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For the two vectors \( \overrightarrow{OA} \) and \( \overrightarrow{OB} \) it is known that \( \overrightarrow{OA} \cdot \overrightarrow{OB} < 0 \) - HSC - SSCE Mathematics Extension 1 - Question 5 - 2021 - Paper 1
Question 5
For the two vectors \( \overrightarrow{OA} \) and \( \overrightarrow{OB} \) it is known that \( \overrightarrow{OA} \cdot \overrightarrow{OB} < 0 \). Which of the ... show full transcript
Worked Solution & Example Answer:For the two vectors \( \overrightarrow{OA} \) and \( \overrightarrow{OB} \) it is known that \( \overrightarrow{OA} \cdot \overrightarrow{OB} < 0 \) - HSC - SSCE Mathematics Extension 1 - Question 5 - 2021 - Paper 1
Step 1
Either, \( \overrightarrow{OA} \) is negative and \( \overrightarrow{OB} \) is positive, or, \( \overrightarrow{OA} \) is positive and \( \overrightarrow{OB} \) is negative.
Answer
This statement does not necessarily hold true. The condition ( \overrightarrow{OA} \cdot \overrightarrow{OB} < 0 ) indicates that the vectors point in opposite directions, but does not definitively imply negativity of one vector and positivity of the other.
Step 2
The angle between \( \overrightarrow{OA} \) and \( \overrightarrow{OB} \) is obtuse.
Answer
This statement must be true because the dot product of two vectors is defined as ( \overrightarrow{OA} \cdot \overrightarrow{OB} = |\overrightarrow{OA}| |\overrightarrow{OB}| \cos(\theta) ). Since ( \overrightarrow{OA} \cdot \overrightarrow{OB} < 0 ), it must be the case that ( \cos(\theta) < 0 ), which occurs when ( \theta ) is obtuse (greater than 90 degrees).
Step 3
The product \( |\overrightarrow{OA}| |\overrightarrow{OB}| \) is negative.
Answer
This statement is false. The magnitudes of the vectors ( |\overrightarrow{OA}| ) and ( |\overrightarrow{OB}| ) are always non-negative. Therefore, their product cannot be negative regardless of the directions or signs of the vectors.
Step 4