Class CBSE Class 12 Mathematics Vector Algebra Q #1419
KNOWLEDGE BASED
REMEMBER
5 Marks 2025 AISSCE(Board Exam) LA
Show that the area of a parallelogram whose diagonals are represented by $\vec{a}$ and $\vec{b}$ is given by $\frac{1}{2}|\vec{a}\times\vec{b}|$. Also find the area of a parallelogram whose diagonals are $2\hat{i}-\hat{j}+\hat{k}$ and $\hat{i}+3\hat{j}-\hat{k}$.
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Detailed Solution

Step 1: Understanding the relationship between diagonals and sides of a parallelogram

Let the sides of the parallelogram be represented by vectors $\vec{p}$ and $\vec{q}$. The diagonals $\vec{a}$ and $\vec{b}$ can be expressed in terms of $\vec{p}$ and $\vec{q}$ as follows: $\vec{a} = \vec{p} + \vec{q}$ $\vec{b} = \vec{p} - \vec{q}$ (or $\vec{q} - \vec{p}$, the result will be the same)

Step 2: Expressing sides in terms of diagonals

From the equations in Step 1, we can express $\vec{p}$ and $\vec{q}$ in terms of $\vec{a}$ and $\vec{b}$: Adding the two equations: $\vec{a} + \vec{b} = 2\vec{p} \implies \vec{p} = \frac{1}{2}(\vec{a} + \vec{b})$ Subtracting the two equations: $\vec{a} - \vec{b} = 2\vec{q} \implies \vec{q} = \frac{1}{2}(\vec{a} - \vec{b})$

Step 3: Finding the area of the parallelogram

The area of the parallelogram is given by the magnitude of the cross product of its sides: Area $= |\vec{p} \times \vec{q}|$ Substituting the expressions for $\vec{p}$ and $\vec{q}$ in terms of $\vec{a}$ and $\vec{b}$: Area $= |\frac{1}{2}(\vec{a} + \vec{b}) \times \frac{1}{2}(\vec{a} - \vec{b})|$ Area $= \frac{1}{4}|(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})|$ Area $= \frac{1}{4}|\vec{a} \times \vec{a} - \vec{a} \times \vec{b} + \vec{b} \times \vec{a} - \vec{b} \times \vec{b}|$ Since $\vec{a} \times \vec{a} = \vec{0}$ and $\vec{b} \times \vec{b} = \vec{0}$, and $\vec{b} \times \vec{a} = -\vec{a} \times \vec{b}$: Area $= \frac{1}{4}|-\vec{a} \times \vec{b} - \vec{a} \times \vec{b}|$ Area $= \frac{1}{4}|-2\vec{a} \times \vec{b}|$ Area $= \frac{1}{2}|\vec{a} \times \vec{b}|$

Step 4: Calculating the cross product of the given diagonals

Given $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} - \hat{k}$, we find their cross product: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{vmatrix} = \hat{i}(1 - 3) - \hat{j}(-2 - 1) + \hat{k}(6 - (-1)) = -2\hat{i} + 3\hat{j} + 7\hat{k}$

Step 5: Finding the magnitude of the cross product

$|\vec{a} \times \vec{b}| = \sqrt{(-2)^2 + (3)^2 + (7)^2} = \sqrt{4 + 9 + 49} = \sqrt{62}$

Step 6: Calculating the area of the parallelogram

Area $= \frac{1}{2}|\vec{a} \times \vec{b}| = \frac{1}{2}\sqrt{62}$ square units.

Final Answer: $\frac{\sqrt{62}}{2}$ square units

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Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because it requires recalling the formula for the area of a parallelogram in terms of its diagonals and applying it to a specific case.
Knowledge Dimension: CONCEPTUAL
Justification: The question requires understanding the relationship between the diagonals and sides of a parallelogram and applying the concept of cross product to find the area.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. It directly tests a standard formula and its application, which is part of the textbook content.

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