公式

向量的坐标运算

坐标加减对应分量运算,模长 √(x²+y²)。

教材原文

a+b=(x1i+y1j)+(x2i+y2j)=x1i+x2i+y1j+y2j=(x1+x2)i+(y1+y2)j,\begin{array}{r l} \boldsymbol {a} + \boldsymbol {b} & = (x _ {1} \boldsymbol {i} + y _ {1} \boldsymbol {j}) + (x _ {2} \boldsymbol {i} + y _ {2} \boldsymbol {j}) \\ & = x _ {1} \boldsymbol {i} + x _ {2} \boldsymbol {i} + y _ {1} \boldsymbol {j} + y _ {2} \boldsymbol {j} \\ & = (x _ {1} + x _ {2}) \boldsymbol {i} + (y _ {1} + y _ {2}) \boldsymbol {j}, \end{array}

a+b=(x1+x2,y1+y2).\boldsymbol {a} + \boldsymbol {b} = (x _ {1} + x _ {2}, y _ {1} + y _ {2}).

同理可得

ab=(x1x2,y1y2).\boldsymbol {a} - \boldsymbol {b} = (x _ {1} - x _ {2}, y _ {1} - y _ {2}).

这就是说,两个向量和(差)的坐标分别等于这两个向量相应坐标的和(差).

如图6.3-12,作向量 OA\overrightarrow{OA}OB\overrightarrow{OB} ,则

AB=OBOA=(x2,y2)(x1,y1)=(x2x1,y2y1).\begin{array}{r l} \overrightarrow {A B} & = \overrightarrow {O B} - \overrightarrow {O A} \\ & = (x _ {2}, y _ {2}) - (x _ {1}, y _ {1}) \\ & = (x _ {2} - x _ {1}, y _ {2} - y _ {1}). \end{array}

因此,一个向量的坐标等于表示此向量的有向线段的终点的坐标减去起点的坐标.

λa=λ(xi+yj)=λxi+λyj,\lambda \boldsymbol {a} = \lambda (x \boldsymbol {i} + y \boldsymbol {j}) = \lambda x \boldsymbol {i} + \lambda y \boldsymbol {j},

λa=(λx,λy).\lambda \boldsymbol {a} = (\lambda x, \lambda y).

这就是说,实数与向量的积的坐标等于用这个实数乘原来向量的相应坐标.

因为 a=x1i+y1j,b=x2i+y2j,a = x_{1}i + y_{1}j, b = x_{2}i + y_{2}j,

ab=(x1i+y1j)(x2i+y2j)=x1x2i2+x1y2ij+y1x2ji+y1y2j2\boldsymbol {a} \cdot \boldsymbol {b} = \left(x _ {1} \boldsymbol {i} + y _ {1} \boldsymbol {j}\right) \cdot \left(x _ {2} \boldsymbol {i} + y _ {2} \boldsymbol {j}\right) = x _ {1} x _ {2} \boldsymbol {i} ^ {2} + x _ {1} y _ {2} \boldsymbol {i} \cdot \boldsymbol {j} + y _ {1} x _ {2} \boldsymbol {j} \cdot \boldsymbol {i} + y _ {1} y _ {2} \boldsymbol {j} ^ {2}

ii=1,jj=1,ij=ji=0,i \cdot i = 1, j \cdot j = 1, i \cdot j = j \cdot i = 0,

所以 ab=x1x2+y1y2.a \cdot b = x_{1}x_{2} + y_{1}y_{2}.

这就是说,两个向量的数量积等于它们对应坐标的乘积的和.

由此可得

(1) 若 a=(x,y)\boldsymbol{a}=(x,y) ,则 a2=x2+y2\left|\boldsymbol{a}\right|^{2}=x^{2}+y^{2} ,或 a=x2+y2\left|\boldsymbol{a}\right|=\sqrt{x^{2}+y^{2}} .

如果表示向量 aa 的有向线段的起点和终点的坐标分别为 (x1,y1)(x_{1},y_{1})(x2,y2)(x_{2},y_{2}) ,那么

a=(x2x1,y2y1),\boldsymbol {a} = (x _ {2} - x _ {1}, y _ {2} - y _ {1}),

a=(x2x1)2+(y2y1)2.\mid \boldsymbol {a} \mid = \sqrt {(x _ {2} - x _ {1}) ^ {2} + (y _ {2} - y _ {1}) ^ {2}}.

(2) 设 a=(x1,y1)\boldsymbol{a}=(x_{1}, y_{1}) , b=(x2,y2)\boldsymbol{b}=(x_{2}, y_{2}) , 则

abx1x2+y1y2=0.\boldsymbol {a} \perp \boldsymbol {b} \Leftrightarrow x _ {1} x _ {2} + y _ {1} y _ {2} = 0.

a,ba, b 都是非零向量,a=(x1,y1)a = (x_1, y_1)b=(x2,y2)b = (x_2, y_2)θ\thetaaabb 的夹角,根据向量数量积的定义及坐标表示可得

cosθ=abab=x1x2+y1y2x12+y12x22+y22.\cos \theta = \frac {\boldsymbol {a} \cdot \boldsymbol {b}}{| \boldsymbol {a} | | \boldsymbol {b} |} = \frac {x _ {1} x _ {2} + y _ {1} y _ {2}}{\sqrt {x _ {1} ^ {2} + y _ {1} ^ {2}} \sqrt {x _ {2} ^ {2} + y _ {2} ^ {2}}}.

公式

a=(x1,y1)\vec{a}=(x_1,y_1)b=(x2,y2)\vec{b}=(x_2,y_2)

a±b=(x1±x2, y1±y2),a=x12+y12.\vec{a}\pm\vec{b}=(x_1\pm x_2,\ y_1\pm y_2),\qquad |\vec{a}|=\sqrt{x_1^2+y_1^2}.

用法

  • 相反向量:b=(x2,y2)-\vec{b}=(-x_2,-y_2)(大小相等、方向相反)。
  • 把物理/几何向量放进坐标系后,一切加减与求模都化为分量算术。

学它之前先会

1 条前置、最深 1 层。源文件只声明直接前置,长链由前置边构建期递归派生(ADR-0016)。

以下内容以本条为前置

学会本条之后能往哪走——由前置边反向派生,无手写清单(ADR-0002)。

4 道题考到本条——由攻略的正向声明反向派生,无手写清单(ADR-0002/0016)。