Q.If $\vec a=\hat i-2\hat j+3\hat k,\ \vec b=2\hat i+\hat j-2\hat k,\ \vec c=3\hat i+2\hat j+\hat k$, find $\vec a\cdot(\vec b\times\vec c)$.

Showing the 12 most recent of 15 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.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

   (a) 0

   (b) -1

   (c) 1

   (d) 3

   ›Reveal solutionSolution

   Evaluate each cross product first using the standard identities for $\hat i,\hat j,\hat k$, then take the dot products.

   Standard identities: $\hat i\times\hat j=\hat k$, $\hat j\times\hat k=\hat i$, $\hat k\times\hat i=\hat j$ (and reversing the order flips the sign).

   Term 1: $\hat{i}\cdot(\hat{j}\times\hat{k}) = \hat i\cdot\hat i = 1$

   Term 2: $\hat{j}\cdot(\hat{i}\times\hat{k})$. Since $\hat i\times\hat k = -(\hat k\times\hat i) = -\hat j$:

   $\hat j\cdot(-\hat j) = -1$

   Term 3: $\hat{k}\cdot(\hat{i}\times\hat{j}) = \hat k\cdot\hat k = 1$

   Sum:

   $1 + (-1) + 1 = 1$

   ✓Final answer

   (c) 1

2. CBSE 2026Set ANNUAL1 markMCQ
   Q.If a vector $\vec{\alpha}$ lies in the plane $\vec{\beta}$ and $\vec{\gamma}$, then

   (a) $\left[\vec{\alpha}, \vec{\beta}, \vec{\gamma}\right]=0$

   (b) $\left[\vec{\alpha}, \vec{\beta}, \vec{\gamma}\right]=1$

   (c) $\left[\vec{\alpha}, \vec{\beta}, \vec{\gamma}\right]=2$

   (d) $\left[\vec{\alpha}, \vec{\beta}, \vec{\gamma}\right]=-1$

   ›Reveal solutionSolution

   A vector lying in the plane of two others can be written as their linear combination, making all three coplanar with zero scalar triple product.

   1. If $\vec\alpha$ lies in the plane of $\vec\beta$ and $\vec\gamma$, then $\vec\alpha=m\vec\beta+n\vec\gamma$ for some scalars $m,n$, i.e. $\vec\alpha,\vec\beta,\vec\gamma$ are linearly dependent.
   2. Geometrically, this means all three vectors lie in the same plane (are coplanar).
   3. The scalar triple product $\left[\vec\alpha,\vec\beta,\vec\gamma\right]=\vec\alpha\cdot(\vec\beta\times\vec\gamma)$ represents the (signed) volume of the parallelepiped formed by the three vectors.
   4. Coplanar vectors form a degenerate parallelepiped with zero height above the plane, so its volume — and hence the scalar triple product — is zero.

   ✓Final answer

   (a) $\left[\vec{\alpha}, \vec{\beta}, \vec{\gamma}\right]=0$

3. CBSE 2025Set ANNUAL1 markMCQ
   Q.The value of $\hat i \cdot (\hat j \times \hat k) + \hat j \cdot (\hat k \times \hat i) + \hat k \cdot (\hat i \times \hat j)$ is

   (a) 0

   (b) 1

   (c) 2

   (d) 3

   ›Reveal solutionSolution

   Each scalar triple product $\hat i\cdot(\hat j\times\hat k)$ etc. equals 1 by the right-hand rule.

   For the standard orthonormal basis $\hat i,\hat j,\hat k$:

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

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

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

   Sum $= 1+1+1 = 3$.

   ✓Final answer

   $3$ — option (d)

4. CBSE 2025Set ANNUAL1 markMCQ
   Q.The volume of the parallelepiped with its edges represented by the vectors $\hat{i}+\hat{j}$, $\hat{i}+2\hat{j}$, $\hat{i}+\hat{j}+\pi\hat{k}$ is :

   (a) $\pi$

   (b) $\dfrac{\pi}{2}$

   (c) $\dfrac{\pi}{4}$

   (d) $\dfrac{\pi}{3}$

   ›Reveal solutionSolution

   The volume of a parallelepiped is the absolute value of the scalar triple product of its edge vectors, computed as a $3\times3$ determinant.

   1. Edge vectors: $\vec a=\hat i+\hat j=(1,1,0)$, $\vec b=\hat i+2\hat j=(1,2,0)$, $\vec c=\hat i+\hat j+\pi\hat k=(1,1,\pi)$.
   2. Volume $=|[\vec a,\vec b,\vec c]|=\left|\det\begin{bmatrix}1&1&0\\1&2&0\\1&1&\pi\end{bmatrix}\right|$.
   3. Expand along the first row: $1\cdot(2\cdot\pi-0\cdot1)-1\cdot(1\cdot\pi-0\cdot1)+0\cdot(1\cdot1-2\cdot1)$.
   4. $=1(2\pi)-1(\pi)+0=2\pi-\pi=\pi$.
   5. Since $\pi>0$, the volume $=|\pi|=\pi$.

   ✓Final answer

   (a) $\pi$

5. CBSE 2024Set ANNUAL1 markMCQ
   Q.The value of $(\hat{i} \times \hat{j}) \cdot \hat{k} + (\hat{j} \times \hat{k}) \cdot \hat{i}$ is

   (a) 1

   (b) 2

   (c) 0

   (d) -1

   ›Reveal solutionSolution

   Use the cyclic cross-product rules of the unit vectors, then dot.

   Recall the standard cross products:

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

   Therefore

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

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

   Adding,

   $(\hat i\times\hat j)\cdot\hat k + (\hat j\times\hat k)\cdot\hat i = 1 + 1 = 2.$

   ✓Final answer

   (b) $2$

6. CBSE 2024Set ANNUAL1 markMCQ
   Q.If a vector $\vec{\alpha}$ lies in the plane of $\vec{\beta}$ and $\vec{\gamma}$, then :

   (a) $\left[\vec{\alpha}, \vec{\beta}, \vec{\gamma}\right]=0$

   (b) $\left[\vec{\alpha}, \vec{\beta}, \vec{\gamma}\right]=1$

   (c) $\left[\vec{\alpha}, \vec{\beta}, \vec{\gamma}\right]=2$

   (d) $\left[\vec{\alpha}, \vec{\beta}, \vec{\gamma}\right]=-1$

   ›Reveal solutionSolution

   A vector lying in the plane spanned by two others is a linear combination of them, making the three coplanar and their scalar triple product zero.

   1. If $\vec\alpha$ lies in the plane of $\vec\beta$ and $\vec\gamma$, then $\vec\alpha=m\vec\beta+n\vec\gamma$ for some scalars $m,n$ — i.e. $\vec\alpha,\vec\beta,\vec\gamma$ are linearly dependent (coplanar).
   2. The scalar triple product $[\vec\alpha,\vec\beta,\vec\gamma]$ measures the volume of the parallelepiped formed by the three vectors; coplanar vectors form a degenerate (flat) parallelepiped of zero volume.
   3. Hence $[\vec\alpha,\vec\beta,\vec\gamma]=0$.

   ✓Final answer

   (a) $\left[\vec{\alpha}, \vec{\beta}, \vec{\gamma}\right]=0$

7. CBSE 2024Set ANNUAL1 markMCQ
   Q.The value of $[\hat{i}+\hat{j}\ \ 2\hat{j}\ \ 3\hat{k}]$ is ................. .

   (a) 0

   (b) 6

   (c) 3

   (d) 5

   ›Reveal solutionSolution

   The scalar triple product $[\bar a\ \bar b\ \bar c]$ equals the determinant of the vectors written as rows.

   The three vectors are $\bar a = \hat i+\hat j = (1,1,0)$, $\bar b = 2\hat j = (0,2,0)$, $\bar c = 3\hat k = (0,0,3)$.

   $[\bar a\ \bar b\ \bar c] = \begin{vmatrix}1 & 1 & 0\\ 0 & 2 & 0\\ 0 & 0 & 3\end{vmatrix}$

   Since this is an upper-triangular determinant, its value is the product of the diagonal entries:

   $1 \times 2 \times 3 = 6$

   ✓Final answer

   $[\hat{i}+\hat{j}\ \ 2\hat{j}\ \ 3\hat{k}] = 6$ (option b).

8. CBSE 2024Set ANNUAL1 markMCQ
   Q.The value of [î ĵ k̂] is -

   (a) 0

   (b) 1

   (c) 2

   (d) 3

   ›Reveal solutionSolution

   The scalar triple product of the unit vectors along the axes equals 1, since they form a right-handed orthonormal system.

   $[\hat{i}\ \hat{j}\ \hat{k}] = \hat{i}\cdot(\hat{j}\times\hat{k})$

   Since $\hat{j}\times\hat{k} = \hat{i}$:

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

   ✓Final answer

   $[\hat{i}\ \hat{j}\ \hat{k}] = 1$ — option (b).

9. CBSE 2024Set ANNUAL1 markMCQ
   Q.If $\vec a=\hat i+\hat j+\hat k$, $\vec b=2\hat i+x\hat j+\hat k$, $\vec c=\hat i-\hat j+4\hat k$ and $\vec a\cdot(\vec b\times\vec c)=70$ then $x$ is equal to:

   (a) $26$

   (b) $5$

   (c) $10$

   (d) $7$

   ›Reveal solutionSolution

   Solving the determinant equation for the scalar triple product gives $x=26$.

   $\vec a\cdot(\vec b\times\vec c)=\begin{vmatrix}1&1&1\\2&x&1\\1&-1&4\end{vmatrix}$

   Expanding along the first row:

   $=1(4x-(-1))-1(8-1)+1(-2-x)=(4x+1)-7+(-2-x)=3x-8.$

   Setting this equal to $70$:

   $3x-8=70\ \Rightarrow\ 3x=78\ \Rightarrow\ x=26.$

   ✓Final answer

   $x=26$ — option (a).

10. CBSE 2023Set ANNUAL1 markMCQ
    Q.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

    (a) 0

    (b) -1

    (c) 1

    (d) 3

    ›Reveal solutionSolution

    Use the standard cross products $\hat j\times\hat k=\hat i$, $\hat i\times\hat k=-\hat j$, $\hat i\times\hat j=\hat k$, then dot each.

    Evaluate each term separately:

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

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

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

    Adding:

    $1+(-1)+1=1.$

    ✓Final answer

    (c) $1$.

11. CBSE 2023Set ANNUAL1 markMCQ
    Q.If $\vec{a}$ and $\vec{b}$ are parallel vectors then $\left[\vec{a}, \vec{c}, \vec{b}\right]$ is equal to :

    (a) $1$

    (b) $2$

    (c) $0$

    (d) $-1$

    ›Reveal solutionSolution

    Parallel vectors are linearly dependent, so any triple containing both is automatically coplanar and its scalar triple product vanishes.

    1. The scalar triple product $[\vec a,\vec c,\vec b]=\vec a\cdot(\vec c\times\vec b)$ represents (up to sign) the volume of the parallelepiped formed by $\vec a,\vec b,\vec c$.
    2. If $\vec a$ and $\vec b$ are parallel, $\vec b=\lambda\vec a$ for some scalar $\lambda$, so $\vec a,\vec b,\vec c$ are linearly dependent (they lie in, at most, a plane).
    3. Three coplanar (linearly dependent) vectors always have zero scalar triple product, since the "parallelepiped" they form is flat (zero volume).
    4. Hence $[\vec a,\vec c,\vec b]=0$.

    ✓Final answer

    (c) $0$

12. CBSE 2022Set ANNUAL1 markMCQ
    Q.If the vectors $2\hat{i}-\hat{j}+3\hat{k}$, $3\hat{i}+2\hat{j}+\hat{k}$, $\hat{i}+m\hat{j}+4\hat{k}$ are coplanar, then the value of m is :

    (a) $2$

    (b) $3$

    (c) $-2$

    (d) $-3$

    ›Reveal solutionSolution

    Coplanarity requires the scalar triple product (determinant of the three vectors) to be zero, which solves to $m=-3$.

    1. The vectors are $\vec a=2\hat i-\hat j+3\hat k$, $\vec b=3\hat i+2\hat j+\hat k$, $\vec c=\hat i+m\hat j+4\hat k$.
    2. Three vectors are coplanar if and only if their scalar triple product $[\vec a\ \vec b\ \vec c]=0$, i.e. $\begin{vmatrix}2&-1&3\\3&2&1\\1&m&4\end{vmatrix}=0$.
    3. Expanding along the first row: $2(2\cdot4-1\cdot m)-(-1)(3\cdot4-1\cdot1)+3(3m-2\cdot1)$.
    4. This is $2(8-m)+1(12-1)+3(3m-2)=(16-2m)+11+(9m-6)$.
    5. Simplifying: $16-2m+11+9m-6=21+7m$.
    6. Setting the determinant to zero: $21+7m=0$, so $7m=-21$, giving $m=-3$.

    ✓Final answer

    $m=-3$ — option (d).