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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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 68 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$|\vec{a}| = 8$
›Reveal solutionSolution
The key is the identity $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$ . Substituting the given magnitudes gives $144 + (\vec{a} \cdot \vec{b})^2 = 576$ , so $|\vec{a} \cdot \vec{b}| = \sqrt{432} = 12\sqrt{3}$ . The correct option is (C).

The problem gives you the magnitudes of two vectors and the magnitude of their cross product, and asks for the magnitude of their dot product. This is a classic setup — it tests a single, powerful relationship that ties the dot product and cross product together.

The core idea: The dot product depends on $\cos\theta$ , and the cross product depends on $\sin\theta$ , where $\theta$ is the angle between the vectors. Since $|\vec{a}|$ and $|\vec{b}|$ are known, you can use the identity $\sin^2\theta + \cos^2\theta = 1$ to eliminate $\theta$ and directly connect the two products.

Write the definitions:

$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta$

$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$

Here $\theta$ is the angle between $\vec{a}$ and $\vec{b}$ , with $0 \le \theta \le \pi$ .

Square both equations:

$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2\theta$

$(\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \cos^2\theta$

Add them together:

$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 (\sin^2\theta + \cos^2\theta) = |\vec{a}|^2 |\vec{b}|^2$
This is the identity you need. It holds for any two vectors in 3D space.
$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$

Plug in the given numbers:

$|\vec{a}| = 8$ , $|\vec{b}| = 3$ , so $|\vec{a}|^2 |\vec{b}|^2 = 64 \times 9 = 576$

$|\vec{a} \times \vec{b}| = 12$ , so $|\vec{a} \times \vec{b}|^2 = 144$

Therefore: …
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CBSE 2026Set 65/2/11 markMCQ
$\vec{a}$
›Reveal solutionSolution
Assertion (A) is a fundamental identity relating the magnitudes of the cross product and dot product, which is true. Reason (R) is also a true relationship between the magnitudes of the cross product and dot product, but it does not explain Assertion (A). The correct option is (B).

To evaluate this assertion-reason question, we need to understand the definitions of the dot product and cross product of two vectors and the geometric meaning of the angle between them. Both the dot product and the cross product are fundamental operations in vector algebra, and their properties are frequently tested.

The dot product (or scalar product) of two vectors $\vec{a}$ and $\vec{b}$ is defined as:

$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$

where $|\vec{a}|$ and $|\vec{b}|$ are the magnitudes of vectors $\vec{a}$ and $\vec{b}$ respectively, and $\theta$ is the angle between them ( $0 \le \theta \le \pi$ ). The result is a scalar.

The cross product (or vector product) of two vectors $\vec{a}$ and $\vec{b}$ results in a vector perpendicular to both $\vec{a}$ and $\vec{b}$ . Its magnitude is defined as:

$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$

where $\theta$ is again the angle between $\vec{a}$ and $\vec{b}$ . The direction of $\vec{a} \times \vec{b}$ is given by the right-hand rule.

Now, let's evaluate the Assertion and Reason.

Evaluate Assertion (A):

The assertion states: $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$ .

Let's substitute the definitions of $|\vec{a} \times \vec{b}|$ and $(\vec{a} \cdot \vec{b})$ into the left-hand side (LHS) of the equation.

LHS $= (|\vec{a}| |\vec{b}| \sin \theta)^2 + (|\vec{a}| |\vec{b}| \cos \theta)^2$

LHS $= |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta + |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta$

We can factor out $|\vec{a}|^2 |\vec{b}|^2$ :

LHS $= |\vec{a}|^2 |\vec{b}|^2 (\sin^2 \theta + \cos^2 \theta)$

Important

Recall the fundamental trigonometric identity: $\sin^2 \theta + \cos^2 \theta = 1$ .

Using this identity:

LHS $= |\vec{a}|^2 |\vec{b}|^2 (1)$

LHS $= |\vec{a}|^2 |\vec{b}|^2$

This matches the right-hand side (RHS) of the assertion.

Therefore, Assertion (A) is True. This identity is often known as Lagrange's Identity for vectors.

Evaluate Reason (R):

The reason states: $|\vec{a} \times \vec{b}| = (\vec{a} \cdot \vec{b}) \tan \theta$ , $\left(\theta \neq \frac{\pi}{2}\right)$ .

Let's substitute the definitions of $|\vec{a} \times \vec{b}|$ and $(\vec{a} \cdot \vec{b})$ into this equation.

LHS: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$

RHS: $(\vec{a} \cdot \vec{b}) \tan \theta = (|\vec{a}| |\vec{b}| \cos \theta) \tan \theta$
$\tan \theta$
Substitute this into the RHS:

RHS $= (|\vec{a}| |\vec{b}| \cos \theta) \left(\frac{\sin \theta}{\cos \theta}\right)$

Since $\theta \neq \frac{\pi}{2}$ , $\cos \theta \neq 0$ , so we can cancel $\cos \theta$ : …
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CBSE 2026Set ANNUAL1 mark
$\vec{a}$
›Reveal solutionSolution

Use $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$ to solve for $\theta$ .

$\cos\theta = \dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|} = \dfrac{\sqrt6}{\sqrt3\cdot2} = \dfrac{\sqrt6}{2\sqrt3}=\dfrac{\sqrt2}{2}=\dfrac{1}{\sqrt2}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$\hat{i} + 3\hat{j} + 3\hat{k}$
(a) $0°$
(b) $45°$
(c) $60°$
(d) $90°$
›Reveal solutionSolution

The two vectors have zero dot product, so they are perpendicular.

Let $\vec{a}=\hat{i}+3\hat{j}+3\hat{k}$ and $\vec{b}=3\hat{i}-2\hat{j}+\hat{k}$ .

$\vec{a}\cdot\vec{b}=1(3)+3(-2)+3(1)=3-6+3=0$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If vector a . vector b = \sqrt3 |vector a \times vector b| then angle between vector a and vector b is:$
(a) π/2
(b) π/6
(c) π/4
(d) π/3
›Reveal solutionSolution

Writing the dot and cross product magnitudes in terms of $\cos\theta$ and $\sin\theta$ turns the given condition into $\tan\theta = \frac{1}{\sqrt3}$ .

We know:

$\vec a \cdot \vec b = |\vec a||\vec b|\cos\theta, \qquad |\vec a \times \vec b| = |\vec a||\vec b|\sin\theta$

Given $\vec a \cdot \vec b = \sqrt3\, |\vec a \times \vec b|$ :

$|\vec a||\vec b|\cos\theta = \sqrt3\,|\vec a||\vec b|\sin\theta$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q.If |\vec a| = 1, |\vec b| = 2 and \vec a \cdot \vec b = 1, then angle between \vec a and \vec b is:$
(a) \pi/2
(b) \pi/6
(c) \pi/3
(d) \pi/4
›Reveal solutionSolution

Use $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$ to solve for the angle $\theta$ .

Working:

$\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$ …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Dot Product of two vectors is defined as$
(a) a·b = |a||b| sin θ
(b) a·b = |a||b| cos θ
(c) a·b = |a||b| sin θ n̂
(d) None of the above
›Reveal solutionSolution

The dot (scalar) product uses cosine of the angle between the vectors; the cross (vector) product uses sine.

For two vectors $\vec a$ and $\vec b$ with angle $\theta$ between them, the dot product is defined as

$\vec a \cdot \vec b = |\vec a||\vec b|\cos\theta,$ …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If the scalar product of two vectors is zero, then the angle between them is:$
(a) 45°
(b) 0°
(c) 180°
(d) 90°
›Reveal solutionSolution

A.B = AB*cos(theta) is zero only when cos(theta) = 0, i.e., when the vectors are perpendicular, theta = 90 degrees.

The scalar product (dot product) of two vectors A and B is defined as

A.B = |A||B|*cos(theta)

where theta is the angle between the two vectors.

For the scalar product to be zero (assuming neither vector is itself the zero vector), we need

cos(theta) = 0

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If P = 4 î - 5 ĵ and Q = 5 î + 4 ĵ, then the angle between the two vectors is$
(a) 0
(b) π
(c) π/2
(d) π/4
›Reveal solutionSolution

P.Q = 0, so the angle between them is 90 degrees = pi/2. Answer (C).

P = 4 i - 5 j and Q = 5 i + 4 j.

Dot product: P.Q = (4)(5) + (-5)(4) = 20 - 20 = 0.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If the angle between two vectors is 90°, then their dot product will be$
(a) 0
(b) 1
(c) − 1
(d) ∞
›Reveal solutionSolution

A.B = |A||B|cos(theta); at 90 degrees this is 0. Answer (A).

The scalar (dot) product of two vectors is A.B = |A||B| cos(theta), where theta is the angle between them.

…

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CBSE 2026Set ANNUAL1 markMCQ
$\vec{a}$
(a) $\dfrac{\pi}{2}$
(b) $\dfrac{\pi}{6}$
(c) $\dfrac{\pi}{3}$
(d) $\dfrac{\pi}{4}$
›Reveal solutionSolution

$|\vec a\times\vec b|=7$ , so $\sin\theta=\tfrac{7}{2\cdot 7}=\tfrac12\Rightarrow\theta=\tfrac{\pi}{6}$ .

Step 1: $|\vec a\times\vec b|=\sqrt{3^{2}+2^{2}+6^{2}}=\sqrt{9+4+36}=\sqrt{49}=7.$

…

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CBSE 20251 markMCQ
$\vec{a} = 3\hat{i} + 2\hat{j} - \hat{k}$
›Reveal solutionSolution
The dot product of $\vec{a}$ and $\vec{b}$ is zero, which means the vectors are perpendicular. The correct option is (B).

The key to deciding whether two vectors are parallel, perpendicular, or neither lies in their dot product. The dot product (also called the scalar product) of two vectors $\vec{a}$ and $\vec{b}$ is defined as:
$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$
where $\theta$ is the angle between them. This single formula tells us everything:

If $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}|$ , then $\cos\theta = 1$ , so $\theta = 0^\circ$ — the vectors are parallel (and pointing in the same direction).

If $\vec{a} \cdot \vec{b} = -|\vec{a}| |\vec{b}|$ , then $\cos\theta = -1$ , so $\theta = 180^\circ$ — the vectors are anti-parallel.

If $\vec{a} \cdot \vec{b} = 0$ , then $\cos\theta = 0$ , so $\theta = 90^\circ$ — the vectors are perpendicular (orthogonal).

So instead of guessing, we just compute the dot product and see which case holds.

Write the vectors in component form.

$\vec{a} = 3\hat{i} + 2\hat{j} - \hat{k}$

$\vec{b} = \hat{i} - \hat{j} + \hat{k}$

Compute the dot product.

The dot product is the sum of the products of corresponding components:

$\vec{a} \cdot \vec{b} = (3)(1) + (2)(-1) + (-1)(1)$ $= 3 - 2 - 1 = 0$

Interpret the result.

Since $\vec{a} \cdot \vec{b} = 0$ , the angle between them is $90^\circ$ . Therefore, $\vec{a}$ is perpendicular to $\vec{b}$ .

Check the other options for completeness.

Option (A) says they are parallel — but that would require the dot product to be $\pm |\vec{a}||\vec{b}|$ , not zero. So (A) is false.

Option (C) says $|\vec{b}| > |\vec{a}|$ . Let's check magnitudes: …
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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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Q.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.)

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