Question

Даны координаты вершин треугольника АВС относительно декартовой системы координат: A=(-1,8,8), B=(-3,4,4), C=(-5,4,6). Запишите каноническое и параметрическое уравнения прямой Z, на которой расположена биссектриса AК треугольника АВС. В ответ введите координаты точки Р пересечения прямой Z с плоскостью z=-22, разделив их точкой с запятой.

Answer 1

не знаю как у вас там в школе обучают,но у нас так:

Answer 2

I will try to answer your question in detail. Given the coordinates of the vertices of triangle ABC relative to the Cartesian coordinate system: A=(-1,8,8), B=(-3,4,4), C=(-5,4,6). Write the canonical and parametric equations of the line Z, on which the bisector AK of triangle ABC is located. In the answer, enter the coordinates of the point P of intersection of the line Z with the plane z=-22, separating them with a semicolon.

To find the bisector AK of triangle ABC, we need to find the point K that divides the side BC in half. The coordinates of K can be found by averaging the coordinates of B and C:

$$K=\left(\frac{x_B+x_C}{2},\frac{y_B+y_C}{2},\frac{z_B+z_C}{2}\right)=\left(-4,4,5\right)$$

Then, the vector equation of the line Z passing through A and K is:

$$\vec{r}=\vec{r_A}+t\vec{AK}=\left(-1,8,8\right)+t\left(-3,-4,-3\right)$$

The parametric equations of the line Z are:

$$x=-1-3t$$ $$y=8-4t$$ $$z=8-3t$$

The canonical equation of the line Z is:

$$\frac{x+1}{-3}=\frac{y-8}{-4}=\frac{z-8}{-3}$$

To find the point P of intersection of the line Z with the plane z=-22, we substitute z=-22 into the parametric equation for z and solve for t:

$$-22=8-3t$$ $$t=10$$

Then, we substitute t=10 into the parametric equations for x and y and get the coordinates of P:

$$x=-1-3(10)=-31$$ $$y=8-4(10)=-32$$ $$z=-22$$

Therefore, the coordinates of P are (-31;-32;-22).