Q.Find $|\vec{a}\times\vec{b}|$, if $\vec{a}=2\hat{i}+\hat{j}+3\hat{k}$ and $\vec{b}=3\hat{i}+5\hat{j}-2\hat{k}$.

Showing the 12 most recent of 15 on this concept.

1. CBSE 20261 markMCQ
   Q.If $|\vec{a}| = 8$, $|\vec{b}| = 3$ and $|\vec{a} \times \vec{b}| = 12$, then the value of $|\vec{a} \cdot \vec{b}|$ is (A) $6\sqrt{3}$ (B) $8\sqrt{3}$ (C) $12\sqrt{3}$ (D) $3\sqrt{12}$

   ›Reveal solutionSolution

   The cross product magnitude gives $\sin\theta$, and the dot product magnitude uses $\cos\theta$. Using $|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}||\sin\theta|$ and $|\vec{a}\cdot\vec{b}| = |\vec{a}||\vec{b}||\cos\theta|$, we find $|\vec{a}\cdot\vec{b}| = 12\sqrt{3}$.

   The key here is the relationship between the dot product, the cross product, and the angle between two vectors. Both products depend on the magnitudes of the vectors and the sine or cosine of the angle between them.

   Given $|\vec{a}| = 8$, $|\vec{b}| = 3$, and $|\vec{a} \times \vec{b}| = 12$, we can find $\sin\theta$ first, then $\cos\theta$, and finally the dot product magnitude.

   1. Use the cross product formula. The magnitude of the cross product is

   $|\vec{a} \times \vec{b}| = |\vec{a}|\,|\vec{b}|\,|\sin\theta|.$

   Substituting the given values:

   $12 = 8 \cdot 3 \cdot |\sin\theta| = 24\,|\sin\theta|.$

   So

   $|\sin\theta| = \frac{12}{24} = \frac{1}{2}.$

   2. Find $|\cos\theta|$ using the identity. We know $\sin^2\theta + \cos^2\theta = 1$. Therefore

   $|\cos\theta| = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}.$

   Note: we take the absolute value because the dot product magnitude uses $|\cos\theta|$, not the signed value.

   3. Compute the dot product magnitude.

   $|\vec{a} \cdot \vec{b}| = |\vec{a}|\,|\vec{b}|\,|\cos\theta| = 8 \cdot 3 \cdot \frac{\sqrt{3}}{2} = 24 \cdot \frac{\sqrt{3}}{2} = 12\sqrt{3}.$

   Watch out

   A common mistake is to forget the absolute value on $\cos\theta$ and assume $\theta$ is acute. The problem asks for $|\vec{a} \cdot \vec{b}|$, so we only need the magnitude — the sign of $\cos\theta$ doesn't matter here.

   Tip

   You can also solve this directly using the identity $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$, which avoids finding $\theta$ explicitly. Plug in: $12^2 + (\vec{a} \cdot \vec{b})^2 = 8^2 \cdot 3^2$, so $144 + (\vec{a} \cdot \vec{b})^2 = 576$, giving $(\vec{a} \cdot \vec{b})^2 = 432$, and $|\vec{a} \cdot \vec{b}| = \sqrt{432} = 12\sqrt{3}$.

   ✓Final answer

   The value is $\boxed{12\sqrt{3}}$, which corresponds to option (C).

2. CBSE 2026Set A1 markMCQ
   Q.If $\vec{a} = \vec{i} - \vec{j} + 2\vec{k}$ and $\vec{b} = 2\vec{i} + 3\vec{j} - 4\vec{k}$ then $|\vec{a} \times \vec{b}| =$

   (a) $\sqrt{174}$

   (b) $\sqrt{87}$

   (c) $\sqrt{93}$

   (d) none of these

   ›Reveal solutionSolution

   Compute the cross product, then its magnitude.

   With $\vec{a} = \vec{i}-\vec{j}+2\vec{k}$, $\vec{b} = 2\vec{i}+3\vec{j}-4\vec{k}$:

   $\vec{a}\times\vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & -1 & 2 \\ 2 & 3 & -4 \end{vmatrix}.$

   • $\vec{i}$: $(-1)(-4) - (2)(3) = 4 - 6 = -2$.
   • $\vec{j}$: $-[(1)(-4) - (2)(2)] = -[-4-4] = 8$.
   • $\vec{k}$: $(1)(3) - (-1)(2) = 3 + 2 = 5$.

   So $\vec{a}\times\vec{b} = -2\vec{i}+8\vec{j}+5\vec{k}$ and

   $|\vec{a}\times\vec{b}| = \sqrt{(-2)^2 + 8^2 + 5^2} = \sqrt{4 + 64 + 25} = \sqrt{93}.$

   ✓Final answer

   (c) $\sqrt{93}$.

3. CBSE 2025Set ANNUAL1 mark
   Q.If vector a = 2i - 3j + k and vector a = 2i - 3j + k, then find vector a x vector b.

   ›Reveal solutionSolution

   The stem names both vectors "vector a" with the same components $2\hat i - 3\hat j + \hat k$, so this is $\vec a \times \vec a$, which is always the zero vector.

   Taking the stem literally, both vectors are $2\hat i - 3\hat j + \hat k$. Using the determinant method for a cross product:

   $\vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & -3 & 1 \\ 2 & -3 & 1 \end{vmatrix}$

   $= \hat i[(-3)(1)-(1)(-3)] - \hat j[(2)(1)-(1)(2)] + \hat k[(2)(-3)-(-3)(2)]$

   $= \hat i(-3+3) - \hat j(2-2) + \hat k(-6+6) = \vec 0$

   This matches the general rule: whenever the two rows of the cross-product determinant are identical, the determinant (and hence the cross product) is zero — a vector is always parallel to itself, and the cross product of parallel (or equal) vectors vanishes.

   ✓Final answer

   $\vec a \times \vec b = \vec 0$, the zero vector.

4. CBSE 2025Set ANNUAL1 markMCQ
   Q.The vectors $\vec a$ and $\vec b$ are such that $|\vec a| = 3$ and $|\vec b| = \dfrac{2}{3}$. Then $\vec a \times \vec b$ is a unit vector if the angle between $\vec a$ and $\vec b$ is

   (a) $\dfrac{\pi}{6}$

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

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

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

   ›Reveal solutionSolution

   Use $|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta$ and set it equal to 1 (unit vector).

   Given $|\vec a|=3$, $|\vec b|=\dfrac{2}{3}$. For $\vec a\times\vec b$ to be a unit vector, $|\vec a\times\vec b|=1$.

   $|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta$

   $1 = 3\times\dfrac{2}{3}\times\sin\theta = 2\sin\theta$

   $\sin\theta = \dfrac{1}{2}$

   $\theta = \dfrac{\pi}{6}$

   ✓Final answer

   $\theta = \dfrac{\pi}{6}$ — option (a)

5. CBSE 2025Set ANNUAL1 mark
   Q.Find the magnitude of a, where a = (î + 3ĵ − 2k̂) × (−î + 3k̂).

   ›Reveal solutionSolution

   Compute the cross product of the two given vectors using the determinant method, then find its magnitude.

   Given: $\vec a = (\hat i + 3\hat j - 2\hat k) \times (-\hat i + 3\hat k)$

   Step 1 — set up the determinant:

   $\vec a = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 3 & -2 \\ -1 & 0 & 3 \end{vmatrix}$

   Step 2 — expand along the first row:

   $\hat i(3\cdot3 - (-2)\cdot0) - \hat j(1\cdot3 - (-2)(-1)) + \hat k(1\cdot0 - 3\cdot(-1))$

   $= \hat i(9-0) - \hat j(3-2) + \hat k(0+3) = 9\hat i - \hat j + 3\hat k$

   Step 3 — magnitude:

   $|\vec a| = \sqrt{9^2 + (-1)^2 + 3^2} = \sqrt{81+1+9} = \sqrt{91}$

   ✓Final answer

   $|\vec a| = \sqrt{91}$

6. CBSE 20241 markMCQ
   Q.Let the vectors $\vec{a}$ and $\vec{b}$ be such that $|\vec{a}| = \sqrt{3}$ and $|\vec{b}| = \frac{2}{\sqrt{3}}$, then $\vec{a} \times \vec{b}$ is a unit vector, if the angle between $\vec{a}$ and $\vec{b}$ is: (A) $\frac{\pi}{3}$ (B) $\frac{\pi}{4}$ (C) $\frac{\pi}{6}$ (D) $\frac{\pi}{2}$

   ›Reveal solutionSolution

   $|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta = 2\sin\theta$. Setting this equal to $1$ gives $\sin\theta = \tfrac{1}{2}$, so $\theta = \tfrac{\pi}{6}$ — option (C).

   The magnitude of a cross product is

   $|\vec{a}\times\vec{b}| = |\vec{a}|\,|\vec{b}|\sin\theta.$

   With $|\vec{a}| = \sqrt{3}$ and $|\vec{b}| = \tfrac{2}{\sqrt{3}}$,

   $|\vec{a}||\vec{b}| = \sqrt{3}\cdot\frac{2}{\sqrt{3}} = 2,$

   so $|\vec{a}\times\vec{b}| = 2\sin\theta$.

   Since $\vec{a}\times\vec{b}$ is a unit vector, $|\vec{a}\times\vec{b}| = 1$:

   $2\sin\theta = 1 \ \Rightarrow\ \sin\theta = \frac{1}{2} \ \Rightarrow\ \theta = \frac{\pi}{6}.$

   ✓Final answer

   The angle between $\vec{a}$ and $\vec{b}$ is $\dfrac{\pi}{6}$ — option (C).

7. CBSE 2024Set ANNUAL1 markMCQ
   Q.Let the vectors $\vec{a}$ and $\vec{b}$ be such that $|\vec{a}| = 3$ and $|\vec{b}| = \dfrac{\sqrt{2}}{3}$, then $\vec{a} \times \vec{b}$ is a unit vector, if angle between $\vec{a}$ and $\vec{b}$ is:

   (a) $\frac{\pi}{6}$

   (b) $\frac{\pi}{4}$

   (c) $\frac{\pi}{3}$

   (d) $\frac{\pi}{2}$

   ›Reveal solutionSolution

   $\theta = \dfrac{\pi}{4}$ — option (b).

   $|\vec a|=3$, $|\vec b|=\dfrac{\sqrt2}{3}$, and $\vec a\times\vec b$ is a unit vector, so $|\vec a\times\vec b|=1$.

   $|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta = 3\cdot\frac{\sqrt2}{3}\sin\theta = \sqrt2\sin\theta$

   Setting this equal to $1$:

   $\sqrt2\sin\theta = 1 \implies \sin\theta=\frac{1}{\sqrt2} \implies \theta=\frac{\pi}{4}$

   ✓Final answer

   $\theta = \dfrac{\pi}{4}$ — option (b).

8. CBSE 2024Set ANNUAL1 mark
   Q.Find the area of parallelogram whose adjacent sides are given by the vectors $\bar{a} = \hat{i}+\hat{j}$ and $\bar{b} = 2\hat{i}+3\hat{k}$.

   ›Reveal solutionSolution

   The area of a parallelogram with adjacent sides $\bar a$, $\bar b$ is $|\bar a \times \bar b|$.

   Given $\bar a = \hat i+\hat j = (1,1,0)$ and $\bar b = 2\hat i+3\hat k = (2,0,3)$.

   $\bar a \times \bar b = \begin{vmatrix}\hat i & \hat j & \hat k\\ 1 & 1 & 0\\ 2 & 0 & 3\end{vmatrix}$

   $= \hat i(1\cdot3-0\cdot0) - \hat j(1\cdot3-0\cdot2) + \hat k(1\cdot0-1\cdot2)$

   $= 3\hat i - 3\hat j - 2\hat k$

   $|\bar a\times\bar b| = \sqrt{3^2+(-3)^2+(-2)^2} = \sqrt{9+9+4} = \sqrt{22}$

   ✓Final answer

   Area $= \sqrt{22}$ square units.

9. CBSE 2022Set FF1 markMCQ
   Q.The area of $\triangle ABC$, whose vertices are $A(1, 1, 1)$, $B(1, 2, 3)$ and $C(2, 3, 1)$ in square units is:

   (a) $\dfrac{\sqrt{21}}{2}$

   (b) $\dfrac{\sqrt{22}}{3}$

   (c) $\dfrac{\sqrt{23}}{3}$

   (d) None of these

   ›Reveal solutionSolution

   Area $=\tfrac12|\vec{AB}\times\vec{AC}|=\dfrac{\sqrt{21}}{2}$ — option (a).

   Concept. The area of a triangle with vertices $A,B,C$ is $\tfrac12|\vec{AB}\times\vec{AC}|$.

   $\vec{AB}=B-A=(0,1,2),\qquad \vec{AC}=C-A=(1,2,0).$

   $\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat i&\hat j&\hat k\\0&1&2\\1&2&0\end{vmatrix}=\hat i(1\cdot0-2\cdot2)-\hat j(0\cdot0-2\cdot1)+\hat k(0\cdot2-1\cdot1)=(-4,\,2,\,-1).$

   $|\vec{AB}\times\vec{AC}|=\sqrt{16+4+1}=\sqrt{21}.$

   $\text{Area}=\frac12\sqrt{21}=\frac{\sqrt{21}}{2}.$

   ✓Final answer

   (a) $\dfrac{\sqrt{21}}{2}$ square units.

10. CBSE 2022Set ANNUAL1 markMCQ
    Q.$5\vec{j} \times 4\vec{i} =$

    (a) $20$

    (b) $-20$

    (c) $20\vec{k}$

    (d) $-20\vec{k}$

    ›Reveal solutionSolution

    $\hat j\times\hat i=-\hat k$, giving $-20\hat k$.

    The cyclic rule gives $\hat i\times\hat j=\hat k$, so $\hat j\times\hat i=-\hat k$.

    $5\hat j\times4\hat i = 20(\hat j\times\hat i) = 20(-\hat k) = -20\hat k$.

    ✓Final answer

    (d) $-20\vec{k}$.

11. CBSE 2021Set NC1 mark
    Q.Find $\vec{a} \times \vec{b}$ where $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$. OR Find the vector equation of the straight line joining the points $(1, 2, 3)$ and $(2, 1, 4)$.

    ›Reveal solutionSolution

    Compute the cross product using the determinant formula with $\vec a,\vec b$ as the second and third rows.

    $\vec a=\hat i-2\hat j+3\hat k$, $\vec b=\hat i+2\hat j-\hat k$.

    $\vec a\times\vec b = \begin{vmatrix}\hat i & \hat j & \hat k\\ 1 & -2 & 3\\ 1 & 2 & -1\end{vmatrix}$

    $= \hat i\big[(-2)(-1)-(3)(2)\big] - \hat j\big[(1)(-1)-(3)(1)\big] + \hat k\big[(1)(2)-(-2)(1)\big]$

    $= \hat i(2-6) - \hat j(-1-3) + \hat k(2+2)$

    $= -4\hat i + 4\hat j + 4\hat k$

    Check (orthogonality): $\vec a\cdot(\vec a\times\vec b) = (1)(-4)+(-2)(4)+(3)(4) = -4-8+12=0$ correct; $\vec b\cdot(\vec a\times\vec b)=(1)(-4)+(2)(4)+(-1)(4)=-4+8-4=0$ correct -- both confirm the cross product is perpendicular to $\vec a$ and $\vec b$.

    ✓Final answer

    $\vec a\times\vec b = -4\hat i+4\hat j+4\hat k$

    Alternative (Or): Find the vector equation of the line through $(1,2,3)$ and $(2,1,4)$.

    A line through a point $\vec a$ with direction $\vec d$ (found from the two given points) has vector equation $\vec r=\vec a+t\vec d$.

    Position vector of the first point: $\vec a=\hat i+2\hat j+3\hat k$ (for $(1,2,3)$); the other given point is $(2,1,4)$.

    Direction vector:

    $\vec d = (2-1)\hat i+(1-2)\hat j+(4-3)\hat k = \hat i-\hat j+\hat k$

    Vector equation of the line:

    $\vec r = (\hat i+2\hat j+3\hat k) + t(\hat i-\hat j+\hat k), \qquad t\in\mathbb R$

    Check: at $t=1$: $\vec r = 2\hat i+\hat j+4\hat k$, i.e. the point $(2,1,4)$ correct -- the second given point lies on the line.

    ✓Final answer

    $\vec r=(\hat i+2\hat j+3\hat k)+t(\hat i-\hat j+\hat k)$

12. CBSE 2021Set ANNUAL1 markMCQ
    Q.Let the vectors $\vec a$ and $\vec b$ be such that $|\vec a|=3$ and $|\vec b|=\dfrac{\sqrt{2}}{3}$, then $\vec a \times \vec b$ is a unit vector, if the angle between $\vec a$ and $\vec b$ is

    (a) $\dfrac{\pi}{6}$

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

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

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

    ›Reveal solutionSolution

    $|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta = 1$ gives $\sin\theta = 1/\sqrt2$, so $\theta=\pi/4$.

    $|\vec a\times\vec b| = |\vec a|\,|\vec b|\sin\theta = 3\cdot\dfrac{\sqrt2}{3}\sin\theta = \sqrt2\sin\theta$

    Setting this equal to 1 (unit vector): $\sqrt2 \sin\theta = 1 \Rightarrow \sin\theta = \dfrac{1}{\sqrt2} \Rightarrow \theta = \dfrac{\pi}{4}$

    ✓Final answer

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