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.
The line $L_1$ passes through the point (2, -1, 1) and is parallel to the vector $\vec{b_1} = \hat{i} + \hat{j} + 3\hat{k}$. Therefore, the equation of $L_1$ is given by: $\vec{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda(\hat{i} + \hat{j} + 3\hat{k})$
The equation of line $L_2$ is given as: $\vec{r} = \hat{i} + (2\mu+1)\hat{j} - (\mu+2)\hat{k} = (\hat{i} + \hat{j} - 2\hat{k}) + \mu(2\hat{j} - \hat{k})$. Thus, $L_2$ passes through the point (1, 1, -2) and is parallel to the vector $\vec{b_2} = 2\hat{j} - \hat{k}$.
Let $\vec{a_1} = 2\hat{i} - \hat{j} + \hat{k}$ be a point on $L_1$ and $\vec{a_2} = \hat{i} + \hat{j} - 2\hat{k}$ be a point on $L_2$. Then, the vector connecting these two points is: $\vec{a_2} - \vec{a_1} = (\hat{i} + \hat{j} - 2\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = -\hat{i} + 2\hat{j} - 3\hat{k}$.
The direction vectors of the lines are $\vec{b_1} = \hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b_2} = 2\hat{j} - \hat{k}$. Their cross product is: $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 3 \\ 0 & 2 & -1 \end{vmatrix} = (-1 - 6)\hat{i} - (-1 - 0)\hat{j} + (2 - 0)\hat{k} = -7\hat{i} + \hat{j} + 2\hat{k}$.
The magnitude of the cross product is: $|\vec{b_1} \times \vec{b_2}| = \sqrt{(-7)^2 + 1^2 + 2^2} = \sqrt{49 + 1 + 4} = \sqrt{54} = 3\sqrt{6}$.
The shortest distance between the lines is given by: $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$. We have $(\vec{a_2} - \vec{a_1}) = -\hat{i} + 2\hat{j} - 3\hat{k}$ and $(\vec{b_1} \times \vec{b_2}) = -7\hat{i} + \hat{j} + 2\hat{k}$. Therefore, $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-1)(-7) + (2)(1) + (-3)(2) = 7 + 2 - 6 = 3$. Thus, the shortest distance is: $d = \frac{|3|}{3\sqrt{6}} = \frac{1}{\sqrt{6}} = \frac{\sqrt{6}}{6}$.
\r\n Final Answer: $\frac{\sqrt{6}}{6}$<\/span>\r\n <\/p>\r\n <\/div>\r\n <\/div>