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Chapter 10 · Vector Algebra

Miscellaneous · Q4: Does a⃗=b⃗+c⃗\vec a=\vec b+\vec c give ∣a⃗∣=∣b⃗∣+∣c⃗∣|\vec a|=|\vec b|+|\vec c|⁠?

Miscellaneous ExerciseTriangle law, magnitudes · Sections 10.4 and 10.6.1
The question

If a⃗=b⃗+c⃗\vec a = \vec b + \vec c⁠, then is it true that ∣a⃗∣=∣b⃗∣+∣c⃗∣|\vec a| = |\vec b| + |\vec c|⁠? Justify your answer.

Exam answer

No, it is not true in general.

Take b⃗=i^\vec b = \hat i and c⃗=j^\vec c = \hat j⁠. Then a⃗=b⃗+c⃗=i^+j^\vec a = \vec b + \vec c = \hat i + \hat j⁠, and

∣a⃗∣=12+12=2,∣b⃗∣+∣c⃗∣=1+1=2\begin{aligned} |\vec a| &= \sqrt{1^2 + 1^2} = \sqrt2, \\ |\vec b| + |\vec c| &= 1 + 1 = 2 \end{aligned}

Since 2≠2\sqrt2 \ne 2⁠, ∣a⃗∣≠∣b⃗∣+∣c⃗∣|\vec a| \ne |\vec b| + |\vec c| here.

In general, let b⃗≠0⃗\vec b \ne \vec 0⁠, c⃗≠0⃗\vec c \ne \vec 0 and let θ\theta be the angle between them. Then

∣a⃗∣2=∣b⃗+c⃗∣2=∣b⃗∣2+∣c⃗∣2+2 b⃗⋅c⃗=∣b⃗∣2+∣c⃗∣2+2∣b⃗∣∣c⃗∣cos⁡θ≤(∣b⃗∣+∣c⃗∣)2\begin{aligned} |\vec a|^2 &= |\vec b + \vec c|^2 \\ &= |\vec b|^2 + |\vec c|^2 + 2\,\vec b\cdot\vec c \\ &= |\vec b|^2 + |\vec c|^2 \\ &\quad + 2|\vec b||\vec c|\cos\theta \\ &\le (|\vec b| + |\vec c|)^2 \end{aligned}

since cos⁡θ≤1\cos\theta \le 1⁠. So ∣a⃗∣≤∣b⃗∣+∣c⃗∣|\vec a| \le |\vec b| + |\vec c|⁠, with equality only when cos⁡θ=1\cos\theta = 1⁠, that is θ=0\theta = 0⁠. (If b⃗\vec b and c⃗\vec c are not along one line, a⃗\vec a⁠, b⃗\vec b⁠, c⃗\vec c are the sides of a triangle and ∣a⃗∣<∣b⃗∣+∣c⃗∣|\vec a| < |\vec b| + |\vec c|⁠.)

∴\therefore No; ∣a⃗∣=∣b⃗∣+∣c⃗∣|\vec a| = |\vec b| + |\vec c| holds only when b⃗\vec b and c⃗\vec c have the same direction (or one of them is 0⃗\vec 0⁠).

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

The one idea this problem is built on

Adding vectors is adding arrows, not lengths. Placed head to tail, b⃗\vec b⁠, c⃗\vec c and their sum a⃗\vec a form a triangle, and a straight side of a triangle is shorter than the path along the other two sides. Lengths add only when there is no bend: b⃗\vec b and c⃗\vec c pointing the same way.

Given, to show, and the plan
Given
a⃗=b⃗+c⃗\vec a = \vec b + \vec c⁠.
To decide
Whether ∣a⃗∣=∣b⃗∣+∣c⃗∣|\vec a| = |\vec b| + |\vec c| must follow, with a reason.
Plan
(1) Test one simple example; one counterexample answers "no". (2) Explain why it fails in general (the triangle). (3) Find exactly when equality does hold, using the dot product.

Basics you need

1Triangle law

If b⃗\vec b runs from A to B and c⃗\vec c from B to C, then b⃗+c⃗\vec b + \vec c runs from A to C (Section 10.4). Lesson: adding arrows, head to tail

2Expanding a dot product

(b⃗+c⃗)⋅(b⃗+c⃗)=∣b⃗∣2+2 b⃗⋅c⃗+∣c⃗∣2(\vec b + \vec c)\cdot(\vec b + \vec c) = |\vec b|^2 + 2\,\vec b\cdot\vec c + |\vec c|^2⁠, just like (x+y)2(x + y)^2⁠. Basics: expanding brackets

3Dot product and angle

b⃗⋅c⃗=∣b⃗∣∣c⃗∣cos⁡θ\vec b\cdot\vec c = |\vec b||\vec c|\cos\theta with 0≤θ≤π0 \le \theta \le \pi (Section 10.6.1). Lesson: the dot product

Solution

  1. One example settles it: the answer is no

    Take b⃗=i^\vec b = \hat i and c⃗=j^\vec c = \hat j⁠. Then a⃗=b⃗+c⃗=i^+j^\vec a = \vec b + \vec c = \hat i + \hat j and

    ∣a⃗∣=12+12=2,∣b⃗∣+∣c⃗∣=1+1=2.\begin{aligned} |\vec a| &= \sqrt{1^2 + 1^2} = \sqrt2, \\ |\vec b| + |\vec c| &= 1 + 1 = 2. \end{aligned}

    Since 2≠2\sqrt2 \ne 2⁠, the statement is not true in general.

    The question asks whether something is always true. One case where a⃗=b⃗+c⃗\vec a = \vec b + \vec c holds but ∣a⃗∣=∣b⃗∣+∣c⃗∣|\vec a| = |\vec b| + |\vec c| fails is enough to show that the first does not force the second.
    ∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

  2. Why it fails: the triangle

    Place b⃗\vec b and c⃗\vec c head to tail: b⃗=AB→\vec b = \overrightarrow{AB}⁠, c⃗=BC→\vec c = \overrightarrow{BC}⁠. By the triangle law a⃗=AC→\vec a = \overrightarrow{AC}⁠. If b⃗\vec b and c⃗\vec c are not along one line, ABC is a genuine triangle and

    AC<AB+BC, that is,∣a⃗∣<∣b⃗∣+∣c⃗∣.\begin{aligned} &\text{AC} < \text{AB} + \text{BC}, \text{ that is,} \\ &|\vec a| < |\vec b| + |\vec c|. \end{aligned}
    A B C b, length 1 c, length 1 a, length √2 √2 ≈ 1.41 < 2 b c a = b + c, length 2 = 1 + 1 same direction: no bend
    Left: b⃗=i^\vec b = \hat i⁠, c⃗=j^\vec c = \hat j head to tail, drawn true to scale; the side a⃗=b⃗+c⃗\vec a = \vec b + \vec c has length 2≈1.41\sqrt2 \approx 1.41⁠, less than 1+1=21 + 1 = 2⁠. Right: when c⃗\vec c points the same way as b⃗\vec b⁠, there is no bend and the lengths add.
    Any side of a triangle is shorter than the other two sides together: the straight route from A to C is shorter than the detour through B. This is the answer key's hint: "take a⃗\vec a⁠, b⃗\vec b and c⃗\vec c to represent the sides of a triangle".
    Triangle law (Section 10.4); triangle inequality, Example 20. Lesson: adding arrows, head to tail

  3. Exactly when equality holds

    Both sides are lengths (never negative), so we may compare their squares:

    ∣a⃗∣2=(b⃗+c⃗)⋅(b⃗+c⃗)=∣b⃗∣2+2 b⃗⋅c⃗+∣c⃗∣2,(∣b⃗∣+∣c⃗∣)2=∣b⃗∣2+2∣b⃗∣∣c⃗∣+∣c⃗∣2.\begin{aligned} |\vec a|^2 &= (\vec b + \vec c)\cdot(\vec b + \vec c) \\ &= |\vec b|^2 + 2\,\vec b\cdot\vec c + |\vec c|^2, \\ (|\vec b| + |\vec c|)^2 &= |\vec b|^2 + 2|\vec b||\vec c| + |\vec c|^2. \end{aligned}

    So ∣a⃗∣=∣b⃗∣+∣c⃗∣|\vec a| = |\vec b| + |\vec c| exactly when b⃗⋅c⃗=∣b⃗∣∣c⃗∣\vec b\cdot\vec c = |\vec b||\vec c|⁠. If b⃗\vec b and c⃗\vec c are non-zero, this means cos⁡θ=1\cos\theta = 1⁠, so θ=0\theta = 0⁠: b⃗\vec b and c⃗\vec c have the same direction. If one of them is 0⃗\vec 0⁠, equality holds trivially.

    For non-negative numbers pp and qq⁠, p=qp = q exactly when p2=q2p^2 = q^2⁠. After cancelling ∣b⃗∣2+∣c⃗∣2|\vec b|^2 + |\vec c|^2⁠, only the middle terms are left. Since b⃗⋅c⃗=∣b⃗∣∣c⃗∣cos⁡θ\vec b\cdot\vec c = |\vec b||\vec c|\cos\theta and cos⁡θ≤1\cos\theta \le 1⁠, the left side is at most the right side, and the gap closes only at θ=0\theta = 0⁠.
    a⃗⋅a⃗=∣a⃗∣2\vec a\cdot\vec a = |\vec a|^2 and the distributive law (Section 10.6.1). Basics: expanding brackets
Result

No. In general ∣a⃗∣=∣b⃗+c⃗∣≤∣b⃗∣+∣c⃗∣|\vec a| = |\vec b + \vec c| \le |\vec b| + |\vec c|⁠, with strict inequality when b⃗\vec b and c⃗\vec c are nonzero and not in the same direction (when they are not along one line, they form a triangle with a⃗\vec a⁠). For example, b⃗=i^\vec b = \hat i⁠, c⃗=j^\vec c = \hat j give ∣a⃗∣=2≠2|\vec a| = \sqrt2 \ne 2⁠. Equality holds only when b⃗\vec b and c⃗\vec c have the same direction (or one is 0⃗\vec 0⁠).

(Textbook answer: No; take a⃗\vec a⁠, b⃗\vec b and c⃗\vec c to represent the sides of a triangle.)

Check it with numbers

2≈1.414214<2\sqrt2 \approx 1.414214 < 2⁠.

Another example: b⃗=3i^\vec b = 3\hat i⁠, c⃗=4j^\vec c = 4\hat j give a⃗=3i^+4j^\vec a = 3\hat i + 4\hat j with ∣a⃗∣=5|\vec a| = 5⁠, while ∣b⃗∣+∣c⃗∣=7|\vec b| + |\vec c| = 7⁠.

An equality case: b⃗=i^+j^\vec b = \hat i + \hat j⁠, c⃗=2i^+2j^\vec c = 2\hat i + 2\hat j (same direction) give a⃗=3i^+3j^\vec a = 3\hat i + 3\hat j⁠, and ∣a⃗∣=32=2+22|\vec a| = 3\sqrt2 = \sqrt2 + 2\sqrt2⁠.

Common mistakes
  1. Treating vectors like numbers. a⃗=b⃗+c⃗\vec a = \vec b + \vec c is an equation between arrows; lengths add only along a straight line.
  2. Answering "yes" from one example where it works. One equality case (same direction) does not make it true in general; one counterexample is enough to answer "no".
  3. Saying it is never true. It is true when b⃗\vec b and c⃗\vec c point the same way.
What to take away
  1. ∣b⃗+c⃗∣≤∣b⃗∣+∣c⃗∣|\vec b + \vec c| \le |\vec b| + |\vec c| (the triangle inequality).
  2. Equality holds only for vectors in the same direction (or a zero vector).
  3. To disprove an "always" statement, one explicit counterexample is enough.

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}Step 1Section 10.5.1 (Basics: use Pythagoras twice)
Triangle law of vector additionStep 2Section 10.4 (Lesson: adding arrows, head to tail)
Triangle inequality ∣a⃗+b⃗∣≤∣a⃗∣+∣b⃗∣|\vec a + \vec b| \le |\vec a| + |\vec b|Step 2, ResultExample 20 (Lesson: Examples 13–21)
a⃗⋅a⃗=∣a⃗∣2\vec a\cdot\vec a = |\vec a|^2⁠; distributive law; b⃗⋅c⃗=∣b⃗∣∣c⃗∣cos⁡θ\vec b\cdot\vec c = |\vec b||\vec c|\cos\thetaStep 3Section 10.6.1 (Lesson: the dot product)
The answerResultNCERT answer key, Miscellaneous Exercise on Chapter 10, Q4 (Lesson: put it all together)
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