Q.If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, then $(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$ is equal to :
(A) $\frac{3}{2}$
(B) $\frac{1}{2}$
(C) $-\frac{1}{2}$
(D) $-\frac{3}{2}$

Showing the 12 most recent of 18 on this concept.

1. CBSE 2026Set V11 mark
   Q.Choose from $[0, 3, -1, 2, -2, 1]$. The value of $\hat{i}\cdot\hat{i}+\hat{j}\cdot\hat{j}=$ ____.

   ›Reveal solutionSolution

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

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2. CBSE 2025Set 65/2/11 markMCQ
   Q.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) $\dfrac{2\pi}{3}$

   ›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.

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

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

   $1 = 2 + 2\cos\alpha$

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

   $-1 = 2\cos\alpha$

   $\cos\alpha = -\frac12$

   3. 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 $120^\circ$). …

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3. CBSE 2025Set IX1 markMCQ
   Q.The value of expression $\hat{i} \cdot \hat{i} - \hat{j} \cdot \hat{j} + \hat{k} \times \hat{k}$ is

   (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).

   …

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4. CBSE 2025Set E1 markMCQ
   Q.$\vec{i} \cdot \vec{i} + \vec{i} \cdot \vec{j} + \vec{j} \cdot \vec{j} + \vec{j} \cdot \vec{k} + \vec{k} \cdot \vec{k} =$

   (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 $\vec{i}\cdot\vec{j}=\vec{j}\cdot\vec{k}=0$.

   …

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5. CBSE 2025Set A1 mark
   Q.Write the value of $(\hat{i} \times \hat{j}) \cdot \hat{k} + (\hat{j} \times \hat{k}) \cdot \hat{i}$.

   ›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:

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

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6. CBSE 20241 markMCQ
   Q.If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, then $(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$ is equal to : (A) $\frac{3}{2}$ (B) $\frac{1}{2}$ (C) $-\frac{1}{2}$ (D) $-\frac{3}{2}$

   ›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.

   1. Start with the given condition.

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

   2. Take the dot product of both sides with itself.

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

   3. 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:

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

   4. Use the unit vector property. For any unit vector, $\vec{a}\cdot\vec{a} = |\vec{a}|^2 = 1$. So: $$$$

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7. CBSE 2024Set D1 markMCQ
   Q.$\vec{j} \cdot \vec{j} =$

   (a) $0$

   (b) $1$

   (c) $-1$

   (d) $\vec{k}$

   ›Reveal solutionSolution

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

   …

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8. CBSE 2024Set A1 mark
   Q.Write the value of $\hat{i}\cdot(\hat{j}\times\hat{k}) + \hat{j}\cdot(\hat{i}\times\hat{k}) + \hat{k}\cdot(\hat{i}\times\hat{j})$.

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

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

   …

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9. CBSE 2023Set E1 markMCQ
   Q.$\vec{i}\cdot\vec{i} =$

   (a) $0$

   (b) $1$

   (c) $-1$

   (d) $\vec{j}$

   ›Reveal solutionSolution

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

   …

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10. CBSE 2023Set ANNUAL1 markMCQ
    Q.If $\vec a, \vec b, \vec c$ are unit vectors such that $\vec a + \vec b + \vec c = \vec 0$, then the value of $\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c \cdot \vec a$ is

    (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, $|\vec a|^2=|\vec b|^2=|\vec c|^2=1$: …

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11. CBSE 2022Set FF1 markMCQ
    Q.The value of the expression $\hat{i}\cdot\hat{i} - \hat{j}\cdot\hat{j} + \hat{k}\cdot\hat{k}$ is:

    (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 …

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12. CBSE 2022Set HE2191 mark
    Q.Write true or false: The value of $\hat{i}\cdot(\hat{j}\times\hat{k}) + \hat{j}\cdot(\hat{i}\times\hat{k}) + \hat{k}\cdot(\hat{i}\times\hat{j})$ is 1.

    ›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.

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

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

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