Chapter 10 · Vector Algebra

Exercise 10.2 · Q18 (MCQ): Which triangle-law statement for triangle ABC is not true

Exercise 10.2Triangle law of vector addition · Sections 10.3 and 10.4

The question

In triangle ABC (Fig 10.18), which of the following is not true:

(A) $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \vec 0$

(B) $\overrightarrow{AB} + \overrightarrow{BC} - \overrightarrow{AC} = \vec 0$

(D) $\overrightarrow{AB} - \overrightarrow{CB} + \overrightarrow{CA} = \vec 0$

Exam answer

By the triangle law, $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC} = -\overrightarrow{CA}$ ⁠, so (A), (B) and (D) are true. But

\begin{aligned} &\overrightarrow{AB} + \overrightarrow{BC} - \overrightarrow{CA} \\ &= \overrightarrow{AC} + \overrightarrow{AC} \\ &= 2\overrightarrow{AC} \ne \vec 0 \quad (\text{A} \ne \text{C}) \end{aligned}

Answer: (C) $\overrightarrow{AB} + \overrightarrow{BC} - \overrightarrow{CA} = \vec 0$

The book prints (B) and (C) alike, as $\overrightarrow{AB} + \overrightarrow{BC} - \overrightarrow{AC} = \vec 0$ ⁠, which is true; the key's answer (C) fits (C) read with $\overrightarrow{CA}$ ⁠, as above.

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

The one idea this problem is built on

Everything follows from the triangle law, $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ ⁠, together with the fact that reversing a vector changes its sign: $\overrightarrow{CA} = -\overrightarrow{AC}$ and $\overrightarrow{CB} = -\overrightarrow{BC}$ ⁠. We simplify each option and see whether it really gives $\vec 0$ ⁠.

Given, to find, and the plan

Given: A triangle ABC, as in Fig 10.18 (arrows $\overrightarrow{AB}$ ⁠, $\overrightarrow{BC}$ and $\overrightarrow{AC}$ ⁠), and the four options as printed. In this reprint, options (B) and (C) are printed identically (checked on the page image: the two lines are the same drawing).
To find: The statement that is not true.
Plan: (1) Write the triangle law and the rule for reversing a vector. (2) Simplify (A) and (D). (3) Simplify (B) and (C) as printed. (4) Examine the reading of (C) that the answer key uses, and decide.

Basics you need

1The triangle law

Going from A to B and then from B to C is the same displacement as going from A to C: $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ (Section 10.4, Figs 10.7 and 10.8). Lesson: adding arrows, head to tail

2The negative of a vector

$\overrightarrow{BA}$ has the length of $\overrightarrow{AB}$ and the opposite direction, so $\overrightarrow{BA} = -\overrightarrow{AB}$ (Section 10.3). Lesson: six kinds of vectors worth naming

3Sides in order give zero

$\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \overrightarrow{AA} = \vec 0$ ⁠: when the sides of a triangle are taken in order, the start and the end coincide (Section 10.4, Fig 10.8 (iii)). Lesson: adding arrows, head to tail

Solution

The two rules we need
$\begin{aligned} \overrightarrow{AB} + \overrightarrow{BC} &= \overrightarrow{AC}, \\ \overrightarrow{CA} &= -\overrightarrow{AC}, \\ \overrightarrow{CB} &= -\overrightarrow{BC}. \end{aligned}$

A triangle ABC drawn like Fig 10.18, with the arrows $\overrightarrow{AB}$ ⁠, $\overrightarrow{BC}$ and $\overrightarrow{AC}$ ⁠. Head to tail, $\overrightarrow{AB}$ then $\overrightarrow{BC}$ ends where $\overrightarrow{AC}$ ends, so $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ ⁠.

The first is the triangle law read off the figure. The other two say that reversing the order of the letters reverses the arrow: same length, opposite direction, so the vector changes sign.

Triangle law (Section 10.4); negative of a vector (Section 10.3). Lesson: adding arrows, head to tail
Options (A) and (D)
$\begin{aligned} \text{(A)}\quad &\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} \\ &= \overrightarrow{AC} + \overrightarrow{CA} = \vec 0, \\ \text{(D)}\quad &\overrightarrow{AB} - \overrightarrow{CB} + \overrightarrow{CA} \\ &= \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \vec 0. \end{aligned}$
Both are true.

(A) is the sides of the triangle taken in order: A to B to C and back to A, a round trip with zero net displacement. In (D), $-\overrightarrow{CB} = \overrightarrow{BC}$ ⁠, so (D) is (A) written differently.

$\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \overrightarrow{AA} = \vec 0$ (Section 10.4, Fig 10.8 (iii)). Lesson: adding arrows, head to tail
Options (B) and (C) as printed
$\overrightarrow{AB} + \overrightarrow{BC} - \overrightarrow{AC} = \overrightarrow{AC} - \overrightarrow{AC} = \vec 0.$
So (B) is true. In this reprint, (C) is printed exactly as (B), so (C) as printed is true as well. Read literally, all four options are true and none answers the question.

Subtracting $\overrightarrow{AC}$ from both sides of the triangle law gives (B). Since the question asks for the statement that is not true, one option must have been meant differently; the natural candidate is the repeated one.

$\vec a + (-\vec a) = \vec 0$ ⁠: $-\vec a$ is the additive inverse of $\vec a$ (Section 10.5). Lesson: adding arrows, head to tail

The reading of (C) that the key uses, and every option side by side

The answer key gives (C). It fits (C) read with $\overrightarrow{CA}$ in the last term, $\overrightarrow{AB} + \overrightarrow{BC} - \overrightarrow{CA} = \vec 0$ ⁠:

\begin{aligned} \overrightarrow{AB} + \overrightarrow{BC} - \overrightarrow{CA} &= \overrightarrow{AC} + \overrightarrow{AC} \\ &= 2\overrightarrow{AC}. \end{aligned}

A and C are different vertices, so $\overrightarrow{AC} \ne \vec 0$ and $2\overrightarrow{AC} \ne \vec 0$ ⁠. This statement is not true.

Option	Simplifies to	True?
(A)	$\overrightarrow{AC} + \overrightarrow{CA} = \vec 0$	Yes
(B)	$\overrightarrow{AC} - \overrightarrow{AC} = \vec 0$	Yes
(C) as printed	the same as (B)	Yes
(C) read with $\overrightarrow{CA}$	$2\overrightarrow{AC} \ne \vec 0$	No
(D)	the same as (A)	Yes

Subtracting

\overrightarrow{CA}

adds

\overrightarrow{AC}

⁠, so instead of cancelling, the two arrows along AC pile up. This is the only reading of the options under which exactly one statement fails, and it is the one the key's answer (C) describes.

-\overrightarrow{CA} = \overrightarrow{AC}

(Section 10.3); a vector is

\vec 0

only when its initial and terminal points coincide (zero vector, Section 10.3). Lesson: six kinds of vectors worth naming

Result

The statement that is not true is (C), read as $\overrightarrow{AB} + \overrightarrow{BC} - \overrightarrow{CA} = \vec 0$ ⁠: the left side is $2\overrightarrow{AC} \ne \vec 0$ ⁠. (Textbook answer: (C).)

As printed in this reprint, (C) is identical to (B), $\overrightarrow{AB} + \overrightarrow{BC} - \overrightarrow{AC} = \vec 0$ ⁠, which is true; with that text all four options would be true. The key's answer (C) matches the reading with $\overrightarrow{CA}$ ⁠.

Check it with numbers

Take A(1, 2, 3), B(4, 0, 1), C(2, 5, $-1$ ⁠). Then $\overrightarrow{AB} = 3\hat i - 2\hat j - 2\hat k$ ⁠, $\overrightarrow{BC} = -2\hat i + 5\hat j - 2\hat k$ ⁠, $\overrightarrow{CA} = -\hat i - 3\hat j + 4\hat k$ and $\overrightarrow{AC} = \hat i + 3\hat j - 4\hat k$ ⁠.

(A): $(3 - 2 - 1)\hat i + (-2 + 5 - 3)\hat j + (-2 - 2 + 4)\hat k = \vec 0$ ⁠. (B): $\overrightarrow{AB} + \overrightarrow{BC} = \hat i + 3\hat j - 4\hat k = \overrightarrow{AC}$ ⁠, so subtracting $\overrightarrow{AC}$ gives $\vec 0$ ⁠. (C) with $\overrightarrow{CA}$ ⁠: $\hat i + 3\hat j - 4\hat k - (-\hat i - 3\hat j + 4\hat k) = 2\hat i + 6\hat j - 8\hat k$ ⁠, of length $2\sqrt{26} \approx 10.198039$ ⁠, not 0.

Common mistakes

Treating $\overrightarrow{AC}$ and $\overrightarrow{CA}$ as the same. They are negatives of each other; mixing them up is exactly what makes (C) false.
Missing that $-\overrightarrow{CB} = \overrightarrow{BC}$ ⁠. Then (D) looks wrong, though it is (A) in disguise.
Assuming an MCQ must have a printed false option. Here the printed (B) and (C) coincide; the key's answer (C) refers to the reading with $\overrightarrow{CA}$ ⁠.

What to take away

Triangle law: $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ ⁠; sides in order: $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \vec 0$ ⁠.
Swapping the letters changes the sign: $\overrightarrow{CA} = -\overrightarrow{AC}$ ⁠.
To test a vector statement, rewrite every arrow in terms of a few, then simplify.

Every fact used, and where it comes from

Fact	Used in	Source
Triangle law $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$	Steps 1, 3, 4	Section 10.4, Figs 10.7 and 10.8 (Lesson: adding arrows)
Negative of a vector, $\overrightarrow{BA} = -\overrightarrow{AB}$	Steps 1, 2, 4	Section 10.3
$\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \vec 0$	Step 2	Section 10.4, Fig 10.8 (iii)
$\vec a + (-\vec a) = \vec 0$ (additive inverse)	Step 3	Section 10.5
Zero vector: initial and terminal points coincide	Step 4	Section 10.3
The printed options and the answer	Result	Exercise 10.2 Q18 and Fig 10.18 (book p355, page image); NCERT answer key (Lesson: components, lengths and unit vectors)

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Exercise 10.2 · Q18 (MCQ): Which triangle-law statement for triangle ABC is not true

Basics you need

Solution

The two rules we need

Options (A) and (D)

Options (B) and (C) as printed

The reading of (C) that the key uses, and every option side by side

Every fact used, and where it comes from