Q.If $\vec a,\vec b,\vec c$ are position vectors of the vertices $A,B,C$ of a triangle $ABC$, show that the area of the triangle $ABC$ is $\dfrac12|\vec a\times\vec b+\vec b\times\vec c+\vec c\times\vec a|$. Also deduce the condition for collinearity of the points $A,B,C$.

Showing the 12 most recent of 20 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.$\vec{i} \times \vec{k} =$

   (a) $1$

   (b) $\vec{k}$

   (c) $\vec{j}$

   (d) $-\vec{j}$

   ›Reveal solutionSolution

   $\vec{i}\times\vec{k} = -\vec{j}$.

   The cyclic order gives $\vec{i}\times\vec{j}=\vec{k}$, $\vec{j}\times\vec{k}=\vec{i}$, $\vec{k}\times\vec{i}=\vec{j}$. Reversing the last one: …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Find the value of $(2\hat{i}+3\hat{j}) \times (\hat{i}+2\hat{j})$

   (a) $\hat{i}$

   (b) $\hat{j}$

   (c) $\hat{k}$

   (d) None of these

   ›Reveal solutionSolution

   Compute the cross product using the determinant formula; since both vectors have zero $\hat k$-component, only the $\hat k$ term survives.

   $(2\hat i+3\hat j)\times(\hat i+2\hat j) = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 3 & 0 \\ 1 & 2 & 0 \end{vmatrix}$

   …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $|\vec{a}| = 13$, $|\vec{b}| = 5$ and $\vec{a}\cdot\vec{b} = 60^\circ$, then $|\vec{a}\times\vec{b}|$ is:

   (a) 45

   (b) 15

   (c) 25

   (d) 35

   ›Reveal solutionSolution

   With $\vec a\cdot\vec b=60$, $|\vec a|=13$, $|\vec b|=5$, we get $\cos\theta=60/65=12/13$, so $\sin\theta=5/13$, giving $|\vec a\times\vec b|=65\times5/13=25$.

   Given $|\vec a|=13$, $|\vec b|=5$, and $\vec a\cdot\vec b=60$ (this is read as the dot-product value between the two vectors, since a dot product cannot equal an angle in degrees).

   Since $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$: $60=13\times5\times\cos\theta=65\cos\theta \Rightarrow \cos\theta=\dfrac{60}{65}=\dfrac{12}{13}$.

   …

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4. CBSE 2025Set E1 markMCQ
   Q.$(\vec{i} \times \vec{j}) + (\vec{i} \times \vec{i}) =$

   (a) $2$

   (b) $1$

   (c) $\vec{k}$

   (d) $-\vec{k}$

   ›Reveal solutionSolution

   The cross product of a vector with itself is zero, so the result is $\vec{k}$.

   By the cyclic rule $\vec{i}\times\vec{j}=\vec{k}$, and any vector crossed with itself is the zero vector, so $\vec{i}\times\vec{i}=\vec{0}$.

   …

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5. CBSE 2025Set ANNUAL1 markMCQ
   Q.If |a⃗|=4, |b⃗|=2√3, |a⃗×b⃗|=12 then the angle between the vectors a⃗ and b⃗ is

   (a) π/3

   (b) π/6

   (c) π/4

   (d) π/2

   ›Reveal solutionSolution

   Use $|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta$ to solve for $\theta$.

   The magnitude of the cross product satisfies $|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta$, where $\theta$ is the angle between the vectors.

   $12 = (4)(2\sqrt3)\sin\theta = 8\sqrt3\,\sin\theta$

   $\sin\theta = \frac{12}{8\sqrt3} = \frac{3}{2\sqrt3} = \frac{\sqrt3}{2}$

   …

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6. CBSE 2024Set D1 markMCQ
   Q.$(\vec{i} + 3\vec{j} - 2\vec{k}) \times (-\vec{i} + 3\vec{k}) =$

   (a) $9\vec{i} - \vec{j} + 3\vec{k}$

   (b) $9\vec{i} + \vec{j} - 3\vec{k}$

   (c) $\vec{i} - \vec{j} + 3\vec{k}$

   (d) $\vec{i} + \vec{j} - 3\vec{k}$

   ›Reveal solutionSolution

   Evaluate the determinant form of the cross product.

   With $\vec{a} = (1, 3, -2)$ and $\vec{b} = (-1, 0, 3)$:

   $\vec{a}\times\vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 3 & -2 \\ -1 & 0 & 3 \end{vmatrix}.$ …

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

   (a) $-\vec{j}$

   (b) $\vec{j}$

   (c) $0$

   (d) $\vec{k}$

   ›Reveal solutionSolution

   By the right-hand rule $\vec{k}\times\vec{j} = -\vec{i}$; none of the four printed options matches this, so the question as printed is defective.

   Using the cyclic rule $\vec{i}\times\vec{j} = \vec{k}$, $\vec{j}\times\vec{k} = \vec{i}$, $\vec{k}\times\vec{i} = \vec{j}$, and the anti-commutativity $\vec{b}\times\vec{a} = -(\vec{a}\times\vec{b})$:

   $\vec{k}\times\vec{j} = -(\vec{j}\times\vec{k}) = -\vec{i}.$ …

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8. CBSE 2024Set ANNUAL1 markMCQ
   Q.$\hat{k} \times \hat{j} =$

   (a) $\hat{i}$

   (b) $-\hat{i}$

   (c) $\hat{j}$

   (d) $0$

   ›Reveal solutionSolution

   The standard cyclic products are î×ĵ=k̂, ĵ×k̂=î, k̂×î=ĵ; reversing any pair flips the sign.

   …

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

   (a) $\vec{k}$

   (b) $-\vec{k}$

   (c) $\vec{0}$

   (d) $1$

   ›Reveal solutionSolution

   $\vec{j}\times\vec{i}=-\vec{k}$.

   The cyclic rule gives $\vec{i}\times\vec{j}=\vec{k}$. Reversing the order changes the sign:

   …

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10. CBSE 2023Set E1 markMCQ
    Q.$(3\vec{k}-7\vec{i})\times 2\vec{k} =$

    (a) $-14\vec{j}$

    (b) $14\vec{j}$

    (c) $11\vec{i}-2\vec{k}$

    (d) $2\vec{k}-11\vec{i}$

    ›Reveal solutionSolution

    Using $\vec{k}\times\vec{k}=0$ and $\vec{i}\times\vec{k}=-\vec{j}$, the product is $14\vec{j}$.

    Distribute the cross product:

    $(3\vec{k}-7\vec{i})\times 2\vec{k}=6(\vec{k}\times\vec{k})-14(\vec{i}\times\vec{k}).$

    …

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11. CBSE 2023Set E1 markMCQ
    Q.$(10\vec{i}+\vec{j}+\vec{k})\times(-4\vec{i}+7\vec{j}-11\vec{k}) =$

    (a) $-18\vec{i}+106\vec{j}+74\vec{k}$

    (b) $18\vec{i}-106\vec{j}-74\vec{k}$

    (c) $18\vec{i}+106\vec{j}+74\vec{k}$

    (d) $5\vec{i}-6\vec{j}-7\vec{k}$

    ›Reveal solutionSolution

    Evaluating the determinant gives $-18\vec{i}+106\vec{j}+74\vec{k}$.

    Compute

    $(10\vec{i}+\vec{j}+\vec{k})\times(-4\vec{i}+7\vec{j}-11\vec{k})=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\10&1&1\\-4&7&-11\end{vmatrix}.$

    $\vec{i}$-component: $(1)(-11)-(1)(7)=-11-7=-18$.

    $\vec{j}$-component: $-\big[(10)(-11)-(1)(-4)\big]=-[-110+4]=106$.

    …

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12. CBSE 2023Set ANNUAL1 markMCQ
    Q.$\hat{i} \times \hat{j} =$

    (a) 0

    (b) $\hat{k}$

    (c) $-\hat{k}$

    (d) none of these

    ›Reveal solutionSolution

    This is one of the fundamental unit-vector cross-product identities.

    …

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