# 根據以上的公式，我們可以知道如果△ABC 如下圖: A B C θ a b 則△ABC 的面積是 1 2|

(1)

(24) 向量的應用

θ

△ABC 的面積是1

2𝐵𝐷 × 𝐴𝐶

𝐵𝐷

𝐴𝐵 = 𝑠𝑖𝑛 𝜃

∴ 𝐵𝐷 = 𝐴𝐵 𝑠𝑖𝑛 𝜃

2(𝐴𝐵 × 𝐴𝐶) 𝑠𝑖𝑛 𝜃

θ

## b

△OAB 的面積是|𝑎⃗||𝑏⃗⃗| 𝑠𝑖𝑛 𝜃

(2)

## b C

α β

∴ ∠𝐴𝑂𝐵 = 𝛼 − 𝛽 = 𝜃

## b C

β

θ

a⃗⃗=(𝑎1, 𝑎2)

b⃗⃗=(𝑏1, 𝑏2)

𝑠𝑖𝑛 𝛼 =𝐴𝐷

|𝑎⃗| = 𝑎2

|𝑎⃗|

𝑐𝑜𝑠 𝛼 =𝑂𝐷

|𝑎⃗| = 𝑎1

|𝑎⃗|

𝑠𝑖𝑛 𝛽 =𝐵𝐸

|𝑏⃗⃗|= 𝑏2

|𝑏⃗⃗|

(3)

### 24-3

𝑐𝑜𝑠 𝛽 =𝑂𝐸

|𝑏⃗⃗| = 𝑏1

|𝑏⃗⃗|

∴ 𝑠𝑖𝑛 𝜃 =𝑎2𝑏1− 𝑎1𝑏2

|𝑎⃗||𝑏⃗⃗|

θ

## b

1

2|𝑎⃗||𝑏⃗⃗| 𝑠𝑖𝑛 𝜃 =1

2|𝑎⃗||𝑏⃗⃗|𝑎2𝑏1− 𝑎1𝑏2

|𝑎⃗||𝑏⃗⃗| =1

2(𝑎2𝑏1− 𝑎1𝑏2)

(1)假設𝐴 = (1,2)，𝐵 = (4,6)，𝐶 = (6,5)

𝑎⃗ = 𝐴𝐵⃗⃗⃗⃗⃗⃗ = (4 − 1, 6 − 2) = (3, 4) = ( 𝑎1, 𝑎2) 𝑎1 = 3，𝑎2 = 4

𝑏⃗⃗ = 𝐴𝐶⃗⃗⃗⃗⃗⃗ = (6 − 1, 5 − 2) = (5, 3) = ( 𝑏1, 𝑏2) 𝑏1 = 5，𝑏2 = 3

∴ △ 𝐴𝐵𝐶的面積 =1

2(𝑎2𝑏1− 𝑎1𝑏2) =1

2(4 × 5 − 3 × 3) =1

2(20 − 9) =11 2

(4)

### 24-4

(2)假設𝐴 = (0,0)，𝐵 = (2,2)，𝐶 = (2,0)

𝑎⃗ = (2 − 0, 2 − 0) = (2, 2) 𝑎1 = 2，𝑎2 = 2

𝑏⃗⃗ = (2 − 0, 0 − 0) = (2, 0) 𝑏1 = 2，𝑏2 = 0

∴ △ 𝐴𝐵𝐶的面積 =1

2(𝑎2𝑏1− 𝑎1𝑏2) =1

2(2 × 2 − 2 × 0) =1

2× 4 = 2

2

2

1

2𝐵𝐶 × 𝐴𝐶 =1

2× 2 × 2 = 2

(5)

(3)請看下圖:

## b

𝑎⃗ = (2 − 0, 2 − 0) = (2, 2) 𝑎1 = 2，𝑎2 = 2

𝑏⃗⃗ = (1 − 0, 0 − 0) = (1, 0) 𝑏1 = 1，𝑏2 = 0

△ABC 的面積是

1

2(𝑎2𝑏1− 𝑎1𝑏2) =1

2(2 × 1 − 2 × 0) = 1

(4)請看下圖:

## ab

𝑎⃗ = (1 − 0, 1 − 0) = (1, 1) 𝑎1 = 1，𝑎2 = 1 𝑏⃗⃗ = (2 − 0, 0 − 0) = (2, 0) 𝑏1 = 2，𝑏2 = 0

△ABC 的面積是 1

2(𝑎2𝑏1− 𝑎1𝑏2) =1

2(1 × 2 − 1 × 0) = 1

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## References

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