Paste your question to our Mathbee AI Mentor below to get an instant step-by-step solution.
Three Dimensional Geometry
NCERT Exemplar Class 12
CBSE
Grade 12
Question:
Find shortest distance between lines $\dfrac{x-1}{1} = \dfrac{y-2}{-1} = \dfrac{z-1}{1}$ and $\dfrac{x-2}{2} = \dfrac{y+1}{1} = \dfrac{z+1}{2}$.
Step-by-Step Solution
Given: Find shortest distance between lines $\dfrac{x-1}{1} = \dfrac{y-2}{-1} = \dfrac{z-1}{1}$ and $\dfrac{x-2}{2} = \dfrac{y+1}{1} = \dfrac{z+1}{2}$. Step 1: Formulate initial mathematical setup / substitution: Given equation / conditions: $$SD$$ [1.0 Mark] Step 2: Perform intermediate algebraic / calculus operations: Executing the step-by-step reduction: $$3/\sqrt{2}.$$ [1.0 Mark] Step 3: Evaluate final values / boundary conditions: Substituting limits or solving the simplified system: $$3/\sqrt{2}.$$ [1.0 Mark] Conclusion: The final evaluated answer is 3/\sqrt{2}..
--- 🎯 Official CBSE Marking Scheme: Formulating mathematical relation: 1.0 Mark Executing algebraic/calculus operations: 1.0 Mark Evaluating final solution/vector: 1.0 Mark
Correct Answer:
Mathbee AI Mentor (Free Demo)
Confused by the solution? Ask the AI to explain a specific step, tell you where you went wrong, or break down the key trap in this question.
Master Three Dimensional Geometry with Mathbee
Practice this topic under real exam conditions with strict timers, or ask our AI Mentor to explain the concepts step-by-step.