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

Showing the 12 most recent of 21 on this concept.

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

   (a) $1$

   (b) $0$

   (c) $-1$

   (d) $\vec{i}$

   ›Reveal solutionSolution

   $\vec{j}\times\vec{k} = \vec{i}$, and $\vec{i}\cdot\vec{i} = 1$.

   Using the right-handed rule, $\vec{j} \times \vec{k} = \vec{i}$. Then

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

   This is the scalar triple product of the unit vectors, which equals $1$.

   ✓Final answer

   (a) $1$.

2. CBSE 2026Set A1 markMCQ
   Q.$\vec{a} \cdot (\vec{a} \times \vec{a}) =$

   (a) $1$

   (b) $0$

   (c) $\vec{a}$

   (d) $-1$

   ›Reveal solutionSolution

   $\vec{a}\times\vec{a}=\vec{0}$, hence $\vec{a}\cdot\vec{0}=0$.

   Any vector crossed with itself is the zero vector: $\vec{a}\times\vec{a} = \vec{0}$. Therefore

   $\vec{a}\cdot(\vec{a}\times\vec{a}) = \vec{a}\cdot\vec{0} = 0.$

   ✓Final answer

   (b) $0$.

3. 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

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

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

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

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

   Sum $=1-1+1=1$.

   ✓Final answer

   (c) $1$.

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

   (a) $0$

   (b) $1$

   (c) $-1$

   (d) $2\vec{i}$

   ›Reveal solutionSolution

   This is a scalar triple product; its value is $-1$.

   Using the right-hand cyclic rule, $\vec{j}\times\vec{k}=\vec{i}$, $\vec{k}\times\vec{i}=\vec{j}$, $\vec{i}\times\vec{j}=\vec{k}$. Reversing the order changes the sign, so

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

   Then

   $(\vec{k}\times\vec{j})\cdot\vec{i}=(-\vec{i})\cdot\vec{i}=-1.$

   ✓Final answer

   (C) $-1$.

5. CBSE 2025Set ANNUAL1 markMCQ
   Q.Value of $\hat{i}\cdot(\hat{k}\times\hat{j})-\hat{j}\cdot(\hat{k}\times\hat{i})+\hat{k}\cdot(\hat{i}\times\hat{j})$ is -

   (a) $1$

   (b) $0$

   (c) $-3$

   (d) $-1$

   ›Reveal solutionSolution

   Evaluate each triple-scalar-product term using the standard identities $\hat{i}\times\hat{j}=\hat{k}$, $\hat{j}\times\hat{k}=\hat{i}$, $\hat{k}\times\hat{i}=\hat{j}$.

   $\hat{k}\times\hat{j}=-(\hat{j}\times\hat{k})=-\hat{i}$, so $\hat{i}\cdot(\hat{k}\times\hat{j})=\hat{i}\cdot(-\hat{i})=-1$.

   $\hat{k}\times\hat{i}=\hat{j}$, so $\hat{j}\cdot(\hat{k}\times\hat{i})=\hat{j}\cdot\hat{j}=1$, hence $-\hat{j}\cdot(\hat{k}\times\hat{i})=-1$.

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

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

   ✓Final answer

   The correct option is (d) $-1$.

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

   (a) $1$

   (b) $0$

   (c) $-1$

   (d) $\vec{i}$

   ›Reveal solutionSolution

   This is the scalar triple product $[\vec{i}\ \vec{j}\ \vec{k}] = 1$.

   $\vec{j}\times\vec{k} = \vec{i}$, so $\vec{i}\cdot(\vec{j}\times\vec{k}) = \vec{i}\cdot\vec{i} = 1.$ (It is the volume of the unit cube.)

   ✓Final answer

   (A) $1$

7. CBSE 2023Set 65/1/11 markMCQ
   Q.The value of $(\hat{i} \times \hat{j}) \cdot \hat{j} + (\hat{j} \times \hat{i}) \cdot \hat{k}$ is: (A) 2 (B) 0 (C) 1 (D) -1

   ›Reveal solutionSolution

   We evaluate the expression by applying the properties of cross and dot products for orthonormal unit vectors. The first term $(\hat{i} \times \hat{j}) \cdot \hat{j}$ simplifies to $0$, and the second term $(\hat{j} \times \hat{i}) \cdot \hat{k}$ simplifies to $-1$. The sum is -1.

   The problem asks us to evaluate an expression involving the cross product and dot product of the standard orthonormal unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$. These vectors represent the directions along the positive x, y, and z axes, respectively, and each has a magnitude of 1.

   The cross product of two vectors results in a vector perpendicular to both original vectors. For $\hat{i}$, $\hat{j}$, $\hat{k}$, they follow a right-hand rule:

   • $\hat{i} \times \hat{j} = \hat{k}$
   • $\hat{j} \times \hat{k} = \hat{i}$
   • $\hat{k} \times \hat{i} = \hat{j}$ The cross product is anti-commutative, meaning reversing the order of the vectors changes the sign of the result: $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$. For example, $\hat{j} \times \hat{i} = -\hat{k}$.

   The dot product of two vectors results in a scalar. It measures the extent to which two vectors point in the same direction.

   • If two vectors are orthogonal (perpendicular), their dot product is 0. For example, $\hat{i} \cdot \hat{j} = 0$.
   • If two vectors are parallel, their dot product is the product of their magnitudes. For unit vectors, $\hat{a} \cdot \hat{a} = |\hat{a}|^2 = 1^2 = 1$.

   Let's apply these properties to evaluate the given expression term by term.

   The expression we need to evaluate is $(\hat{i} \times \hat{j}) \cdot \hat{j} + (\hat{j} \times \hat{i}) \cdot \hat{k}$.

   1. Evaluate the first term: $(\hat{i} \times \hat{j}) \cdot \hat{j}$

      First, we determine the cross product $\hat{i} \times \hat{j}$.

      The cross product of $\hat{i}$ and $\hat{j}$ is $\hat{k}$:

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

      Substituting this into the first term, we get:

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

      Next, we evaluate the dot product $\hat{k} \cdot \hat{j}$. Since $\hat{k}$ and $\hat{j}$ are orthogonal (perpendicular) unit vectors, their dot product is zero.

      The dot product of two orthogonal unit vectors is 0:

      $\hat{a} \cdot \hat{b} = 0 \quad \text{if } \hat{a} \perp \hat{b}$

      Therefore,

      $\hat{k} \cdot \hat{j} = 0$

      So, the first term evaluates to $0$.

      Tip

      This term is a scalar triple product $(\hat{i} \times \hat{j}) \cdot \hat{j}$. A property of the scalar triple product is that if any two vectors are identical, the value is zero. This is because the three vectors would be coplanar, and the volume of the parallelepiped they form would be zero.

   2. Evaluate the second term: $(\hat{j} \times \hat{i}) \cdot \hat{k}$

      First, we determine the cross product $\hat{j} \times \hat{i}$. The cross product is anti-commutative.

      The anti-commutativity property of the cross product states:

      $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$

      Since $\hat{i} \times \hat{j} = \hat{k}$, it follows that:

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

      Now, substitute this result back into the second term:

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

      Next, we evaluate the dot product $(-\hat{k}) \cdot \hat{k}$. We can factor out the scalar constant:

      $(-\hat{k}) \cdot \hat{k} = -(\hat{k} \cdot \hat{k})$

      The dot product of a unit vector with itself is its magnitude squared, which is 1.

      The dot product of a unit vector with itself is 1:

      $\hat{a} \cdot \hat{a} = |\hat{a}|^2 = 1^2 = 1$

      Therefore,

      $\hat{k} \cdot \hat{k} = 1$

      So, the second term evaluates to:

      $-(\hat{k} \cdot \hat{k}) = -(1) = -1$

      Watch out

      A common mistake is to assume $\hat{j} \times \hat{i} = \hat{k}$, ignoring the anti-commutativity. This would lead to $\hat{k} \cdot \hat{k} = 1$, resulting in an incorrect final answer.

   3. Sum the results of the two terms

      The total expression is the sum of the values from Step 1 and Step 2:

      $(\hat{i} \times \hat{j}) \cdot \hat{j} + (\hat{j} \times \hat{i}) \cdot \hat{k} = 0 + (-1)$

      $= -1$

   The final value of the expression is $-1$.

   ✓Final answer

   The value of the expression is $\boxed{-1}$.

8. 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 unit-vector cross products $\hat{j}\times\hat{k}=\hat{i}$, $\hat{i}\times\hat{k}=-\hat{j}$, $\hat{i}\times\hat{j}=\hat{k}$.

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

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

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

   Sum $=1-1+1=1$

   ✓Final answer

   (c) $1$.

9. CBSE 2022Set ANNUAL1 mark
   Q.$[2\hat i,\ 3\hat i,\ \hat j] =$ ____ (scalar triple product). Choices given: $[-6,\ 0,\ 5,\ 6]$

   ›Reveal solutionSolution

   A scalar triple product is zero whenever two of the three vectors are parallel (scalar multiples of each other).

   $[2\hat i,\,3\hat i,\,\hat j]=2\hat i\cdot(3\hat i\times\hat j)$.

   $3\hat i\times\hat j=3\hat k$ (since $\hat i\times\hat j=\hat k$), so the product is $2\hat i\cdot3\hat k=6(\hat i\cdot\hat k)=6\cdot0=0$.

   (Equivalently: $2\hat i$ and $3\hat i$ are parallel, so the determinant/triple product is automatically $0$.)

   ✓Final answer

   $0$.

10. CBSE 2022Set ANNUAL1 markMCQ
    Q.$(2\vec{i} + 3\vec{k}) \cdot (\vec{i} + \vec{j} + 4\vec{k}) \times (3\vec{i} + \vec{j} + 7\vec{k}) =$

    (a) $0$

    (b) $112$

    (c) $126$

    (d) $192$

    ›Reveal solutionSolution

    The scalar triple product evaluates to $0$ (the three vectors are coplanar).

    First, $(\hat i+\hat j+4\hat k)\times(3\hat i+\hat j+7\hat k)$:

    $\hat i:(1)(7)-(4)(1)=3$; $\hat j:-[(1)(7)-(4)(3)]=5$; $\hat k:(1)(1)-(1)(3)=-2$.

    So the cross product is $3\hat i+5\hat j-2\hat k$.

    Now dot with $(2\hat i+0\hat j+3\hat k)$: $(2)(3)+(0)(5)+(3)(-2)=6-6=0$.

    ✓Final answer

    (a) $0$.

11. CBSE 2022Set ANNUAL1 markMCQ
    Q.$(\vec{i} + \vec{j} + \vec{k}) \cdot (\vec{i} - \vec{j} - \vec{k}) \times (\vec{i} + 2\vec{j} - \vec{k}) =$

    (a) $0$

    (b) $2$

    (c) $4$

    (d) $6$

    ›Reveal solutionSolution

    The scalar triple product equals $6$.

    First compute $(\hat i-\hat j-\hat k)\times(\hat i+2\hat j-\hat k)$:

    $\hat i:((-1)(-1)-(-1)(2))=1+2=3$; $\hat j:-((1)(-1)-(-1)(1))=-(-1+1)=0$; $\hat k:((1)(2)-(-1)(1))=2+1=3$.

    Cross $=(3,0,3)$. Dot with $(1,1,1)$: $3+0+3=6$.

    ✓Final answer

    (d) $6$.

12. CBSE 2021Set I1 markMCQ
    Q.$\vec{j}\cdot(\vec{k}\times\vec{i}) =$

    (a) $0$

    (b) $1$

    (c) $-1$

    (d) $\vec{j}$

    ›Reveal solutionSolution

    $\vec k\times\vec i=\vec j$ and $\vec j\cdot\vec j=1$.

    Using the right-handed cyclic rule, $\vec k\times\vec i=\vec j$.

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

    Equivalently, this is the scalar triple product $[\vec j\ \vec k\ \vec i]$, a cyclic permutation of $[\vec i\ \vec j\ \vec k]=1$.

    ✓Final answer

    (b) $1$.