Q.Find a unit vector perpendicular to each of the vectors $\vec{a} = 5\hat{i} + 6\hat{j} - 2\hat{k}$ and $\vec{b} = 7\hat{i} + 6\hat{j} + 2\hat{k}$.

Showing the 12 most recent of 25 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

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

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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 …

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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

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

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

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

   …

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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 …

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5. CBSE 2025Set ANNUAL1 markMCQ
   Q.Which vector is normal to both $\hat i+\hat k$ and $\hat i+\hat j$?

   (i) $\hat i-\hat j+\hat k$

   (ii) $-\hat i+\hat j-\hat k$

   (iii) $\hat i+\hat j+\hat k$

   (iv) $\hat i-\hat j-\hat k$

   ›Reveal solutionSolution

   A vector normal to both given vectors is (a scalar multiple of) their cross product.

   Let $\vec p = \hat i+\hat k=(1,0,1)$ and $\vec q=\hat i+\hat j=(1,1,0)$.

   $\vec p \times \vec q = \begin{vmatrix}\hat i & \hat j & \hat k \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{vmatrix} = \hat i(0\cdot0-1\cdot1) - \hat j(1\cdot0-1\cdot1) + \hat k(1\cdot1-0\cdot1)$

   $= -\hat i + \hat j + \hat k$

   Any nonzero scalar multiple of this is also normal to both vectors, including its negative: …

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

   …

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7. CBSE 2024Set 65/2/11 markMCQ
   Q.The unit vector perpendicular to both vectors $\hat{i} + \hat{k}$ and $\hat{i} - \hat{k}$ is: (A) $2\hat{j}$ (B) $\hat{j}$ (C) $\dfrac{\hat{i} - \hat{k}}{\sqrt{2}}$ (D) $\dfrac{\hat{i} + \hat{k}}{\sqrt{2}}$

   ›Reveal solutionSolution

   To find a vector perpendicular to two given vectors, we use their cross product. Normalizing this resulting vector gives the unit vector. The unit vector perpendicular to $\hat{i} + \hat{k}$ and $\hat{i} - \hat{k}$ is $\boxed{\hat{j}}$.

   Concept and Intuition

   When you're asked to find a vector that is perpendicular to two other vectors simultaneously, the most direct and fundamental tool in vector algebra is the cross product.

   Imagine two non-parallel vectors originating from the same point. They define a unique plane in space. The cross product of these two vectors yields a new vector that is perpendicular to this entire plane. This means the resulting vector is perpendicular to both of the original vectors.

   The cross product of two vectors $\vec{A}$ and $\vec{B}$ is given by $\vec{A} \times \vec{B} = (A_y B_z - A_z B_y)\hat{i} + (A_z B_x - A_x B_z)\hat{j} + (A_x B_y - A_y B_x)\hat{k}$.

   A more convenient way to compute this is using a determinant:

   $\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}$

   Once we have a vector that is perpendicular to both, the problem asks for a unit vector. A unit vector is simply a vector with a magnitude of 1, pointing in the same direction as the original vector. To convert any non-zero vector $\vec{V}$ into a unit vector $\hat{V}$, we divide it by its own magnitude: $\hat{V} = \frac{\vec{V}}{|\vec{V}|}$.

   Step-by-step Solution

   1. Identify the given vectors.

      Let the two given vectors be $\vec{A}$ and $\vec{B}$.

      $\vec{A} = \hat{i} + \hat{k}$

      $\vec{B} = \hat{i} - \hat{k}$

      We can write these in component form as:

      $\vec{A} = 1\hat{i} + 0\hat{j} + 1\hat{k}$

      $\vec{B} = 1\hat{i} + 0\hat{j} - 1\hat{k}$

   2. Calculate the cross product $\vec{A} \times \vec{B}$.

      This will give us a vector perpendicular to both $\vec{A}$ and $\vec{B}$.

   $\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 1 \\ 1 & 0 & -1 \end{vmatrix}$

   Expand the determinant:
   $= \hat{i}((0)(-1) - (1)(0)) - \hat{j}((1)(-1) - (1)(1)) + \hat{k}((1)(0) - (0)(1))$
   $= \hat{i}(0 - 0) - \hat{j}(-1 - 1) + \hat{k}(0 - 0)$
   $= 0\hat{i} - \hat{j}(-2) + 0\hat{k}$
   $= 2\hat{j}$
   
   Let's call this resulting vector $\vec{P} = 2\hat{j}$. This vector $\vec{P}$ is perpendicular to both $\hat{i} + \hat{k}$ and $\hat{i} - \hat{k}$.
   
   > [!TIP]
   > You can quickly verify perpendicularity by checking the dot product. If $\vec{P} \cdot \vec{A} = 0$ and $\vec{P} \cdot \vec{B} = 0$, then $\vec{P}$ is indeed perpendicular to both.
   > $\vec{P} \cdot \vec{A} = (0\hat{i} + 2\hat{j} + 0\hat{k}) \cdot (1\hat{i} + 0\hat{j} + 1\hat{k}) = (0)(1) + (2)(0) + (0)(1) = 0$. …
   

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8. 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 $[\vec{i}\ \vec{j}\ \vec{k}] = 1$.

   …

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9. CBSE 2024Set ANNUAL1 mark
   Q.Write the unit vector which is perpendicular to both $\hat{j}-\hat{k}$ and $\hat{i}+\hat{j}$.

   ›Reveal solutionSolution

   A vector perpendicular to both given vectors is found via their cross product; normalizing it gives the unit vector.

   Let $\vec a = \hat j - \hat k = (0,1,-1)$ and $\vec b = \hat i + \hat j = (1,1,0)$.

   A vector perpendicular to both is $\vec a \times \vec b$:

   $\vec a\times\vec b = \begin{vmatrix}\hat i & \hat j & \hat k\\ 0 & 1 & -1\\ 1 & 1 & 0\end{vmatrix} = \hat i(1\cdot0-(-1)\cdot1) - \hat j(0\cdot0-(-1)\cdot1) + \hat k(0\cdot1-1\cdot1)$

   $= \hat i(1) - \hat j(1) + \hat k(-1) = \hat i - \hat j - \hat k$

   …

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10. 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: …

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11. 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 $\hat{j}\times\hat{k}=\hat{i}$, $\hat{i}\times\hat{k}=-\hat{j}$, $\hat{i}\times\hat{j}=\hat{k}$.

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

    …

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12. 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).

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

    …

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