal vectors need the same direction too.
    Equal vectors: same magnitude and same direction (Section 10.3); equality by components (Section 10.5.1 (iii)). Lesson: six kinds of vectors worth naming
Result

Yes, ∣a⃗∣=∣b⃗∣=5|\vec a| = |\vec b| = \sqrt5⁠. But a⃗≠b⃗\vec a \ne \vec b⁠: their corresponding components are different, so their directions are different.

(Textbook answer: ∣a⃗∣=5|\vec a| = \sqrt5 and ∣b⃗∣=5|\vec b| = \sqrt5⁠, so ∣a⃗∣=∣b⃗∣|\vec a| = |\vec b|⁠; but the two vectors are not equal since their corresponding components are distinct.)

Check it with numbers

5≈2.236068\sqrt5 \approx 2.236068 for both. The difference is a⃗−b⃗=−i^+j^≠0⃗\vec a - \vec b = -\hat i + \hat j \ne \vec 0⁠, of length 2≈1.414214\sqrt2 \approx 1.414214⁠, which is the distance between A(1, 2) and B(2, 1).

Directions: a⃗\vec a makes an angle with the x-axis whose tangent is 21\frac21⁠, about 63.435∘63.435^\circ⁠; b⃗\vec b makes an angle whose tangent is 12\frac12⁠, about 26.565∘26.565^\circ⁠. The two angles add to 90∘90^\circ⁠: the arrows are mirror images in the line y=xy = x⁠.

Common mistakes
  1. Equal magnitudes, so equal vectors. Magnitude is only half of a vector; the directions must also agree.
  2. Comparing the list of numbers, not the components. a⃗\vec a and b⃗\vec b use the same numbers 1 and 2, but on different unit vectors: 1i^1\hat i is not 2i^2\hat i⁠.
  3. Forgetting to square. ∣i^+2j^∣|\hat i + 2\hat j| is 12+22=5\sqrt{1^2 + 2^2} = \sqrt5⁠, not 1+2=31 + 2 = 3⁠.
What to take away
  1. a⃗=b⃗\vec a = \vec b implies ∣a⃗∣=∣b⃗∣|\vec a| = |\vec b|⁠, but not the other way round.
  2. To test equality, compare components; to compare lengths, compare magnitudes.
  3. Swapping the components of xi^+yj^x\hat i + y\hat j keeps the length and changes the direction (unless x=yx = y⁠).

Every fact used, and where it comes from

FactUsed inSource
∣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)
Magnitude is the length of the vector, a numberStep 3Section 10.2
Equal vectors have the same magnitude and directionStep 4Section 10.3 (Lesson: six kinds of vectors)
a⃗=b⃗\vec a = \vec b if and only if corresponding components are equalStep 4Section 10.5.1 (iii)
The book's working and answerResultExample 5, Section 10.5.1 (Lesson: Examples 4–10)
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