Let and be unit vectors with . If a vector satisfies , find the maximum possible value of .

Triangle Solving Solution – Problem 21: find the maximum possible value of

Question

Let $\vec{a}$ and $\vec{b}$ be unit vectors with $\vec{a}\cdot\vec{b}=0$ . If a vector $\vec{c}$ satisfies $|\vec{c}-\vec{a}-2\vec{b}|=1$ , find the maximum possible value of $|\vec{c}|$ .

Step-by-step solution

Step 1. Since $\vec{a}\cdot\vec{b}=0$ and $|\vec{a}|=|\vec{b}|=1$ , choose coordinates so that $\vec{a}=(1,0)$ and $\vec{b}=(0,1)$ .

Step 2. Let $\vec{c}=(x,y)$ . The condition $|\vec{c}-\vec{a}-2\vec{b}|=1$ becomes $(x-1)^{2}+(y-2)^{2}=1$ .

Step 3. Thus the point $(x,y)$ lies on the circle centered at $(1,2)$ with radius 1.

Step 4. The maximum of $|\vec{c}|$ equals the distance from the origin to the center plus the radius: $|\vec{c}|_{\max}=\sqrt{1^{2}+2^{2}}+1=\sqrt{5}+1.$

Final answer

$\sqrt{5}+1$

Marking scheme

1. Checkpoints (max 7 pts total)

Chain A: Combined Law of Sines and Cosines approach

Set up side-angle relations [2 pts]: States and correctly advances the key derivation steps
Substitute and simplify [2 pts]: Substitutes correctly and simplifies accurately
Handle multiple cases / admissibility [1 pt]: Considers branches and rejects invalid cases
Conclusion and verification [1 pt]: States the conclusion and checks against constraints
Final answer [1 pt]: Gives the correct final result (for multiple-choice, include the option letter)

2. Zero-credit items

Copies formulas without concrete substitution or derivation
Guesses the answer / provides only a conclusion with no reasoning
Uses an approach incompatible with the problem conditions, leading to an invalid conclusion

3. Deductions

Computation error [-1]: Incorrect algebraic/trigonometric manipulation
Logical gap [-1]: Missing a key equivalence step or a necessary condition check
Nonstandard final statement [-1]: Missing units/range/option letter or wrong answer format

Triangle Solving – Problem 21: find the maximum possible value of