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 – 1st PUC 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.

COMEDK 2025Set 2025-A1 markMCQ
$\frac{\pi}{5}$
›Reveal solutionSolution
The problem reduces to solving for the angle between two equal vectors given that the square of their resultant equals three times their product. Using the law of cosines, we find the angle is $\pi/3$ , so the correct option is (D).

The key here is to recall how two vectors add. When you have two vectors of equal magnitude, their resultant’s magnitude depends only on the angle between them. The phrase “square of resultant” means the square of the magnitude of the sum, and “product” means the product of the magnitudes of the two vectors. Since the vectors are equal, that product is just the square of one vector’s magnitude. This is a classic setup for using the parallelogram law of vector addition.

Let’s work through it step by step.

Set up the notation.

Let each electric field vector have magnitude $E$ . So $|\vec{E}_1| = |\vec{E}_2| = E$ .

The resultant vector is $\vec{R} = \vec{E}_1 + \vec{E}_2$ .

The square of the resultant is $|\vec{R}|^2$ , and the product of the two vectors’ magnitudes is $E \cdot E = E^2$ .

Translate the given condition into an equation.

The problem says: “The square of resultant is three times their product.”

So:

$|\vec{R}|^2 = 3 \cdot (E \cdot E) = 3E^2.$

Use the formula for the magnitude of the sum of two vectors. For any two vectors with magnitudes $a$ and $b$ and angle $\theta$ between them:

$|\vec{a} + \vec{b}|^2 = a^2 + b^2 + 2ab \cos\theta.$
Here $a = b = E$ , so:
$|\vec{R}|^2 = E^2 + E^2 + 2E^2 \cos\theta = 2E^2 (1 + \cos\theta).$

Set this equal to the given value.

$2E^2 (1 + \cos\theta) = 3E^2.$
Since $E^2 \neq 0$ , divide both sides by $E^2$ :
$2(1 + \cos\theta) = 3.$

Solve for $\cos\theta$ . …
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COMEDK 2025Set 2025-E1 markMCQ
$\vec{a}$
›Reveal solutionSolution
The condition $\vec{a} \cdot \vec{b} = |\vec{a} \times \vec{b}|$ forces the angle $\theta$ between the vectors to satisfy $\tan\theta = 1$ , so $\theta = \pi/4$ . The correct option is (A).

The key idea is to express both the dot product and the magnitude of the cross product in terms of the magnitudes of the vectors and the angle between them. The dot product is $|\vec{a}||\vec{b}|\cos\theta$ , and the magnitude of the cross product is $|\vec{a}||\vec{b}|\sin\theta$ . Setting them equal gives a simple trigonometric equation.

Write the given condition using standard formulas. For any two vectors $\vec{a}$ and $\vec{b}$ , with angle $\theta$ between them ( $0 \le \theta \le \pi$ ), we have:

$\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\,\cos\theta$
and
$|\vec{a} \times \vec{b}| = |\vec{a}|\,|\vec{b}|\,\sin\theta.$
The problem states:
$|\vec{a}|\,|\vec{b}|\,\cos\theta = |\vec{a}|\,|\vec{b}|\,\sin\theta.$

Simplify the equation. Assuming neither vector is zero (if either were zero, the angle is undefined, but the condition would hold trivially; however, the problem implies non-zero vectors), we can divide both sides by $|\vec{a}|\,|\vec{b}|$ (which is positive), obtaining:

$\cos\theta = \sin\theta.$

Solve the trigonometric equation. The equation $\cos\theta = \sin\theta$ can be rewritten as $\tan\theta = 1$ (provided $\cos\theta \neq 0$ ; if $\cos\theta = 0$ , then $\sin\theta = 1$ or $-1$ , which doesn't satisfy equality). The principal solution in the range $0 \le \theta \le \pi$ is: $\theta = \frac{\pi}{4}.$ …
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COMEDK 2024Set 2024-A1 markMCQ
$\text { The angle between } \hat{\imath}-\hat{\jmath} ~\&~ \hat{\jmath}-\hat{k} \text { is }$
›Reveal solutionSolution
The angle between two vectors is found using the dot product formula; for $\vec{a} = \hat{\imath} - \hat{\jmath}$ and $\vec{b} = \hat{\jmath} - \hat{k}$ , the cosine is $-\frac12$ , so the angle is $\frac{2\pi}{3}$ , which corresponds to option (B).

The key idea: The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ satisfies $\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}$ . This works because the dot product measures how much the vectors point in the same direction, scaled by their lengths. Here, we just compute the dot product and magnitudes carefully.

Write the vectors in component form.

$\vec{a} = \hat{\imath} - \hat{\jmath}$ means components $(1, -1, 0)$ .

$\vec{b} = \hat{\jmath} - \hat{k}$ means components $(0, 1, -1)$ .

Compute the dot product.

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

Compute the magnitudes.

$|\vec{a}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{1+1} = \sqrt{2}$ .

$|\vec{b}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{0+1+1} = \sqrt{2}$ .

Find $\cos\theta$ .

$\cos\theta = \frac{-1}{\sqrt{2}\cdot\sqrt{2}} = \frac{-1}{2}$ .

Determine the angle. …
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COMEDK 2024Set 2024-A1 markMCQ
$\alpha, \beta$
›Reveal solutionSolution

The scalar triple product $(\vec Q\times\vec R)\cdot\vec Q$ has a repeated vector, so it vanishes identically.

The vector $\vec Q\times\vec R$ is perpendicular to both $\vec Q$ and $\vec R$ . A dot product of two perpendicular vectors is zero:

$(\vec Q\times\vec R)\cdot\vec Q=0.$ …
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COMEDK 2023Set 2023-M1 markMCQ
$\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$
›Reveal solutionSolution

$\cos\theta=\dfrac{\mathbf a\cdot\mathbf b}{|\mathbf a||\mathbf b|}=\dfrac{1}{9}$ , hence $\theta=\cos^{-1}(1/9)$ .

$\mathbf a=\hat i+2\hat j+2\hat k,\ \mathbf b=\hat i+2\hat j-2\hat k$ .

$\mathbf a\cdot\mathbf b=1\cdot1+2\cdot2+2\cdot(-2)=1+4-4=1.$

$|\mathbf a|=\sqrt{1+4+4}=3,\quad |\mathbf b|=\sqrt{1+4+4}=3.$ …
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COMEDK 2022Set 20221 markMCQ
$\theta$
›Reveal solutionSolution

cos(theta) = 4 / (3 * 7) = 4/21

Concept: cos(theta) = (a . b) / (|a| |b|)

a = 2i + 2j - k , b = 6i - 3j + 2k

Dot product:

a . b = (2)(6) + (2)(-3) + (-1)(2) = 12 - 6 - 2 = 4

Magnitudes:

|a| = sqrt(4 + 4 + 1) = sqrt(9) = 3

|b| = sqrt(36 + 9 + 4) = sqrt(49) = 7 …
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KCET 2021Set A-11 markMCQ
$\left(\dfrac{\sqrt{3}}{4},\, \dfrac{1}{4},\, \dfrac{\sqrt{3}}{2}\right)$
›Reveal solutionSolution
The angle between two lines is found using the dot product of their direction vectors. The cosine of the angle is $-\frac{1}{2}$ , so the angle is $\frac{2\pi}{3}$ , which is not among the given options — but the acute angle between them is $\frac{\pi}{3}$ , option (C).

The key idea: direction cosines are the components of a unit vector along the line. So the given triples are already unit vectors. The angle between the lines is simply the angle between these two unit vectors, found via their dot product.

A common pitfall: the dot product formula gives $\cos\theta$ , and $\theta$ is taken as the acute angle between lines (since lines have no direction, the angle between them is defined as the smaller angle, between $0$ and $\frac{\pi}{2}$ ). If $\cos\theta$ turns out negative, the acute angle is $\pi - \theta$ .

Let the two direction vectors be:
$\mathbf{a} = \left(\frac{\sqrt{3}}{4},\ \frac{1}{4},\ \frac{\sqrt{3}}{2}\right), \quad \mathbf{b} = \left(\frac{\sqrt{3}}{4},\ \frac{1}{4},\ -\frac{\sqrt{3}}{2}\right)$

Verify they are unit vectors For $\mathbf{a}$ :

$|\mathbf{a}|^2 = \left(\frac{\sqrt{3}}{4}\right)^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{16} + \frac{1}{16} + \frac{3}{4} = \frac{3}{16} + \frac{1}{16} + \frac{12}{16} = \frac{16}{16} = 1$
For $\mathbf{b}$ :
$|\mathbf{b}|^2 = \frac{3}{16} + \frac{1}{16} + \frac{3}{4} = 1$
Both are indeed unit vectors, so the given triples are valid direction cosines.

Compute the dot product

$\mathbf{a} \cdot \mathbf{b} = \frac{\sqrt{3}}{4} \cdot \frac{\sqrt{3}}{4} + \frac{1}{4} \cdot \frac{1}{4} + \frac{\sqrt{3}}{2} \cdot \left(-\frac{\sqrt{3}}{2}\right)$ $= \frac{3}{16} + \frac{1}{16} - \frac{3}{4} = \frac{3}{16} + \frac{1}{16} - \frac{12}{16} = -\frac{8}{16} = -\frac{1}{2}$

Find the angle …
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KCET 2019Set A-11 markMCQ
$\frac{x-5}{2} = \frac{y+1}{-1} = \frac{z+4}{1}$
›Reveal solutionSolution

Use $\sin\theta=\dfrac{|\vec b\cdot\vec n|}{|\vec b||\vec n|}$ for a line-and-plane angle, then convert the result into the $\cos^{-1}$ form that the options are written in.

Step 1 — Extract the vectors.

The line $\dfrac{x-5}{2}=\dfrac{y+1}{-1}=\dfrac{z+4}{1}$ has direction vector
$\vec b = 2\hat i - \hat j + \hat k .$
The plane $3x-4y-z+5=0$ has normal
$\vec n = 3\hat i - 4\hat j - \hat k .$
Step 2 — Use the correct formula.

The angle $\theta$ between a line and a plane is measured from the plane, so it is the complement of the angle between $\vec b$ and $\vec n$ :
$\sin\theta = \frac{|\vec b\cdot\vec n|}{|\vec b|\,|\vec n|}.$
Step 3 — Compute.
$\vec b\cdot\vec n = (2)(3)+(-1)(-4)+(1)(-1) = 6+4-1 = 9$ $|\vec b| = \sqrt{2^2+(-1)^2+1^2} = \sqrt6,\qquad |\vec n| = \sqrt{3^2+(-4)^2+(-1)^2} = \sqrt{26}$ $\sin\theta = \frac{9}{\sqrt6\,\sqrt{26}} = \frac{9}{\sqrt{156}}$
Step 4 — Convert to the $\cos^{-1}$ form used in the options. …
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KCET 2018Set A-11 markMCQ
$\vec{a}$
›Reveal solutionSolution
Since $\vec{a}$ and $\vec{b}$ are perpendicular unit vectors, their dot products simplify to $\vec{a}\cdot\vec{a}=1$ , $\vec{b}\cdot\vec{b}=1$ , and $\vec{a}\cdot\vec{b}=0$ . Expanding the given expression yields $15 - 12 = 3$ .

The key idea here is that mutually perpendicular unit vectors give us two clean facts: each vector has length 1, and their dot product is zero. This turns what looks like a messy expansion into simple arithmetic.

When you see a dot product of two linear combinations of perpendicular unit vectors, you never need to guess coordinates — just expand using the distributive property and apply the three basic rules:

$\vec{a}\cdot\vec{a} = |\vec{a}|^2 = 1$

$\vec{b}\cdot\vec{b} = |\vec{b}|^2 = 1$

$\vec{a}\cdot\vec{b} = 0$

Let’s work through it.

Expand the dot product using the distributive law (FOIL works here):

$(3\vec{a} + 2\vec{b})\cdot(5\vec{a} - 6\vec{b}) = (3\vec{a})\cdot(5\vec{a}) + (3\vec{a})\cdot(-6\vec{b}) + (2\vec{b})\cdot(5\vec{a}) + (2\vec{b})\cdot(-6\vec{b})$

Pull out the scalar coefficients — dot product is bilinear, so constants factor out:

$= 15(\vec{a}\cdot\vec{a}) - 18(\vec{a}\cdot\vec{b}) + 10(\vec{b}\cdot\vec{a}) - 12(\vec{b}\cdot\vec{b})$

Apply the unit/perpendicular facts:

$\vec{a}\cdot\vec{a} = 1$

$\vec{b}\cdot\vec{b} = 1$

$\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a} = 0$

So the expression becomes: …
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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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