The dot product makes it easy to determine whether two vectors are perpendicular.
Perpendicularity Condition
For →a ≠ →0 and →b ≠ →0, →a ⊥ →b ⇔ →a・→b = 0.
In components, a₁b₁ + a₂b₂ = 0.
Example ①
Find k such that →a=(2, 3) and →b=(k, 4) are perpendicular.
Solution: 2k + 12 = 0 → k=−6
Find k such that →a=(2, 3) and →b=(k, 4) are perpendicular.
Solution: 2k + 12 = 0 → k=−6
Perpendicular Unit Vectors
A vector perpendicular to →a=(a₁, a₂) is (−a₂, a₁) or (a₂, −a₁). The unit vectors are ±(−a₂, a₁)/|→a|.
Example ②
Find the unit vectors perpendicular to →a=(3, 4).
Solution: |→a|=5. Perpendicular vectors are (−4, 3) or (4, −3). The unit vectors are ±(−4/5, 3/5).
Find the unit vectors perpendicular to →a=(3, 4).
Solution: |→a|=5. Perpendicular vectors are (−4, 3) or (4, −3). The unit vectors are ±(−4/5, 3/5).
Key Points
- Perpendicularity condition: →a・→b = 0
- Perpendicular to (a₁, a₂): (−a₂, a₁) or (a₂, −a₁)
- There are two perpendicular unit vectors
Practice Problems
- Find k such that →a=(4, −1) and →b=(2, k) are perpendicular.
- Find the unit vectors perpendicular to →a=(−5, 12).
- When |→a|=2, |→b|=3, and →a⊥→b, find (→a + 2→b)・(3→a − →b).
Answers and Explanations
- Answer: k=8
Explanation: 8 − k = 0 → k=8. - Answer: ±(12/13, 5/13)
Explanation: |→a|=13, and perpendicular vectors are (12, 5) or (−12, −5). - Answer: −6
Explanation: 3|→a|²+5→a・→b−2|→b|²=12+0−18=−6.