Let be \((O, \overrightarrow{i}, \overrightarrow{j}, \overrightarrow{k})\) an orthonormal coordinate system in space.
Parametric equation of a straight line
The parametric equation of a straight line \(\mathcal{D}\) in space, passing through a point \(A(x_0, y_0, z_0)\) and directed by a vector \(\vec{u}\begin{pmatrix} a\\ b \\c \end{pmatrix} \) (with \(a, b, c \) all three non-zero) is:
$$ \forall (x, y, z) \in \hspace{0.05em}\mathbb{R}^3, $$
$$ M(x, y, z) \in \mathcal{D}(A, \vec{u}) \hspace{0.1em} \Longleftrightarrow \hspace{0.1em} \exists^{\infty} t \in \mathbb{R}, \enspace \begin{Bmatrix} x = at + x_0 \\ y = bt + y_0 \\z = ct + z_0 \end{Bmatrix} $$
$$ with \enspace \Biggl \{ \begin{align*} (x_0, y_0, z_0) \in \hspace{0.05em} \mathbb{R}^3 \\ (a, b, c) \in \hspace{0.05em} \mathbb{R}^3 \ three \ real\ numbers\ which \ are \ non-zero \ simultaneously \end{align*} $$
The equation of a place \((\mathcal{P})\) in space, passing through a point \(A(x_0, y_0, z_0)\) and orthogonal to a vector \(\vec{n}\begin{pmatrix} a\\ b \\c \end{pmatrix}\) (with \(a, b, c \) all three non-zero) is:
$$ \forall (x, y, z) \in \hspace{0.05em}\mathbb{R}^3, $$
$$ M(x, y, z) \in \mathcal{P}(A, \vec{n}) \hspace{0.1em} \Longleftrightarrow \hspace{0.1em} ax + by + cz + d = 0$$
$$ with \enspace \Biggl \{ \begin{align*} (a, b, c) \in \hspace{0.05em} \mathbb{R}^3 \ three \ real\ numbers\ which \ are \ non-zero \ simultaneously \\ d = -ax_0 - by_0 -c z_0 \end{align*} $$
Distance from a point to a plane
Let \(\mathcal{P}\) be a plane orthogonal to a vector \(\vec{n}\begin{pmatrix} a\\ b \\c \end{pmatrix}\) (with \(a, b, c \) all three non-zero), having as equation:
$$ \forall (x, y, z) \in \hspace{0.05em}\mathbb{R}^3, \ ax + by + cz + d = 0$$
The distance from a point \(A(x_0, y_0, z_0)\) in relation to this plane \((\mathcal{P})\) orthogonally projecting on this plane at point \(H(x, y, z)\) is worth:
$$ d(A, \mathcal{P}) = \frac{\Bigl | -ax_0 - by_0 -c z_0 -d \Bigr |}{\sqrt{a^2 + b^2 + c^2}} $$
$$ with \enspace \Biggl \{ \begin{align*} (x_0, y_0, z_0) \in \hspace{0.05em} \mathbb{R}^3 \\ (a, b, c) \in \hspace{0.05em} \mathbb{R}^3 \ three \ real\ numbers\ which \ are \ non-zero \ simultaneously \\ d = -ax - by -c z \end{align*} $$
Projection of a sum of vectors upon a plane
The projection of a sum of vectors is the sum of each vector's projection
$$ \forall n \in \mathbb{N}, \ \forall (\vec{u_1}, \vec{u_2}, \ ..., \vec{u_n}), $$
$$ proj\left(\sum_{k=0}^n \overrightarrow{ u_k} \right) = \sum_{k=0}^n proj(\overrightarrow{u_k})$$
The sphere \((\mathcal{S})\) having a radius \(R\) and centered at point \(A(x_0, y_0, z_0)\) has for equation in space:
$$ \forall (x, y, z) \in \hspace{0.05em}\mathbb{R}^3, $$
$$ M \in \mathcal{S}(A, R) \hspace{0.1em} \Longleftrightarrow \hspace{0.1em} R^2 = (x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2 $$
$$ (with \enspace (x_0, y_0, z_0) \in \hspace{0.05em} \mathbb{R}^3) $$
The cylinder\((\mathcal{C})\) having a radius \(r\) and centered at point \(A(x_0, y_0, z_0)\) has for equation in space:
$$ \forall (x, y) \in \hspace{0.05em}\mathbb{R}^2, $$
$$ M \in \mathcal{C}(A, r) \hspace{0.1em} \Longleftrightarrow \hspace{0.1em} r^2 = (x-x_0)^2 + (y-y_0)^2 $$
$$ (with \enspace (x_0, y_0) \in \hspace{0.05em} \mathbb{R}^2) $$
The cone \((\mathcal{C})\) having a half-angle \( \theta\), and centered at point \(A(x_0, y_0, z_0)\) has for equation in space:
$$ \forall (x, y, z) \in \hspace{0.05em}\mathbb{R}^3, $$
$$ M \in \mathcal{C}(A, r) \hspace{0.1em} \Longleftrightarrow \hspace{0.1em} (x- x_0)^2 + (y- y_0)^2 = k(z-z_0)^2 $$
$$ with \enspace \Biggl \{ \begin{align*} (x_0, y_0, z_0) \in \hspace{0.05em} \mathbb{R}^3 \\ k = tan^2(\theta) \end{align*} $$
From cartesian to spherical coordinates
$$ \Biggl \{ \begin{align*} x = R \ cos(\varphi) \ cos(\theta) \\ y = R \ cos(\varphi) \ sin(\theta) \\ z = R \ sin(\varphi) \end{align*} \qquad (\theta : longitude- \varphi : latitude) $$
$$with \enspace \left \{ \begin{align*} R =\sqrt{x^2 +y^2 +z^2 } \\ \theta = arctan \left( \frac{y}{x} \right) \\ \varphi = arcsin \left( \frac{z}{R} \right) \end{align*} \right \}$$
$$ \Biggl \{ \begin{align*} x = R \ sin(\psi) \ cos(\theta) \\ y = R \ sin(\psi) \ sin(\theta) \\ z = R \ cos(\psi) \end{align*} \qquad (\theta : longitude- \psi : colatitude) $$
$$with \enspace \left \{ \begin{align*} R =\sqrt{x^2 +y^2 +z^2 } \\ \theta = arctan \left( \frac{y}{x} \right) \\ \psi = arccos \left( \frac{z}{R} \right) \end{align*} \right \}$$
Let \((\mathcal{D})\) be a straight line in space, passing through a point \(A(x_0, y_0, z_0)\) and directed by a vector \(\vec{u}\begin{pmatrix} a\\ b \\c \end{pmatrix} \) (with \((a, b, c) \in \hspace{0.05em} \mathbb{R}^3 \) three real numbers which are non-zero simultaneously).
As \(M \in \mathcal{D}\), so the vectors \(\vec{u}\) and \(\overrightarrow{AM}\) are collinear. Thus:
$$ M(x, y, z) \in \mathcal{D}(A, \vec{u}) \hspace{0.1em} \Longleftrightarrow \hspace{0.1em} \overrightarrow{AM} = t \times \overrightarrow{u}$$
$$ \Longleftrightarrow \exists^{\infty} t \in \mathbb{R}, \enspace \begin{pmatrix} x -x_0 \\ y - y_0 \\z - z_0 \end{pmatrix} = \hspace{0.1em} \begin{pmatrix} ta \\ tb \\tc \end{pmatrix} $$
$$ \exists^{\infty} t \in \mathbb{R}, \enspace \begin{Bmatrix} x -x_0 = at\\ y - y_0 = bt \\z - z_0 = ct \end{Bmatrix} $$
The parametric equation of a straight line \(\mathcal{D}\) in space, passing through a point \(A(x_0, y_0, z_0)\) and directed by a vector \(\vec{u}\begin{pmatrix} a\\ b \\c \end{pmatrix} \) (with \((a, b, c) \in \hspace{0.05em} \mathbb{R}^3 \) three real numbers which are non-zero simultaneously) is:
$$ \forall (x, y, z) \in \hspace{0.05em}\mathbb{R}^3, $$
$$ M(x, y, z) \in \mathcal{D}(A, \vec{u}) \hspace{0.1em} \Longleftrightarrow \hspace{0.1em} \exists^{\infty} t \in \mathbb{R}, \enspace \begin{Bmatrix} x = at + x_0 \\ y = bt + y_0 \\z = ct + z_0 \end{Bmatrix} $$
$$ with \enspace \Biggl \{ \begin{align*} (x_0, y_0, z_0) \in \hspace{0.05em} \mathbb{R}^3 \\ (a, b, c) \in \hspace{0.05em} \mathbb{R}^3 \ three \ real\ numbers\ which \ are \ non-zero \ simultaneously \end{align*} $$
The parameter \(t\) is a free parameter that can take any value.
In fact, the line \((\mathcal{D})\) extending to infinity, we obtain a formation of this latter by varying \(t\), leading to the intersection of three planes corresponding respectively to the three equations; which forms a series of points in space.
Let \((\mathcal{P})\) be a plane in space, passing through a point \(A(x_0, y_0, z_0)\) and orthogonal to a vector \(\vec{n}\begin{pmatrix} a\\ b \\c \end{pmatrix}\) (with \((a, b, c) \in \hspace{0.05em} \mathbb{R}^3 \) three real numbers which are non-zero simultaneously).
So, any point \(M(x, y, z)\) belonging to this plane is orthogonal to \(\vec{n}\).
$$ M(x, y, z) \in \mathcal{P}(A, \vec{n}) \hspace{0.1em} \Longleftrightarrow \hspace{0.1em} \overrightarrow{MA} \perp \vec{n}$$
Two orthogonal vectors have their scalar product worthing zero.
$$ \Longleftrightarrow \hspace{0.1em} \overrightarrow{MA} .\vec{n} = 0$$
$$ \Longleftrightarrow \begin{pmatrix} x -x_0 \\ y - y_0 \\z - z_0 \end{pmatrix} . \hspace{0.1em} \begin{pmatrix} a \\ b \\c \end{pmatrix} = 0$$
$$ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 $$
$$ ax - ax_0 + by - by_0 + cz -c z_0 = 0 $$
$$ ax + by + cz \hspace{0.1em} \underbrace{-ax_0 - by_0 -c z_0} _\text{\( (d \hspace{0.1em} \in \hspace{0.05em} \mathbb{R})\)} \hspace{0.1em} = 0 $$
$$ ax + by + cz + d = 0 $$
The equation of a place \((\mathcal{P})\) in space, passing through a point \(A(x_0, y_0, z_0)\) and orthogonal to a vector \(\vec{n}\begin{pmatrix} a\\ b \\c \end{pmatrix}\) is:
$$ \forall (x, y, z) \in \hspace{0.05em}\mathbb{R}^3, $$
$$ M(x, y, z) \in \mathcal{P}(A, \vec{n}) \hspace{0.1em} \Longleftrightarrow \hspace{0.1em} ax + by + cz + d = 0$$
$$ with \enspace \Biggl \{ \begin{align*} (a, b, c) \in \hspace{0.05em} \mathbb{R}^3 \ three \ real\ numbers\ which \ are \ non-zero \ simultaneously \\ d = -ax_0 - by_0 -c z_0 \end{align*} $$
Let \((\mathcal{P})\) be a plane in space, orthogonal to a vector \(\vec{n}\begin{pmatrix} a\\ b \\c \end{pmatrix}\)(with \((a, b, c) \in \hspace{0.05em} \mathbb{R}^3 \) three real numbers which are non-zero simultaneously).
Let \(A(x_0, y_0, z_0)\) be a point orthogonally projecting on \((\mathcal{P})\) at point \(H(x, y, z)\).
We are trying to estimate the distance \(AH\), shortest distance from the point \(A\) to the plane \((\mathcal{P})\).
By the definition of the scalar product, we do have:
$$ \overrightarrow{AH} .\vec{n} = ||\overrightarrow{AH} || \times ||\overrightarrow{n} || \times cos(\overrightarrow{AH}, \overrightarrow{n}) $$
$$ \overrightarrow{AH} .\vec{n} = AH \times \sqrt{a^2 + b^2 + c^2} \times (\pm 1) \qquad (1)$$
On the other hand, with the calculation of the scalar product by the coordinates product, we do have:
$$ \overrightarrow{AH} .\vec{n} = a(x-x_0) + b(y-y_0) + c(z-z_0) $$
$$ \overrightarrow{AH} .\vec{n} = \hspace{0.1em} \underbrace{ax + by + c z} _\text{\( -d \)} \ -ax_0 - by_0 -c z_0 $$
$$ \overrightarrow{AH} .\vec{n} = -ax_0 - by_0 -c z_0 -d \qquad (2) $$
Given that \((1)\) and \((2)\) are equal, being the smae scalar product, we now have:
$$ AH \times \sqrt{a^2 + b^2 + c^2} \times (\pm 1) = -ax_0 - by_0 -c z_0 -d $$
Calculant une distance, on peut passer en valeur absolue :
$$ \Bigl | AH \times \sqrt{a^2 + b^2 + c^2} \times (\pm 1) \Bigr | = \Bigl | -ax_0 - by_0 -c z_0 -d \Bigr | $$
$$AH = \frac{\Bigl | -ax_0 - by_0 -c z_0 -d \Bigr |}{\Bigl |\sqrt{a^2 + b^2 + c^2} \times (\pm 1) \Bigr |} $$
$$AH = \frac{\Bigl | -ax_0 - by_0 -c z_0 -d \Bigr |}{\sqrt{a^2 + b^2 + c^2}} $$
The distance from a point \(A(x_0, y_0, z_0)\) in relation to this plane \((\mathcal{P})\) orthogonally projecting on this plane at point \(H(x, y, z)\) is worth:
$$ d(A, \mathcal{P}) = \frac{\Bigl | -ax_0 - by_0 -c z_0 -d \Bigr |}{\sqrt{a^2 + b^2 + c^2}} $$
$$ with \enspace \Biggl \{ \begin{align*} (x_0, y_0, z_0) \in \hspace{0.05em} \mathbb{R}^3 \\ (a, b, c) \in \hspace{0.05em} \mathbb{R}^3 \ three \ real\ numbers\ which \ are \ non-zero \ simultaneously \\ d = -ax - by -c z \end{align*} $$
In this part, the term projection will always means orhtogonal projection.
Let \((\mathcal{P})\) be a plane having as normal vector \(\vec{n}\begin{pmatrix} a\\ b \\c \end{pmatrix}\), and two non-null vectors being non-collinear to this plane, \(\vec{u}\begin{pmatrix} x_1\\ y_1 \\z_1 \end{pmatrix}\) and \(\vec{v}\begin{pmatrix} x_2\\ y_2 \\z_2\end{pmatrix}\).
Let us call \(\vec{u'}\) and \(\vec{v'}\) the respective projections of the two vectors \(\vec{u}\) and \(\vec{v}\) upon this plane.
For the sake of simplicity, we placed the points \(A, B, C\) as well as their respective projection \(A', B', C'\).
Now, let \(\vec{w} = \vec{u} + \vec{v}\), the sum of vectors \(\vec{u}\) and \(\vec{v}\), and the vector \(\vec{w'}\) being the projection of \( \vec{w}\) on this plane.
In the case of the initial points \(A, B, C\) and those of their respective projection \(A', B', C'\), the Chasles relation applied:
$$ \Biggl \{ \begin{align*} \vec{u} + \vec{v} = \vec{w} \\ \vec{u}' + \vec{v}' = \ \vec{w}' \end{align*} $$
Therefore,
The projection of a sum of vectors is the sum of each vector's projection
$$ \forall n \in \mathbb{N}, \ \forall (\vec{u_1}, \vec{u_2}, \ ..., \vec{u_n}), $$
$$ proj\left(\sum_{k=0}^n \overrightarrow{ u_k} \right) = \sum_{k=0}^n proj(\overrightarrow{u_k})$$
Let \((\mathcal{S})\) be a sphere having a radius \(R\) and centered at point \(A(x_0, y_0, z_0)\).
On this sphere, every point every poin \(M(x, y, z)\) is equidistant from the point \(A\), which is worth the length of the radius \(R\).
We know that using the the Pythagorean theorem, that the distance \( AB\) in a three-dimensional space is worth:
$$\forall (A, B) \in \hspace{0.05em} (O, \vec{x}, \vec{y}, \vec{z})^2, $$
$$AB =\sqrt{(x_b - x_a)^2 + (y_b - y_a)^2 + (z_b - z_a)^2 }$$
So,
$$ M \in \mathcal{S}(A, R) \hspace{0.1em} \Longleftrightarrow \hspace{0.1em} AM = \sqrt{(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2 }$$
$$ R = \sqrt{(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2 }$$
$$ R^2 = (x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2 $$
The sphere \((\mathcal{S})\) of radius \(R\) and centered at point \(A(x_0, y_0, z_0)\) has for equation in space:
$$ M \in \mathcal{S}(A, R) \hspace{0.1em} \Longleftrightarrow \hspace{0.1em} R^2 = (x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2 $$
$$ (with \enspace (x_0, y_0, z_0) \in \hspace{0.05em} \mathbb{R}^3) $$
Let \((\mathcal{C})\) be a vertical cylinder of radius \(r\) and centered at point \(A(x_0, y_0, z_0)\).
In the plane \((O, \overrightarrow{i}, \overrightarrow{j})\), the cylinder describes a circle.
We know that using the the Pythagorean theorem, that the distance \( AB\) in a plane is worth:
$$\forall (A, B) \in \hspace{0.05em} (O, \vec{x}, \vec{y})^2, $$
$$AB = \sqrt{ (x_b - x_a)^2 + (y_b - y_a)^2} $$
We then have,
$$ M \in \mathcal{C}(A, r) \hspace{0.1em} \Longleftrightarrow \hspace{0.1em} AM = \sqrt{(x-x_0)^2 + (y-y_0)^2 }$$
$$ r = \sqrt{(x-x_0)^2 + (y-y_0)^2 }$$
$$ r^2 = (x-x_0)^2 + (y-y_0)^2 $$
The vertical axis \(z\) can take any value, so it is a free variable.
The cylinder \((\mathcal{C})\) having a radius \(r\) and centered at point \(A(x_0, y_0, z_0)\) has for equation in space:
$$ M \in \mathcal{C}(A, r) \hspace{0.1em} \Longleftrightarrow \hspace{0.1em} r^2 = (x-x_0)^2 + (y-y_0)^2 $$
$$ (with \enspace (x_0, y_0) \in \hspace{0.05em} \mathbb{R}^2) $$
Let \((\mathcal{C})\) a vertical cone have as half-angle \( \theta\), and centered at point \(A(x_0, y_0, z_0)\).
In the figure below, we can see that:
$$ M \in \mathcal{C}(A, \theta) \hspace{0.1em} \Longleftrightarrow \hspace{0.1em} tan(\theta) = \frac{AM'}{AM} $$
$$ \Longleftrightarrow \hspace{0.1em} tan(\theta) = \frac{\sqrt{ (x- x_0)^2 + (y- y_0)^2} }{|z-z_0|} $$
$$ tan^2(\theta) = \ \frac{ (x- x_0)^2 + (y- y_0)^2 }{(z-z_0)^2} $$
$$ (x- x_0)^2 + (y- y_0)^2 = (z-z_0)^2 tan^2(\theta) $$
The cone \((\mathcal{C})\) have as half-angle \( \theta\), and centered at point \(A(x_0, y_0, z_0)\) \(A(x_0, y_0, z_0)\) has for equation in space:
$$ M \in \mathcal{C}(A, r) \hspace{0.1em} \Longleftrightarrow \hspace{0.1em} (x- x_0)^2 + (y- y_0)^2 = k(z-z_0)^2 $$
$$ with \enspace \Biggl \{ \begin{align*} (x_0, y_0, z_0) \in \hspace{0.05em} \mathbb{R}^3 \\ k = tan^2(\theta) \end{align*} $$
In some cases, it can be useful to use spherical coordinates (in two dimensions : polar coordinates), especially when working with spheres.
In a usual orthonormal reference frame in space \((O, \overrightarrow{i}, \overrightarrow{j}, \overrightarrow{k})\), we represented a point \(M(x, y, z)\).
Let \(R\) representing the distance from this point to the origin and \(R'\) its projection on the horizontal plane \((O, \overrightarrow{i}, \overrightarrow{j})\).
Likewise, we called \(\theta\) the angle formed by the abscissa axis \((O, \overrightarrow{i})\) and \(R'\), and also \(\varphi\) the angle formed \(R'\) and \(R\).
$$ \Biggl \{ \begin{align*} \theta = (\overrightarrow{Ox}, \overrightarrow{R'}) \\ \varphi= (\overrightarrow{R'}, \overrightarrow{R})\end{align*} $$
In the following figure, we have represented the point \(M'\), projection of point \(M\) on the horizontal plane \((O, \overrightarrow{i}, \overrightarrow{j})\).
With the classic trigonometry rules, we easily see that:
$$ \Biggl \{ \begin{align*} z = R \ sin(\varphi) \\ R' = R \ cos(\varphi) \end{align*} $$
Now, working with a front view of the horizontal plane \((O, \overrightarrow{i}, \overrightarrow{j})\), we can calculate the coordinates corresponding to \(x\) and \(y\).
We do have :
$$ \Biggl \{ \begin{align*} x = R \ cos(\varphi) \ cos(\theta) \\ y = R \ cos(\varphi) \ sin(\theta) \end{align*} $$
Finally, the distance \(R\) is easily calculated from cartesian coordinates \((x, y, z)\):
$$R =\sqrt{x^2 +y^2 +z^2 }$$
As well, we obtain both angles \(\theta\) and \(\varphi\) by:
$$ \Biggl \{ \begin{align*} \theta = arctan \left( \frac{y}{x} \right) \\ \varphi = arcsin \left( \frac{z}{R} \right) \end{align*} $$
So,
$$ \Biggl \{ \begin{align*} x = R \ cos(\varphi) \ cos(\theta) \\ y = R \ cos(\varphi) \ sin(\theta) \\ z = R \ sin(\varphi) \end{align*} \qquad (\theta : longitude- \varphi : latitude) $$
$$with \enspace \left \{ \begin{align*} R =\sqrt{x^2 +y^2 +z^2 } \\ \theta = arctan \left( \frac{y}{x} \right) \\ \varphi = arcsin \left( \frac{z}{R} \right) \end{align*} \right \}$$
here is another way to represent spherical coordinates; we can use colatitude instead of latitude.
We then have a new angle \(\psi\) which starts from the vertical axis \(\overrightarrow{Oz}\) towards \(R\) in the antitrigonometric direction.
$$ \Biggl \{ \begin{align*} \theta = (\overrightarrow{Ox}, \overrightarrow{R'}) \\ \varphi= (\overrightarrow{Oz}, \overrightarrow{R})\end{align*} $$
Which gives us, applying the same reasoning:
$$ \Biggl \{ \begin{align*} x = R \ sin(\psi) \ cos(\theta) \\ y = R \ sin(\psi) \ sin(\theta) \\ z = R \ cos(\psi) \end{align*} \qquad (\theta : longitude- \psi : colatitude) $$
$$with \enspace \left \{ \begin{align*} R =\sqrt{x^2 +y^2 +z^2 } \\ \theta = arctan \left( \frac{y}{x} \right) \\ \psi = arccos \left( \frac{z}{R} \right) \end{align*} \right \}$$
Let us find the intersection \((\mathcal{P} \cap \mathcal{P}') \) between the two planes \((\mathcal{P}) \) and \((\mathcal{P}') \).
$$ (\mathcal{S}) \ \Biggl \{ \begin{align*} \ \ 3x \ - \ y \ + z \hspace{1.8em}= 0 \qquad (\mathcal{P}) \\ -2x +2y + z + 1 = 0 \qquad (\mathcal{P}') \end{align*} $$
We scale the system \((\mathcal{S})\) performing the linear combination \( 2(\mathcal{P}) \) + \( 3(\mathcal{P}') \).
$$ (\mathcal{S}) \ \Biggl \{ \begin{align*} 3x - y \ + z = 0 \hspace{5em} (\mathcal{P}) \\ \ \ \ \ 4y +5 z = -3 \qquad \qquad (2(\mathcal{P}) + 3(\mathcal{P}') ) \end{align*} $$
The system is of the rank \(2\), then the intersection is a straight line which we will note \( (\mathcal{D})\). Moreover, the parameter \( z \) is a free variable and:
$$z = \frac{4y-3}{5} \Longleftrightarrow y = \frac{-5z-3}{4} $$
So, we solve the system by going backwards and we find:
$$ \left \{ \begin{align*} x = \frac{-3z-1}{4} \\ y = \frac{-5z-3}{4} \\ z= z \end{align*} \right \} $$
We choose as value \( z = 0 \) to fix a point on the line \( (\mathcal{D})\). So \( A\left(-\frac{1}{4},-\frac{3}{4}, 0 \right) \in \mathcal{D}\).
Furthermore, we see that the vector \(\vec{u}\begin{pmatrix} -\frac{3}{4} \\ - \frac{5}{4} \\ \hspace{1em} 1 \end{pmatrix}\) leads the straight line \( (\mathcal{D})\), then the latter has the equation:
$$ M(x, y, z) \in \mathcal{D}(A, \vec{u}) \hspace{0.1em} \Longleftrightarrow \hspace{0.1em} \exists^{\infty} t \in \mathbb{R}, \enspace \begin{Bmatrix} x + \frac{1}{4} = -\frac{3}{4}t\\ y + \frac{3}{4} = -\frac{5}{4}t \\z = t \end{Bmatrix} $$
So,
$$(\mathcal{P} \cap \mathcal{P}')=(\mathcal{D}) \Longleftrightarrow \exists^{\infty} t \in \mathbb{R}, \enspace \begin{Bmatrix} x = -\frac{3}{4}t - \frac{1}{4}\\ y = -\frac{5}{4}t -\frac{3}{4} \\z = t \end{Bmatrix} $$
Let us find the intersection \((\mathcal{D} \cap \mathcal{P}) \) between a line \((\mathcal{D}) \) and a plane \((\mathcal{P}) \).
$$ (\mathcal{D}) \ \Biggl \{ \begin{align*} x = t-1 \\ y = -2t + 3 \\ z= -t+5 \end{align*} $$
$$ 2x + y -3z + 6 = 0 \qquad (\mathcal{P})$$
We inject the coordinates of \((\mathcal{D}) \) in those of \((\mathcal{P}) \), and we solve \((L) \) :
$$ 2(t-1) -2t + 3 -3(-t+5) + 6 = 0 \qquad (L)$$
$$ 2t-2-2t+3+3t -15 +6 = 0 \qquad (L)$$
$$ 3t-8 = 0 \Longleftrightarrow t = \frac{8}{3} \qquad (L)$$
As \(t \) is unique, there is a point of intersection, which we will note \(A\).
We thus determine this point using the parametric equation of the line \((\mathcal{D}) \).
Then this point of intersection is:
$$ (\mathcal{D} \cap \mathcal{P}) = A\left(\frac{5}{3}; -\frac{7}{3}; \frac{7}{3} \right)$$
Let us find the intersection \((\mathcal{D} \cap \mathcal{D}') \) between two lines \((\mathcal{D}) \) and \((\mathcal{D}') \).
$$ (\mathcal{D}) \ \Biggl \{ \begin{align*} x = 2t-1 \\ y = -t-2 \\ z= t+2 \end{align*} $$
$$ (\mathcal{D}') \ \Biggl \{ \begin{align*} x = s+1 \\ y = 3s-1\\ z= 2s+1\end{align*} $$
We do the following system:
$$ (\mathcal{D} \cap \mathcal{D}') \ \Longleftrightarrow \Biggl \{ \begin{align*} 2t-1 = s+1 \\ -t-2 = 3s-1\\ t +2 = 2s+1\end{align*} $$
$$ (\mathcal{D} \cap \mathcal{D}') \ \Longleftrightarrow \ \Biggl \{ \begin{align*} 2t-s = 5 \ \ \qquad (L_1) \\ -t-3s = 1 \qquad (L_2) \\ t-2s = -1 \qquad (L_3) \end{align*} $$
We first solve the first two equations. We carry out \( (L_1 + 2 L_2)\):
$$-7s = 7 \qquad (L_1 + 2 L_2) \ \Longrightarrow \ s = -1 \ \Longrightarrow \ t = 2 $$
This solution is not suitable for the third one \( (L_3)\) because:
$$2 - 2\times(-1) = 4\neq -1 $$
So, as there is no pair \( (t, s) \in \hspace{0.05em} \mathbb{R}^2\) which are suitable for the two parametric equations of the two lines \((\mathcal{D}) \) and \((\mathcal{D}') \), they have no point of intersection.
$$(\mathcal{D} \cap \mathcal{D}') = \emptyset $$
Let us find the intersection \((\mathcal{P} \cap \mathcal{S}) \) between a plane\((\mathcal{P}) \) and a sphere \((\mathcal{S}) \).
$$ z = \frac{1}{2} \qquad (\mathcal{P}) $$
$$ x^2 + y^2 + z^2 = 1 \qquad (\mathcal{S}) $$
We now inject \((\mathcal{P}) \) into \((\mathcal{S}) \):
$$ x^2 + y^2 = \frac{3}{4} $$
So, we get an equation for the coordinates \((x,y) \in \hspace{0.05em} \mathbb{R}^2 \). On the other hand, it is the equation of a cone of radius \(R = \frac{\sqrt{3}}{2}\), so it is then necessary to keep the equation of the plane \((\mathcal{P}) \) to fix the cone and therefore obtain a circle.
So, the intersection between \((\mathcal{P}) \) and \((\mathcal{S}) \) is modelled by the double equation:
$$ (\mathcal{P} \cap \mathcal{S}) \ \Longleftrightarrow \ \Biggl \{ \begin{align*} x^2 + y^2 = \frac{3}{4} \\ z = \frac{1}{2} \end{align*} $$
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