For two vectors $\vec{A}$ and $\vec{B}$, $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$ is always true when (Note: more than one of the given options may be correct.) (a) $|\vec{A}| = |\vec{B}| \neq 0$ (b) $\vec{A} \perp \vec{B}$ (c) $|\vec{A}| = |\vec{B}| \neq 0$ and $\vec{A}$ and $\vec{B}$ are parallel or anti parallel (d) when either $|\vec{A}|$ or $|\vec{B}|$ is zero.

> [!TLDR] > The condition $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$ boils down to $\vec{A} \cdot \vec{B} = 0$, i.e., the vectors are perpendicular. This holds when $\vec{A} \perp \vec{B}$ **or** when either vector is zero (since a zero vector is trivially perpendicular to any vector). So the correct options are **(B)** and **(D)**. The key is to avoid memorizing — instead, square both magnitudes and see what the equality forces. **Why the Triangle Inequality idea?** The magnitudes $|\vec{A} + \vec{B}|$ and $|\vec{A} - \vec{B}|$ are the lengths of the diagonals of the parallelogram formed by $\vec{A}$ and $\vec{B}$. For these diagonals to be equal, the parallelogram must be a rectangle — meaning the sides are perpendicular. That’s the geometric intuition. Algebraically, squaring removes the square root and gives a clean dot-product condition. 1. **Square both sides** Since magnitudes are non-negative, $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$ is equivalent to $$|\vec{A} + \vec{B}|^2 = |\vec{A} - \vec{B}|^2.$$ 2. **Expand using the dot product** Recall $|\vec{V}|^2 = \vec{V} \cdot \vec{V}$. So: $$(\vec{A} + \vec{B})\cdot(\vec{A} + \vec{B}) = (\vec{A} - \vec{B})\cdot(\vec{A} - \vec{B}).$$ Expanding: $$\vec{A}\cdot\vec{A} + 2\vec{A}\cdot\vec{B} + \vec{B}\cdot\vec{B} = \vec{A}\cdot\vec{A} - 2\vec{A}\cdot\vec{B} + \vec{B}\cdot\vec{B}.$$ 3. **Cancel common terms** $|\vec{A}|^2$ and $|\vec{B}|^2$ appear on both sides, so they cancel, leaving: $$2\vec{A}\cdot\vec{B} = -2\vec{A}\cdot\vec{B}.$$ This simplifies to $4\vec{A}\cdot\vec{B} = 0$, i.e., $$\vec{A}\cdot\vec{B} = 0.$$ > [!FORMULA] > $$|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}| \iff \vec{A}\cdot\vec{B} = 0$$ 4. **Interpret the dot product condition** $\vec{A}\cdot\vec{B} = 0$ means the vectors are **perpendicular** (orthogonal). But there’s a special case: if either $\vec{A}$ or $\vec{B}$ is the zero vector, then $\vec{A}\cdot\vec{B} = 0$ holds trivially

For two vectors vecA and vecB, |vecA + vecB| = |ve – Class 11 Physics

NCERT Exemplar · Q15

\vec{A}

(a)

|\vec{A}| = |\vec{B}| \neq 0

(b)

\vec{A} \perp \vec{B}

(c)

|\vec{A}| = |\vec{B}| \neq 0

and

\vec{A}

and

\vec{B}

are parallel or anti parallel

(d) when either

|\vec{A}|

|\vec{B}|

is zero.

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Concept understanding — Dot Product Angle

Finding the Angle Between Vectors

Suppose you have two arrows drawn from the same point. One question is unavoidable in geometry, physics, and mechanics: what is the angle between them? You could measure it with a protractor on paper, but that fails the moment the vectors live in 3D. The dot product gives you the angle by pure calculation.

The Core Idea

The scalar (dot) product of two vectors has two faces that describe the same number:

\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \qquad \text{(components)}

\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta \qquad \text{(geometry)}

The first is easy to compute from coordinates; the second hides the angle $\theta$ (with $0 \le \theta \le \pi$ ) between the vectors. Setting them equal and solving for $\cos\theta$ gives the master formula.

\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|}, \qquad \theta = \cos^{-1}\!\left(\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}\right)

Why It Works

Both vectors have a fixed length, so the only thing the dot product can "vary" with is how aligned they are. When they point the same way, $\cos\theta = 1$ and the dot product is as large as possible, $|\vec{a}||\vec{b}|$ . When they are perpendicular, $\cos\theta = 0$ and the dot product vanishes. When they point opposite ways, $\cos\theta = -1$ . Dividing $\vec{a}\cdot\vec{b}$ by the two lengths simply strips away the size information and leaves behind a pure measure of alignment — exactly $\cos\theta$ .

Note

The sign of the dot product tells you the type of angle at a glance: positive $\Rightarrow$ acute, zero $\Rightarrow$ right angle, negative $\Rightarrow$ obtuse.

Using the Formula

For $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 0\hat{j} + 0\hat{k}$ :

\vec{a}\cdot\vec{b} = 1, \quad |\vec{a}| = 3, \quad |\vec{b}| = 1

$\cos\theta = \frac{1}{3\cdot 1} = \frac{1}{3} \;\Rightarrow\; \theta = \cos^{-1}\tfrac{1}{3} \approx 70.5^\circ$ …

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The condition $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$ boils down to $\vec{A} \cdot \vec{B} = 0$ , i.e., the vectors are perpendicular. This holds when $\vec{A} \perp \vec{B}$ or when either vector is zero (since a zero vector is trivially perpendicular to any vector). So the correct options are (B) and (D).

The key is to avoid memorizing — instead, square both magnitudes and see what the equality forces.

Why the Triangle Inequality idea?

The magnitudes $|\vec{A} + \vec{B}|$ and $|\vec{A} - \vec{B}|$ are the lengths of the diagonals of the parallelogram formed by $\vec{A}$ and $\vec{B}$ . For these diagonals to be equal, the parallelogram must be a rectangle — meaning the sides are perpendicular. That’s the geometric intuition. Algebraically, squaring removes the square root and gives a clean dot-product condition.

Square both sides Since magnitudes are non-negative, $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$ is equivalent to

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

Expand using the dot product Recall $|\vec{V}|^2 = \vec{V} \cdot \vec{V}$ . So:

(\vec{A} + \vec{B})\cdot(\vec{A} + \vec{B}) = (\vec{A} - \vec{B})\cdot(\vec{A} - \vec{B}).

Expanding:

\vec{A}\cdot\vec{A} + 2\vec{A}\cdot\vec{B} + \vec{B}\cdot\vec{B} = \vec{A}\cdot\vec{A} - 2\vec{A}\cdot\vec{B} + \vec{B}\cdot\vec{B}.

Cancel common terms $|\vec{A}|^2$ and $|\vec{B}|^2$ appear on both sides, so they cancel, leaving:

2\vec{A}\cdot\vec{B} = -2\vec{A}\cdot\vec{B}.

This simplifies to $4\vec{A}\cdot\vec{B} = 0$ , i.e.,

\vec{A}\cdot\vec{B} = 0.

|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}| \iff \vec{A}\cdot\vec{B} = 0

Interpret the dot product condition

$\vec{A}\cdot\vec{B} = 0$ means the vectors are perpendicular (orthogonal). But there’s a special case: if either $\vec{A}$ or $\vec{B}$ is the zero vector, then $\vec{A}\cdot\vec{B} = 0$ holds trivially (since $0 \cdot \vec{B} = 0$ ). A zero vector has no direction, so it’s considered perpendicular to every vector by convention.
Check each option …

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Concept: Equal Diagonals Mean a Rectangle -- a Classical Synthetic-Geometry Fact, No Coordinates or Dot Products

Method: The Parallelogram Law of Geometry (Diagonals Equal $\iff$ Rectangle)

Rather than expanding $|\vec A+\vec B|^2$ and $|\vec A-\vec B|^2$ via the dot product, this method uses a classical fact from Euclidean geometry about parallelograms directly: the two diagonals of a parallelogram are equal in length if and only if the parallelogram is a rectangle (equivalently, iff its adjacent sides are perpendicular). No components or dot products appear anywhere.

Setting up the parallelogram

Place $\vec A$ and $\vec B$ tail-to-tail at a common point $O$ , and complete the parallelogram $OPQR$ they generate (with $\vec A = \overrightarrow{OP}$ , $\vec B=\overrightarrow{OR}$ , and $Q$ the fourth vertex). The two diagonals of this parallelogram are the well-known vector-addition results:

\overrightarrow{OQ} = \vec A + \vec B \qquad(\text{the "long" diagonal, sum of adjacent sides})

\overrightarrow{RP} = \vec A - \vec B \qquad(\text{the "short" diagonal, difference of adjacent sides})

Applying the classical theorem

The given condition $|\vec A+\vec B|=|\vec A-\vec B|$ is exactly the statement that these two diagonals have equal length. By the theorem above, this happens if and only if $OPQR$ is a rectangle, i.e. if and only if the adjacent sides $\vec A$ and $\vec B$ meet at a right angle:

|\vec A+\vec B| = |\vec A-\vec B| \iff \vec A \perp \vec B

Checking the special (degenerate) case

A "rectangle" with one side of zero length is a degenerate limiting case (it collapses to a line segment) -- but the geometric theorem still holds trivially there: if $\vec B=\vec 0$ (or $\vec A=\vec 0$ ), the parallelogram itself degenerates, and both diagonals collapse to the same segment $\vec A$ (or $\vec B$ ), which are trivially equal in length. So the perpendicularity theorem, extended to this boundary case, also covers "either vector is zero."

Checking why (A) and (C) fail, using the same picture …

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Q.For two vectors A⃗\vec{A}A and B⃗\vec{B}B, ∣A⃗+B⃗∣=∣A⃗−B⃗∣|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|∣A+B∣=∣A−B∣ is always true when (Note: more than one of the given options may be correct.)

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