Q.Prove that for any three vectors $\bar{a}, \bar{b}, \bar{c}$, $[\bar{b}+\bar{c}\ \ \bar{c}+\bar{a}\ \ \bar{a}+\bar{b}] = 2[\bar{a}\ \bar{b}\ \bar{c}]$.

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

   …

   🔒 Start your 14-day free trial to see the full solution →
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). …

   🔒 Start your 14-day free trial to see the full solution →
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 $\hat i,\hat j,\hat k$:

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

   🔒 Start your 14-day free trial to see the full solution →
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 $=|[\vec a,\vec b,\vec c]|=\left|\det\begin{bmatrix}1&1&0\\1&2&0\\1&1&\pi\end{bmatrix}\right|$. …

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

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

   🔒 Start your 14-day free trial to see the full solution →
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). …

   🔒 Start your 14-day free trial to see the full solution →
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 $\bar a = \hat i+\hat j = (1,1,0)$, $\bar b = 2\hat j = (0,2,0)$, $\bar c = 3\hat k = (0,0,3)$.

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

   …

   🔒 Start your 14-day free trial to see the full solution →
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 $\hat{j}\times\hat{k} = \hat{i}$:

   …

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

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

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

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

    🔒 Start your 14-day free trial to see the full solution →
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 $\vec a,\vec b,\vec c$. …

    🔒 Start your 14-day free trial to see the full solution →
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. $\begin{vmatrix}2&-1&3\\3&2&1\\1&m&4\end{vmatrix}=0$. …

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