Maths Olympiad Prep

Track / Stage 3 / 209 of 260 #209 of 1964

Problem 209

AMC 10/12, early questions
Algebra Difficulty 3.6 Multiple choice

Given the vectors a→=(x−3,2) \overrightarrow{a}=(x-3,2) and b→=(1,1) \overrightarrow{b}=(1,1) , the statement " x>1 x > 1 " is a ______ condition for "the angle between a→ \overrightarrow{a} and b→ \overrightarrow{b} is acute".

Pick one

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Official solution

Let's calculate the dot product of vectors a→ \overrightarrow{a} and b→ \overrightarrow{b} :

a→⋅b→=(x−3)⋅1+2⋅1=x−1 \overrightarrow{a} \cdot \overrightarrow{b} = (x-3) \cdot 1 + 2 \cdot 1 = x - 1

For the vectors a→ \overrightarrow{a} and b→ \overrightarrow{b} to be collinear with the same direction, we must have b→=λa→ \overrightarrow{b} = \lambda \overrightarrow{a} , with λ>0 \lambda > 0 .

Setting up the equations:

{x−3=λ2=λ \begin{cases} x - 3 = \lambda \\ 2 = \lambda \end{cases}

we find that x=5 x = 5 . When x=5 x = 5 , the condition x>1 x > 1 is satisfied, but in this case, the angle between the two vectors is 0∘ 0^{\circ} , so the angle being acute is not true.

If a→ \overrightarrow{a} and b→ \overrightarrow{b} form an acute angle, then their dot product must be positive:

a→⋅b→=x−1>0 \overrightarrow{a} \cdot \overrightarrow{b} = x - 1 > 0

Thus, x>1 x > 1 must be true. This implies that " x>1 x > 1 " is a necessary but not sufficient condition for the angle between a→ \overrightarrow{a} and b→ \overrightarrow{b} to be acute.

Therefore, the correct answer is:
A \boxed{ A }

This problem assesses the understanding of necessary and sufficient conditions along with the relationship between the angle of vectors and their dot product.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.