Q.If $\vec a=2\hat i+2\hat j+\hat k$, $|\vec b|=5$ and the angle between $\vec a$ and $\vec b$ is $\dfrac\pi6$, then the area of the triangle formed by these two vectors as two sides is

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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