如何找到两个向量之间的角度？

具有方向和大小的物理量是向量。大小等于 1、方向等于 1 的向量是单位向量“û.”。那是带有“帽子”抑扬符的小写字母。这样，向量用箭头来描述，有始点和终点，经过了 200 年的发展。向量可以用来表示物理量，如位移、速度、加速度等。

两个向量之间的角度

夹角为 θ 的向量

一个向量在它的尾部之间的角度等于它在两个向量之间的角度。它可以使用点积（标量积）或叉积（向量积）来获得。请注意，两个向量之间的角度保持在 0° 和 180° 之间。可以使用两种方法找到向量之间的角度。但最常用的求两个向量之间夹角的公式涉及到标量积。

使用标量（点）积寻找角度

两个向量组合成一个标量积会给你一个数字。标量积可用于定义能量和功之间的关系。在数学中，标量积用于表示力（为向量）在分散（为向量）对象时所做的功。标量积由点 (.) 表示。让，

点积 be (ab)

向量 a = |a| 的大小

矢量 b 的大小 = |b|

向量之间的角度为 θ = Cos ^-1 [(a · b) / (|a| |b|)]

当两个向量通过点积连接时，角度 ፀ 的方向无关紧要。由于 Cos ፀ = Cos (-ፀ) = Cos (2π – ፀ)，因此角度 ፀ 可以通过任一向量之间的差来测量。

使用叉（矢量）积求角度

叉积也可以称为向量积。它是一种向量乘法形式，发生在具有不同种类或性质的两个向量之间。当两个向量相乘，得到的乘积也是一个向量时，得到的向量称为两个向量的叉积或向量积。两个向量相乘产生方向垂直于每个向量的向量积。让，

叉积 be (a × b)

向量 a = |a| 的大小

矢量 b 的大小 = |b|。

|a × b| = |一个| |b|罪θ

向量之间的角度为 θ = Sin ^-1 [|a × b| / (|a| |b|)]

示例问题

问题 1：求两个向量 a = {4, 5} 和 b = {5, 4} 之间的角度。

解决方案：

1. Finding dot product ( a.b) = 4 × 5 + 5 × 4 = 40.
2. Finding vectors magnitude, |a| = $\sqrt{((4×4)+(5×5))}$ = √41, |b| = $\sqrt{((5×5)+(4×4))}$ = √41.
3. Angle between vectors, θ = Cos^-1 [(a · b) / (|a| |b|)] , θ = Cos^-1 [(40) / (√41 × √41)]

Angle between a and b, 

θ = Cos^-1 [(40) / (41)]

问题 2：求两个向量 a = {2, 2} 和 b = {1, 1} 之间的角度。

解决方案：

1. Finding dot product (a.b) = 2 × 1 + 2 × 1 = 4.
2. Finding vectors magnitude, |a| = $\sqrt{((2×2)+(2×2))}$ = √8, |b| = $\sqrt{((1×1)+(1×1))}$ = √2.
3. Angle between vectors, θ = Cos^-1 [(a · b) / (|a| |b|)], θ = Cos^-1 [(4) / (√8 × √2)].

Angle between a and b,

θ = Cos^-1 [(4) / (4)] 

θ = Cos^-1 [1] = 0°.

This means both vectors are overlapping each other and are in same direction.

问题 3：求两个向量 a = i + 2j – k 和 b = 2i + 4j – 2k 之间的角度。

解决方案：

1. Finding dot product (a.b) = 1 × 2 + 2 × 4 + (-1) × (-2) = 2 + 8 + 2 = 12.
2. Finding vectors magnitude, |a| = $\sqrt{((1×1)+(2×2)+(-1×-1))}$ = √6, |b| = $\sqrt{((2×2)+(4×4)+(-2×-2))}$ = √24.
3. Angle between vectors, θ = Cos^-1 [(a · b) / (|a| |b|)] , θ = Cos^-1 [(12) / (√6 × √24)].

Angle between a and b,

θ = Cos^-1 [(12) / (12)]

θ = Cos^-1 [1] = 0°.

This means both vectors are overlapping each other and are in same direction.

问题 4：求两个向量 a = i + 2j – k 和 b = 4j – 2k 之间的角度。

解决方案：

Vector b can be written as, b = 0i + 4j – 2k.

1. Finding dot product (a.b) = 1 × 0 + 2 × 4 + (-1) × (-2) = 0 + 8 + 2 = 10.
2. Finding vectors magnitude, |a| = $\sqrt{((1×1)+(2×2)+(-1×-1))}$ = √6, |b| = $\sqrt{((0×0)+(4×4)+(-2×-2))}$ = √20.
3. Angle between vectors, θ = Cos^-1 [(a · b) / (|a| |b|)] , θ = Cos^-1 [(10) / (√6 × √20)].

Angle between a and b,

θ = Cos^-1 [(10) / (√120)].

问题 5：求两个向量 a = {1, -3} 和 b = {-3, 1} 之间的角度。

解决方案：

1. Finding cross product magnitude |a × b| = √(0)² + (0)² + (-8)² = 8.
2. Finding vectors magnitude, | a| = $\sqrt{((1×1)+(-3×-3))}$ = √10, |b| = $\sqrt{((-3×-3)+(1×1))}$ = √10.
3. Angle between vectors, θ = Sin^-1 [(|a × b|) / (|a| |b|)], θ = Sin^-1 [(8) / (√10 × √10)]

Angle between a and b,

θ = Sin^-1 [(8) / (10)] 

问题 6：求两个向量 a = -3i + j 和 b = -3i + j 之间的角度。

解决方案：

1. Finding dot product (a.b) = (-3) × (-3) + 1 × 1 = 10.
2. Finding vectors magnitude, |a| = $\sqrt{(((-3)×(-3))+(1×1))}$ = √10, |b| = $\sqrt{(((-3)×(-3))+(1×1))}$ = √10
3. Angle between vectors, θ = Cos^-1 [(a · b) / (|a| |b|)], θ = Cos^-1 [(10) / (√10 × √10)].

Angle between a and b,

θ = Cos^-1 [(10) / (10)]

θ = Cos^-1 [1] = 0°

As both vectors are having same value and direction so they are same vectors hence having angle 0 between them.