Question

scrivi le equazioni parametriche e cartesiane delle rette passanti per p e con vettore di direzione v.
p( - 3;0;0), v( - 2;1;2).
p(2;0;0), v(1; - 1;3).
p(0;5; - 1), v(0;2; - 2).
p(2; - 4;1), v(3;5;1).

Explanation:

Step1: Recall parametric line equations

The parametric equations of a line passing through a point $P(x_0,y_0,z_0)$ and having direction vector $\vec{v}=(a,b,c)$ are given by $x = x_0+at$, $y=y_0 + bt$, $z=z_0+ct$.

Step2: For the first - case $P(-3,0,0)$ and $\vec{v}=(-2,1,2)$

Substitute $x_0=-3,y_0 = 0,z_0 = 0,a=-2,b = 1,c = 2$ into the parametric equations. We get $x=-3-2t$, $y=t$, $z = 2t$.
To find the Cartesian equations, from $t=y$, substitute $t$ into the other two equations. We have $x=-3-2y$ and $z = 2y$.

Step3: For the second - case $P(2,0,0)$ and $\vec{v}=(1,-1,3)$

Substitute $x_0 = 2,y_0=0,z_0 = 0,a = 1,b=-1,c = 3$ into the parametric equations. We get $x=2+t$, $y=-t$, $z = 3t$. From $t=-y$, substitute $t$ into the other two equations. We have $x=2 - y$ and $z=-3y$.

Step4: For the third - case $P(0,5,-1)$ and $\vec{v}=(0,2,-2)$

The parametric equations are $x = 0+0t=0$, $y=5 + 2t$, $z=-1-2t$. From $t=\frac{y - 5}{2}$, substitute into the $z$ - equation: $z=-1-(y - 5)=-y + 4$.

Step5: For the fourth - case $P(2,-4,1)$ and $\vec{v}=(3,5,1)$

The parametric equations are $x=2+3t$, $y=-4 + 5t$, $z=1+t$. From $t=z - 1$, substitute into the $x$ and $y$ equations: $x=2+3(z - 1)=3z-1$ and $y=-4 + 5(z - 1)=5z-9$.

Answer:

For $P(-3,0,0),\vec{v}=(-2,1,2)$: Parametric: $x=-3-2t,y=t,z = 2t$; Cartesian: $x=-3-2y,z = 2y$.
For $P(2,0,0),\vec{v}=(1,-1,3)$: Parametric: $x=2+t,y=-t,z = 3t$; Cartesian: $x=2 - y,z=-3y$.
For $P(0,5,-1),\vec{v}=(0,2,-2)$: Parametric: $x = 0,y=5 + 2t,z=-1-2t$; Cartesian: $x = 0,z=-y + 4$.
For $P(2,-4,1),\vec{v}=(3,5,1)$: Parametric: $x=2+3t,y=-4 + 5t,z=1+t$; Cartesian: $x=3z-1,y=5z-9$.