Let $\vec{p}$ and $\vec{q}$ be two unit vectors and $\alpha$ the angle between them. Then $(\vec{p}+\vec{q})$ will be a unit vector if the value of $\alpha$ is: (A) $\dfrac{\pi}{4}$ (B) $\dfrac{\pi}{3}$ (C) $\dfrac{\pi}{2}$ (D) $\dfrac{2\pi}{3}$

Let vecp and vecq be two unit vectors and α the angle between them. Then (vecp+vecq) will be a unit vector if

Question

\vec{p}

CBSECBSE Class XII Board 2025MCQ· 1mImportance★★★★★

🔒 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 — Unit Vector Dot Product Sum

Sum of Dot Products of Unit Vectors

A classic Class-12 result asks: if several unit vectors add up to the zero vector, what is the sum of their pairwise dot products? The trick is always the same — square the sum — and it turns a hard-looking problem into one line of algebra.

The Master Move: Square the Magnitude

For any vector $\vec{v}$ , $|\vec{v}|^2 = \vec{v}\cdot\vec{v}$ . Applying this to a sum expands like an ordinary algebraic square, because the dot product distributes and is commutative:

|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2\big(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}\big).

That isolates exactly the quantity we want — the sum of the pairwise dot products.

The Standard Result

Suppose $\vec{a}, \vec{b}, \vec{c}$ are unit vectors (so each squared length is $1$ ) and $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ . Then the left side is $|\vec{0}|^2=0$ , giving

0 = 1+1+1 + 2\big(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}\big).

Solving:

\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a} = -\frac{3}{2}

Why the Method Always Works

The whole technique rests on two facts: $|\vec{v}|^2=\vec{v}\cdot\vec{v}$ turns a magnitude condition into dot products, and for a unit vector $\vec{v}\cdot\vec{v}=1$ . Whatever constraint you are given (the vectors sum to zero, or to a known vector), squaring both sides produces an equation in the unknown dot-product sum.

A Variation

If instead $|\vec{a}+\vec{b}+\vec{c}| = k$ with the same three unit vectors, the same expansion gives

$\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a} = \frac{k^2-3}{2}.$ …

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

For two unit vectors, the magnitude of their sum is $|\vec{p}+\vecq}| = \sqrt{2(1+\cos\alpha)}$ . Setting this equal to 1 gives $\cos\alpha = -\frac12$ , so $\alpha = \frac{2\pi}{3}$ . The correct option is (D).

The key idea here is that the magnitude of the sum of two vectors depends on the angle between them through the dot product. When you add two vectors, the length of the result isn't just the sum of their lengths — it's governed by the parallelogram law.

For any two vectors $\vec{p}$ and $\vec{q}$ , the magnitude of their sum is:

|\vec{p}+\vec{q}|^2 = |\vec{p}|^2 + |\vec{q}|^2 + 2\vec{p}\cdot\vec{q}

Since $\vec{p}$ and $\vec{q}$ are unit vectors, $|\vec{p}| = |\vec{q}| = 1$ . And the dot product of two unit vectors is simply $\vec{p}\cdot\vec{q} = \cos\alpha$ , where $\alpha$ is the angle between them.

So the condition " $(\vec{p}+\vec{q})$ is a unit vector" means $|\vec{p}+\vec{q}| = 1$ . Let's work through it.

Write the magnitude-squared condition. We need $|\vec{p}+\vec{q}|^2 = 1^2 = 1$ . Using the formula above:

1 = 1^2 + 1^2 + 2\cos\alpha

1 = 2 + 2\cos\alpha

Solve for $\cos\alpha$ . Subtract 2 from both sides:

-1 = 2\cos\alpha

\cos\alpha = -\frac12

Find the angle $\alpha$ in the given options. The angle between two vectors is conventionally taken between $0$ and $\pi$ . $\cos\alpha = -\frac12$ gives $\alpha = \frac{2\pi}{3}$ (or $120^\circ$ ). …

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

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 18 on this concept.

CBSE 2026Set V11 mark
$[0, 3, -1, 2, -2, 1]$
›Reveal solutionSolution

Each unit vector dotted with itself is $1$ , so $\hat i\cdot\hat i+\hat j\cdot\hat j=1+1=2$ . …

Start your 14-day free trial to see the full solution →
CBSE 2025Set 65/2/11 markMCQ
$\vec{p}$
›Reveal solutionSolution
For two unit vectors, the magnitude of their sum is $|\vec{p}+\vecq}| = \sqrt{2(1+\cos\alpha)}$ . Setting this equal to 1 gives $\cos\alpha = -\frac12$ , so $\alpha = \frac{2\pi}{3}$ . The correct option is (D).

The key idea here is that the magnitude of the sum of two vectors depends on the angle between them through the dot product. When you add two vectors, the length of the result isn't just the sum of their lengths — it's governed by the parallelogram law.

For any two vectors $\vec{p}$ and $\vec{q}$ , the magnitude of their sum is:
$|\vec{p}+\vec{q}|^2 = |\vec{p}|^2 + |\vec{q}|^2 + 2\vec{p}\cdot\vec{q}$
Since $\vec{p}$ and $\vec{q}$ are unit vectors, $|\vec{p}| = |\vec{q}| = 1$ . And the dot product of two unit vectors is simply $\vec{p}\cdot\vec{q} = \cos\alpha$ , where $\alpha$ is the angle between them.

So the condition " $(\vec{p}+\vec{q})$ is a unit vector" means $|\vec{p}+\vec{q}| = 1$ . Let's work through it.

Write the magnitude-squared condition. We need $|\vec{p}+\vec{q}|^2 = 1^2 = 1$ . Using the formula above:

$1 = 1^2 + 1^2 + 2\cos\alpha$ $1 = 2 + 2\cos\alpha$

Solve for $\cos\alpha$ . Subtract 2 from both sides:

$-1 = 2\cos\alpha$ $\cos\alpha = -\frac12$

Find the angle $\alpha$ in the given options. The angle between two vectors is conventionally taken between $0$ and $\pi$ . $\cos\alpha = -\frac12$ gives $\alpha = \frac{2\pi}{3}$ (or $120^\circ$ ). …
Start your 14-day free trial to see the full solution →
CBSE 2025Set IX1 markMCQ
$\hat{i} \cdot \hat{i} - \hat{j} \cdot \hat{j} + \hat{k} \times \hat{k}$
(a) $0$
(b) $1$
(c) $2$
(d) $3$
›Reveal solutionSolution

$1-1+\vec 0=0$ ; option (a).

Concept. For unit orthogonal vectors, $\hat a\cdot\hat a=1$ (dot of a vector with itself is its squared length) and $\hat a\times\hat a=\vec 0$ (cross product of a vector with itself is zero).

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set E1 markMCQ
$\vec{i} \cdot \vec{i} + \vec{i} \cdot \vec{j} + \vec{j} \cdot \vec{j} + \vec{j} \cdot \vec{k} + \vec{k} \cdot \vec{k} =$
(a) $5$
(b) $4$
(c) $3$
(d) $2$
›Reveal solutionSolution

Same unit vectors give $1$ , perpendicular ones give $0$ ; the sum is $3$ .

For the standard orthonormal basis, $\vec{i}\cdot\vec{i}=\vec{j}\cdot\vec{j}=\vec{k}\cdot\vec{k}=1$ and any pair of different unit vectors is perpendicular, so $\vec{i}\cdot\vec{j}=\vec{j}\cdot\vec{k}=0$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set A1 mark
$(\hat{i} \times \hat{j}) \cdot \hat{k} + (\hat{j} \times \hat{k}) \cdot \hat{i}$
›Reveal solutionSolution

Use the standard cyclic cross-product identities $\hat i\times\hat j=\hat k$ , $\hat j\times\hat k=\hat i$ , then dot with the unit vectors.

Using the standard right-handed unit vector identities:

$\hat i \times \hat j = \hat k, \qquad \hat j \times \hat k = \hat i$

So:

$(\hat i\times\hat j)\cdot\hat k = \hat k\cdot\hat k = 1$ …

Start your 14-day free trial to see the full solution →
CBSE 20241 markMCQ
$\vec{a}, \vec{b}$
›Reveal solutionSolution
The key idea is to square the given vector sum condition and use the fact that each vector is a unit vector. The sum of the dot products equals $-\frac{3}{2}$ , which corresponds to option (D).

We have three unit vectors $\vec{a}, \vec{b}, \vec{c}$ — each has magnitude $1$ . They satisfy $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ . This means the three vectors form a closed triangle when placed head-to-tail. The question asks for the sum of their pairwise dot products.

The direct approach: take the dot product of the sum with itself. Since the sum is zero, its magnitude squared is zero. But the square of the sum expands into a sum of squares and cross terms — exactly what we need.

Start with the given condition.

$\vec{a} + \vec{b} + \vec{c} = \vec{0}$ .

Take the dot product of both sides with itself.

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

Expand the left-hand side.

The dot product distributes:

$\vec{a}\cdot\vec{a} + \vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} + \vec{b}\cdot\vec{a} + \vec{b}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} + \vec{c}\cdot\vec{b} + \vec{c}\cdot\vec{c} = 0.$
Since dot product is commutative ( $\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}$ ), each cross term appears twice. So:
$\vec{a}\cdot\vec{a} + \vec{b}\cdot\vec{b} + \vec{c}\cdot\vec{c} + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0.$

Use the unit vector property. For any unit vector, $\vec{a}\cdot\vec{a} = |\vec{a}|^2 = 1$ . So:
Start your 14-day free trial to see the full solution →
CBSE 2024Set D1 markMCQ
$\vec{j} \cdot \vec{j} =$
(a) $0$
(b) $1$
(c) $-1$
(d) $\vec{k}$
›Reveal solutionSolution

$\vec{j}$ is a unit vector, so $\vec{j}\cdot\vec{j} = |\vec{j}|^2 = 1$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set A1 mark
$\hat{i}\cdot(\hat{j}\times\hat{k}) + \hat{j}\cdot(\hat{i}\times\hat{k}) + \hat{k}\cdot(\hat{i}\times\hat{j})$
›Reveal solutionSolution

$\hat i\cdot(\hat j\times\hat k)+\hat j\cdot(\hat i\times\hat k)+\hat k\cdot(\hat i\times\hat j)=1$ .

Using $\hat i\times\hat j=\hat k,\ \hat j\times\hat k=\hat i,\ \hat i\times\hat k=-\hat j$ :

$\hat i\cdot(\hat j\times\hat k)=\hat i\cdot\hat i=1$

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set E1 markMCQ
$\vec{i}\cdot\vec{i} =$
(a) $0$
(b) $1$
(c) $-1$
(d) $\vec{j}$
›Reveal solutionSolution

$\vec{i}\cdot\vec{i}=1$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$\vec a, \vec b, \vec c$
(a) $\dfrac{3}{2}$
(b) $-\dfrac{3}{2}$
(c) $\dfrac{1}{2}$
(d) $-\dfrac{1}{3}$
›Reveal solutionSolution

Square the given vector equation and use $|\hat a|=|\hat b|=|\hat c|=1$ .

Given $\vec a + \vec b + \vec c = \vec 0$ , take the dot product of both sides with itself:

$(\vec a + \vec b + \vec c)\cdot(\vec a + \vec b + \vec c) = 0$

Expanding:

$|\vec a|^2 + |\vec b|^2 + |\vec c|^2 + 2(\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a) = 0$

Since $\vec a,\vec b,\vec c$ are unit vectors, $|\vec a|^2=|\vec b|^2=|\vec c|^2=1$ : …

Start your 14-day free trial to see the full solution →
CBSE 2022Set FF1 markMCQ
$\hat{i}\cdot\hat{i} - \hat{j}\cdot\hat{j} + \hat{k}\cdot\hat{k}$
(a) 0
(b) 1
(c) 2
(d) 3
›Reveal solutionSolution

$\hat i\cdot\hat i=\hat j\cdot\hat j=\hat k\cdot\hat k=1$ , so the expression $=1-1+1=1$ — option (b).

Concept. For the standard orthonormal basis, a unit vector dotted with itself equals its squared magnitude, $=1$ . The …

Start your 14-day free trial to see the full solution →
CBSE 2022Set HE2191 mark
$\hat{i}\cdot(\hat{j}\times\hat{k}) + \hat{j}\cdot(\hat{i}\times\hat{k}) + \hat{k}\cdot(\hat{i}\times\hat{j})$
›Reveal solutionSolution

Use $\hat i\times\hat j=\hat k$ , $\hat j\times\hat k=\hat i$ , $\hat i\times\hat k=-\hat j$ and evaluate each dot product.

$\hat i\cdot(\hat j\times\hat k) = \hat i\cdot\hat i = 1$

$\hat j\cdot(\hat i\times\hat k) = \hat j\cdot(-\hat j) = -1$ …

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

🎓Unlock everything free for 14 days

Unlock everything free for 14 days