Q.Find $|\vec{a} \times \vec{b}|$, if $\vec{a}=\hat{i}-7\hat{j}+7\hat{k}$ and $\vec{b}=3\hat{i}-2\hat{j}+2\hat{k}$.

Showing the 12 most recent of 15 on this concept.

1. GUJCET 2026Set x1 markMCQ
   Q.The area of the triangle with vertices $A(1,1,2)$, $B(2,3,5)$ and $C(1,5,5)$ is ______ (A) $\sqrt{43}$ (B) $\frac{\sqrt{43}}{2}$ (C) $\sqrt{61}$ (D) $\frac{\sqrt{61}}{2}$

   ›Reveal solutionSolution

   Area $=\frac12|\vec{AB}\times\vec{AC}|=\frac{\sqrt{61}}{2}$.

   Form two edge vectors:

   $\vec{AB}=(1,2,3),\qquad \vec{AC}=(0,4,3)$

   Cross product:

   $\vec{AB}\times\vec{AC}=\big(2\cdot3-3\cdot4,\;-(1\cdot3-3\cdot0),\;1\cdot4-2\cdot0\big)=(-6,-3,4)$

   Magnitude:

   $|\vec{AB}\times\vec{AC}|=\sqrt{(-6)^2+(-3)^2+4^2}=\sqrt{36+9+16}=\sqrt{61}$

   Area of the triangle:

   $A=\frac{1}{2}\sqrt{61}$

   ✓Final answer

   Option (D) $\frac{\sqrt{61}}{2}$

   ANSWER: (D)

2. GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
   Q.If the position vectors of points $A$ and $B$ from the origin are $\vec a=2\hat i-3\hat j+2\hat k$ and $\vec b=2\hat i+3\hat j+\hat k$ respectively, then the area of triangle $OAB =$ ____.

   (a) $\sqrt{229}$

   (b) $\sqrt{157}$

   (c) $\frac12\sqrt{229}$

   (d) $\frac12\sqrt{157}$

   ›Reveal solutionSolution

   Area of $\triangle OAB=\dfrac12|\vec a\times\vec b|$.

   $\vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-3&2\\2&3&1\end{vmatrix}=\hat i(-3-6)-\hat j(2-4)+\hat k(6+6)=-9\hat i+2\hat j+12\hat k$.

   $|\vec a\times\vec b|=\sqrt{81+4+144}=\sqrt{229}$. Area $=\dfrac12\sqrt{229}$.

   ✓Final answer

   The correct option is (c) $\dfrac12\sqrt{229}$.

3. GUJCET 2025Set 031 markMCQ
   Q.Area of a rectangle having vertices $A$, $B$, $C$ and $D$ with position vectors $-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$, $\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$, $\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$ and $-\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$, respectively is _____. (A) 4 (B) 1 (C) 2 (D) $\frac{1}{2}$

   ›Reveal solutionSolution

   [!TLDR]

   The emf is maximum at $t = \dfrac{\pi}{2\omega}$.

   Concept

   In an AC generator the induced emf is sinusoidal: $\varepsilon = \varepsilon_0 \sin(\omega t + \phi)$. The condition $\varepsilon = 0$ at $t = 0$ fixes the phase $\phi = 0$, so $\varepsilon = \varepsilon_0 \sin(\omega t)$.

   Solution

   $\varepsilon$ is maximum when $\sin(\omega t) = 1$:

   $\omega t = \frac{\pi}{2} \;\Rightarrow\; t = \frac{\pi}{2\omega}.$

   This is a quarter of the period after the zero crossing.

   [!ANSWER]

   (D)

4. GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
   Q.The area of the triangle with vertices $A(1,1,2)$, $B(2,3,5)$ and $C(1,5,5)$ = ____.

   (a) $\dfrac{\sqrt{41}}{2}$

   (b) $\dfrac{\sqrt{61}}{2}$

   (c) $\dfrac{\sqrt{51}}{2}$

   (d) $\dfrac{\sqrt{71}}{2}$

   ›Reveal solutionSolution

   Area of a triangle with vertices $A,B,C$ is $\dfrac12|\vec{AB}\times\vec{AC}|$.

   $\vec{AB}=(1,2,3)$, $\vec{AC}=(0,4,3)$.

   $\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat i&\hat j&\hat k\\1&2&3\\0&4&3\end{vmatrix}=\hat i(6-12)-\hat j(3-0)+\hat k(4-0)=-6\hat i-3\hat j+4\hat k$.

   $|\vec{AB}\times\vec{AC}|=\sqrt{36+9+16}=\sqrt{61}$. Area $=\dfrac{\sqrt{61}}{2}$.

   ✓Final answer

   The correct option is (b) $\dfrac{\sqrt{61}}{2}$.

5. GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
   Q.If $\vec{a} = \hat{i} - 7\hat{j} + 7\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + 2\hat{k}$, then $|\vec{a}\times\vec{b}|$ = ____.

   (a) $19\sqrt{2}$

   (b) $0$

   (c) $19$

   (d) $38$

   ›Reveal solutionSolution

   Compute the cross product directly and take its magnitude.

   $\vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-7&7\\3&-2&2\end{vmatrix}=\hat i(-14+14)-\hat j(2-21)+\hat k(-2+21)=0\hat i+19\hat j+19\hat k$.

   $|\vec a\times\vec b|=\sqrt{0+361+361}=\sqrt{722}=19\sqrt2$.

   ✓Final answer

   The correct option is (a) $19\sqrt{2}$.

6. GUJCET 2024Set 131 markMCQ
   Q.The area of a parallelogram, whose adjacent sides are given by the vectors $\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b} = -\hat{j} - 2\hat{k}$, is __________. (A) $2\sqrt{3}$ (B) $\sqrt{6}$ (C) $24$ (D) $2\sqrt{6}$

   ›Reveal solutionSolution

   Parallelogram area $=|\vec a\times\vec b|$.

   Steps. $\vec a=(2,3,4)$, $\vec b=(0,-1,-2)$.

   $\vec a\times\vec b = \big(3(-2)-4(-1),\,-(2(-2)-4\cdot0),\,2(-1)-3\cdot0\big) = (-2,\,4,\,-2).$

   $|\vec a\times\vec b| = \sqrt{4+16+4} = \sqrt{24} = 2\sqrt{6}.$

   ✓Final answer

   Option (D) $2\sqrt{6}$

   ANSWER: (D)

7. GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
   Q.The area of a parallelogram whose adjacent sides are $\vec{a} = \hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j}$ is ______.

   (a) $2\sqrt{21}$

   (b) $\sqrt{42}$

   (c) $\sqrt{21}$

   (d) $\frac{1}{2}\sqrt{21}$

   ›Reveal solutionSolution

   The area of a parallelogram equals $|\vec a\times\vec b|$.

   $\vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\0&1&2\\1&2&0\end{vmatrix}=\hat i(0-4)-\hat j(0-2)+\hat k(0-1)=-4\hat i+2\hat j-\hat k$.

   $|\vec a\times\vec b|=\sqrt{16+4+1}=\sqrt{21}$.

   ✓Final answer

   The correct option is (c) $\sqrt{21}$.

8. GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
   Q.For the vectors $\vec{a}$ and $\vec{b}$, $|\vec{a}| = \frac{2}{3}$, $|\vec{b}| = 3$ and $|\vec{a} \times \vec{b}| = 1$, then the 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

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

   $1=\frac23\cdot3\cdot\sin\theta=2\sin\theta \Rightarrow \sin\theta=\frac12 \Rightarrow \theta=\frac{\pi}{6}$.

   ✓Final answer

   The correct option is (a) $\frac{\pi}{6}$.

9. GUJCET 2022Set 081 markMCQ
   Q.The area of a triangle having the points $A(1,1,1)$, $B(1,2,3)$ and $C(2,3,1)$ as its vertices is ______. (A) $\frac{\sqrt{19}}{2}$ (B) $\frac{\sqrt{21}}{2}$ (C) $\frac{19}{2}$ (D) $\frac{21}{2}$

   ›Reveal solutionSolution

   Triangle area from vertices = ½ of the magnitude of the cross product of two edge vectors.

   Concept. $\text{Area}=\tfrac12\,\big|\vec{AB}\times\vec{AC}\big|$.

   Solution. $\vec{AB}=(0,1,2)$, $\vec{AC}=(1,2,0)$.

   $\vec{AB}\times\vec{AC}=\big(1\cdot0-2\cdot2,\;2\cdot1-0\cdot0,\;0\cdot2-1\cdot1\big)=(-4,\,2,\,-1)$

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

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

   ✓Final answer

   (B) $\frac{\sqrt{21}}{2}$

   ANSWER: (B)

10. GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
    Q.The area of a parallelogram whose adjacent sides are given by the vectors $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ is ___.

    (a) $\sqrt{21}$

    (b) $42$

    (c) $\sqrt{42}$

    (d) $21$

    ›Reveal solutionSolution

    The area of the parallelogram is $|\vec a\times\vec b|$.

    $\vec a\times\vec b = \begin{vmatrix}\hat i&\hat j&\hat k\\3&1&4\\1&-1&1\end{vmatrix} = \hat i(1+4) - \hat j(3-4) + \hat k(-3-1) = 5\hat i + \hat j - 4\hat k.$

    $|\vec a\times\vec b| = \sqrt{25+1+16} = \sqrt{42}.$

    ✓Final answer

    (c) $\sqrt{42}$.

11. GUJCET 2021Set 151 markMCQ
    Q.The area of parallelogram whose adjacent sides are determined by the vectors $\vec{a} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$ is $\underline{\hspace{2cm}}$ (A) $15\sqrt{2}$ (B) $15$ (C) $\frac{15}{\sqrt{2}}$ (D) $\frac{15}{2}$

    ›Reveal solutionSolution

    Parallelogram area = |axb|.

    Concept. $\vec a\times\vec b=(a_2b_3-a_3b_2,\ a_3b_1-a_1b_3,\ a_1b_2-a_2b_1)$.

    Solution. $\vec a=(1,-1,3),\ \vec b=(2,-7,1)$:

    $\vec a\times\vec b=(-1+21,\ 6-1,\ -7+2)=(20,5,-5).$

    $|\vec a\times\vec b|=\sqrt{400+25+25}=\sqrt{450}=15\sqrt2.$

    ✓Final answer

    (A) $15\sqrt{2}$

    ANSWER: (A)

12. GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQ
    Q.Find the area of a parallelogram whose adjacent sides are given by the vectors $\vec a=3\hat\imath+5\hat\jmath-2\hat k$ and $\vec b=2\hat\imath+\hat\jmath+3\hat k$.

    (a) $\sqrt{507}$

    (b) $\sqrt{387}$

    (c) $\dfrac12\sqrt{507}$

    (d) 25

    ›Reveal solutionSolution

    Compute the cross product $\vec a\times\vec b$, then its magnitude gives the parallelogram area directly.

    $\vec a=3\hat\imath+5\hat\jmath-2\hat k,\ \vec b=2\hat\imath+\hat\jmath+3\hat k$.

    $\vec a\times\vec b=\begin{vmatrix}\hat\imath&\hat\jmath&\hat k\\3&5&-2\\2&1&3\end{vmatrix}=\hat\imath(15+2)-\hat\jmath(9+4)+\hat k(3-10)=17\hat\imath-13\hat\jmath-7\hat k$.

    $|\vec a\times\vec b|=\sqrt{17^2+13^2+7^2}=\sqrt{289+169+49}=\sqrt{507}$.

    ✓Final answer

    (a) $\sqrt{507}$.