Use the dot product to calculate the area of a triangle and properties of geometric figures.
Area of a Triangle
When →OA=→a and →OB=→b,
S = (1/2)√(|→a|²|→b|² − (→a・→b)²)
Using components, S = (1/2)|a₁b₂ − a₂b₁|.
Example ①
Find the area of triangle OAB with O(0,0), A(3, 1), and B(2, 4).
Solution: S=(1/2)|3×4−1×2|=5
Find the area of triangle OAB with O(0,0), A(3, 1), and B(2, 4).
Solution: S=(1/2)|3×4−1×2|=5
Use in Geometric Proofs
Example ②
When |→a|=|→b|, show that (→a + →b) ⊥ (→a − →b).
Solution: (→a+→b)・(→a−→b)=|→a|²−|→b|²=0. Therefore, they are perpendicular (a property of the diagonals of a rhombus).
When |→a|=|→b|, show that (→a + →b) ⊥ (→a − →b).
Solution: (→a+→b)・(→a−→b)=|→a|²−|→b|²=0. Therefore, they are perpendicular (a property of the diagonals of a rhombus).
Key Points
- Area of a triangle: (1/2)|a₁b₂ − a₂b₁|
- |→a±→b|² = |→a|² ± 2→a・→b + |→b|²
- (→a+→b)・(→a−→b) = |→a|² − |→b|²
Practice Problems
- Find the area of triangle OAB with O(0,0), A(2, 3), and B(5, 1).
- When |→a|=3, |→b|=2, and →a・→b=4, find |→a − →b|.
- When |→a|=4, |→b|=4, and the angle between them is 60°, find (→a+→b)・(→a−→b).
Answers and Explanations
- Answer: 13/2
Explanation: (1/2)|2−15|=13/2. - Answer: √5
Explanation: 9−8+4=5. - Answer: 0
Explanation: Since |→a|=|→b|, |→a|²−|→b|²=0.