A comprehensive platform for Teachers to create standard question papers and Students to practice Case-Based, Assertion-Reason, and Critical Thinking questions.
Create professional PDF/Word papers with logo, instructions, and mixed question types in minutes.
Explore our repository by Class and Topic. Filter by "Knowledge" or "Competency" levels.
For Students. Take timed MCQ tests to check your understanding. Get instant feedback.
According to NEP 2020, rote learning is out. The focus has shifted to assessing a student's ability to apply concepts in real-life situations.
Questions derived from real-world passages to test analytical skills.
Testing the logic behind concepts, not just the definition.
Open-ended scenarios that require thinking beyond the textbook.
We provide complete AI-Powered Explanations for every question.
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)
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: $\vec{a} - \vec{b} = 2\vec{q} \implies \vec{q} = \frac{1}{2}(\vec{a} - \vec{b})$
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 $= \frac{1}{2}|\vec{a} \times \vec{b}|$
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: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{vmatrix} = \hat{i}(1 - 3) - \hat{j}(-2 - 1) + \hat{k}(6 - (-1)) = -2\hat{i} + 3\hat{j} + 7\hat{k}$
$|\vec{a} \times \vec{b}| = \sqrt{(-2)^2 + (3)^2 + (7)^2} = \sqrt{4 + 9 + 49} = \sqrt{62}$
Area $= \frac{1}{2}|\vec{a} \times \vec{b}| = \frac{1}{2}\sqrt{62}$ square units.
Final Answer: $\frac{\sqrt{62}}{2}$ square units