Q.Find the area of the triangle with vertices $A(1, 1, 2)$, $B(2, 3, 5)$ and $C(1, 5, 5)$.

Showing the 12 most recent of 15 on this concept.

1. CBSE 2020Set 65/2/11 markMCQ
   Q.The area of a triangle formed by vertices O, A and B, where $\vec{OA} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{OB} = -3\hat{i} - 2\hat{j} + \hat{k}$ is (A) $3\sqrt{5}$ sq. units (B) $5\sqrt{5}$ sq. units (C) $6\sqrt{5}$ sq. units (D) $4$ sq. units

   ›Reveal solutionSolution

   The area of a triangle formed by two vectors $\vec{a}$ and $\vec{b}$ originating from the same vertex is given by half the magnitude of their cross product, i.e., $\frac{1}{2} |\vec{a} \times \vec{b}|$. For the given vectors, the area is $\mathbf{3\sqrt{5}}$ sq. units.

   The problem asks for the area of a triangle formed by the origin O and two points A and B, given their position vectors $\vec{OA}$ and $\vec{OB}$. This is a classic application of the vector cross product.

   Concept and Intuition: Why the Cross Product?

   The cross product of two vectors, say $\vec{a}$ and $\vec{b}$, is another vector whose magnitude is defined as $|\vec{a}||\vec{b}|\sin\theta$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$. Geometrically, this magnitude, $|\vec{a} \times \vec{b}|$, represents the area of the parallelogram formed by $\vec{a}$ and $\vec{b}$ when they originate from the same point.

   Consider a parallelogram with adjacent sides represented by vectors $\vec{a}$ and $\vec{b}$. If we take $|\vec{a}|$ as the base, the perpendicular height of the parallelogram is $|\vec{b}|\sin\theta$. The area of the parallelogram is thus base $\times$ height $= |\vec{a}|(|\vec{b}|\sin\theta) = |\vec{a}||\vec{b}|\sin\theta$. This is precisely the magnitude of the cross product.

   Now, a triangle formed by these two vectors (sharing the same origin) is exactly half the area of the parallelogram formed by them. Therefore, the area of such a triangle is $\frac{1}{2} |\vec{a} \times \vec{b}|$.

   In this problem, $\vec{OA}$ and $\vec{OB}$ are the two vectors originating from the common vertex O, forming two sides of the triangle OAB. Thus, we can directly apply this formula.

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   Here's the step-by-step solution:

   1. Identify the vectors forming the sides of the triangle.

      We are given the position vectors of points A and B with respect to the origin O:

      $\vec{OA} = \hat{i} + 2\hat{j} + 3\hat{k}$

      $\vec{OB} = -3\hat{i} - 2\hat{j} + \hat{k}$

      These vectors represent two sides of the triangle OAB, both originating from the vertex O.

   2. Calculate the cross product of these two vectors.

      The cross product $\vec{OA} \times \vec{OB}$ is calculated using the determinant form:

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

   Expanding the determinant:
   

   $\vec{OA} \times \vec{OB} = \hat{i}((2)(1) - (3)(-2)) - \hat{j}((1)(1) - (3)(-3)) + \hat{k}((1)(-2) - (2)(-3))$

   $\vec{OA} \times \vec{OB} = \hat{i}(2 - (-6)) - \hat{j}(1 - (-9)) + \hat{k}(-2 - (-6))$

   $\vec{OA} \times \vec{OB} = \hat{i}(2 + 6) - \hat{j}(1 + 9) + \hat{k}(-2 + 6)$

   $$ \vec{OA} \times \vec{OB} = 8\hat{i} - 10\hat{j} + 4\hat{k} $$ …
   

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

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

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

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

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

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5. 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:

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

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

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

   …

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6. 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, $|\vec a\times\vec b|=1$.

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

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7. 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:

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

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8. 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 $1$: …

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9. 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 $\bar b = 2\hat i+3\hat k = (2,0,3)$.

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

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

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10. 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 $\tfrac12|\vec{AB}\times\vec{AC}|$.

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

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11. 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 $\hat j\times\hat i=-\hat k$.

    …

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12. 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 $\vec b$.

    …

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