It is found that $|\vec{A} + \vec{B}| = |\vec{A}|$. This necessarily implies, (a) $\vec{B} = 0$ (b) $\vec{A}$, $\vec{B}$ are antiparallel (c) $\vec{A}$, $\vec{B}$ are perpendicular (d) $\vec{A} \cdot \vec{B} \leq 0$

It is found that |vecA + vecB| = |vecA|. This necessarily implies, (a) vecB = 0 (b) vecA, vecB are antiparalle

NCERT Exemplar · Q10

|\vec{A} + \vec{B}| = |\vec{A}|

(a)

\vec{B} = 0

(b)

\vec{A}

\vec{B}

are antiparallel

(c)

\vec{A}

\vec{B}

are perpendicular

(d)

\vec{A} \cdot \vec{B} \leq 0

Punjab PsebMCQ· 1mImportance★★★★★est

60% · 41/68 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Triangle Inequality

Triangle Inequality (Bounding the Resultant of Two Vectors)

When you combine two displacements, two velocities, or two forces in "Motion in a Plane," you add them as vectors using the triangle law: place the tail of the second vector at the head of the first, and the resultant runs from the start to the finish. The triangle inequality is simply the statement that this resultant can never be longer than the two vectors laid end to end, and can never be shorter than their difference.

The Intuition

Suppose you walk 3 m in one direction, then 4 m in some other direction. Could you end up 8 m from where you started? No — the farthest you can possibly get is $3+4=7$ m, and that only happens if both walks point the same way, so there's no bend at all (a "flat" triangle). The moment the second walk points in a different direction, a real corner appears, and cutting across that corner (the direct path) is always shorter than going via the corner. That's the geometric heart of every triangle: any one side is shorter than the sum of the other two, unless the triangle collapses onto a straight line.

The Precise Statement

For two vectors $\vec{A}$ and $\vec{B}$ added by the triangle law, the magnitude of the resultant $\vec{R} = \vec{A} + \vec{B}$ is bounded on both sides:

|\vec{A}| - |\vec{B}|\big| \;\le\; |\vec{A}+\vec{B}| \;\le\; |\vec{A}| + |\vec{B}|

The upper bound $|\vec{A}+\vec{B}| \le |\vec{A}|+|\vec{B}|$ is reached only when $\vec{A}$ and $\vec{B}$ point in exactly the same direction (the angle between them is $0^\circ$ ) — the triangle flattens out.
The lower bound $|\vec{A}+\vec{B}| \ge \big||\vec{A}|-|\vec{B}|\big|$ is reached only when $\vec{A}$ and $\vec{B}$ point in exactly opposite directions (the angle is $180^\circ$ ).
For any angle in between, the resultant magnitude lies strictly between these two limits.

This is the vector form of the ordinary triangle inequality $|x+y| \le |x|+|y|$ you may already know for numbers: here $x$ and $y$ become vectors, and "the sides of a triangle" become "a vector, another vector, and their sum."

Why It Matters in Kinematics

This bound is genuinely useful when combining physical quantities in the plane:

Relative velocity: if a boat has speed 5 m/s relative to water and the river flows at 3 m/s, the boat's speed relative to the ground must lie between $|5-3|=2$ m/s and $5+3=8$ m/s, depending on the angle the boat is steered — it can never be less than 2 or more than 8.
Combining forces or displacements: if you know only the magnitudes of two vectors, not the angle between them, this inequality instantly tells you the range of possible resultant magnitudes without doing any trigonometry. …

Start your 14-day free trial to unlock the full solution →

When $|\vec{A} + \vec{B}| = |\vec{A}|$ , squaring both sides reveals that $\vec{B}$ must either be zero or point in a direction that makes a non-acute angle with $\vec{A}$ , meaning $\vec{A} \cdot \vec{B} \leq 0$ . The answer is (D).

The triangle inequality tells us that when we add two vectors, the magnitude of the sum depends on both the individual magnitudes and the angle between them. The condition $|\vec{A} + \vec{B}| = |\vec{A}|$ is saying that adding $\vec{B}$ to $\vec{A}$ doesn't change the length at all—a very special constraint.

To understand what this means, we need to examine the relationship between the vectors algebraically. The key insight is that magnitudes are always non-negative, so we can square both sides of an equation involving magnitudes without losing information.

Step-by-step analysis

Square both sides of the given condition:

|\vec{A} + \vec{B}|^2 = |\vec{A}|^2

Expand the left side using the dot product: The magnitude squared of a vector sum is:

(\vec{A} + \vec{B}) \cdot (\vec{A} + \vec{B}) = \vec{A} \cdot \vec{A} + 2\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{B}

This gives us:

|\vec{A}|^2 + 2\vec{A} \cdot \vec{B} + |\vec{B}|^2 = |\vec{A}|^2

Simplify by canceling $|\vec{A}|^2$ from both sides:

2\vec{A} \cdot \vec{B} + |\vec{B}|^2 = 0

Rearrange to isolate the dot product:

\vec{A} \cdot \vec{B} = -\frac{|\vec{B}|^2}{2}

Interpret the result: Since $|\vec{B}|^2 \geq 0$ (a magnitude squared is never negative), we have:

\vec{A} \cdot \vec{B} = -\frac{|\vec{B}|^2}{2} \leq 0

The dot product is zero when $\vec{B} = \vec{0}$ , and strictly negative when $\vec{B} \neq \vec{0}$ .

Why the other options fail

Let's check each option against our result: …

Start your 14-day free trial to unlock the full solution →

Concept: Align the Axes with $\vec A$ First, Then Read the Constraint Off Components

Method: WLOG Coordinate Alignment (Put $\vec A$ Along the $x$ -axis), Not Abstract Dot-Product Manipulation

Instead of manipulating $\vec A\cdot\vec B$ as an abstract symbol, this method exploits the freedom to choose coordinates: since the condition $|\vec A+\vec B|=|\vec A|$ doesn't care which way our axes point, align the $x$ -axis along $\vec A$ itself. The constraint on $\vec B$ then becomes a concrete statement about its components, which is easier to interpret geometrically.

Steps

Choose the $x$ -axis along $\vec A$ , without loss of generality. Then $\vec A = A\hat i$ (with $A=|\vec A|>0$ ), and write $\vec B = B_x\hat i + B_y\hat j$ in this frame -- $B_x, B_y$ are just numbers, to be constrained by the given condition.
Write $\vec A+\vec B$ in this frame:

\vec A + \vec B = (A+B_x)\hat i + B_y\hat j

Impose $|\vec A+\vec B|=|\vec A|=A$ and square both sides:

(A+B_x)^2 + B_y^2 = A^2

A^2 + 2AB_x + B_x^2 + B_y^2 = A^2

2AB_x + (B_x^2+B_y^2) = 0 \quad\Rightarrow\quad 2AB_x + |\vec B|^2 = 0 \quad\Rightarrow\quad B_x = -\frac{|\vec B|^2}{2A}

Read off the sign of $B_x$ directly. Since $A>0$ and $|\vec B|^2\geq0$ , we get $B_x \leq 0$ always (equal to $0$ only if $\vec B=\vec 0$ ). So in this special frame, $\vec B$ 's component along $\vec A$ 's own direction can never be positive.
Connect this coordinate fact back to the dot product. In this frame, $\vec A\cdot\vec B = (A\hat i)\cdot(B_x\hat i+B_y\hat j) = AB_x$ . Since $A>0$ and $B_x\leq0$ from Step 4:

$\vec A\cdot\vec B = AB_x \leq 0$ …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ9 — Three vectors vecA, vecB and vecC add up to zero. Find which is false.…→Next questionQ11 — Two particles are projected in air with speed vo at angles θ₁ and θ₂ (…

Q.It is found that ∣A⃗+B⃗∣=∣A⃗∣|\vec{A} + \vec{B}| = |\vec{A}|∣A+B∣=∣A∣. This necessarily implies,

Step-by-step analysis

Why the other options fail

🎓Unlock everything free for 14 days

Q.It is found that $|\vec{A} + \vec{B}| = |\vec{A}|$ . This necessarily implies,

Unlock everything free for 14 days