Answer to Question #213617 in Analytic Geometry for Tshego

Question

1.1 find parametric equations of the line that passes through the point P =(2, 0,-1) and is parallel to the vector n =<2, 1, 3>

1.2 find paramedic equations of the line that passes through the points A= (1, 2, - 3) and B =(7, 2, - 4).

1.3 find paramedic equations for the line of intersection of the planes - 5x + y - 2z =3 and 2x - 3y + 5z =-7

Answer 1

QUESTION 1

Let $p.M\left(x,y,z\right)$ another point on the straight line, the equation of which we are looking for.

Then, $\overrightarrow{PM}||\overrightarrow{n}, \,\,\,\overrightarrow{n}=\left(2,1,3\right)-\text{specified by condition}$ .

$\left\{\begin{array}{l} p.P\left(2,0,-1\right)\\[0.3cm] p.M\left(x,y,z\right) \end{array}\right.\rightarrow\overrightarrow{PM}\left(x-2,y,z+1\right)\\[0.3cm] \overrightarrow{PM}||\overrightarrow{n}\rightarrow\frac{x-2}{2}=\frac{y}{1}=\frac{z+1}{3}=t\\[0.3cm] \left\{\begin{array}{l} x-2=2t\\[0.3cm] y=t\\[0.3cm] z+1=3t \end{array}\right.\longrightarrow \left\{\begin{array}{l} x=2+2t\\[0.3cm] y=t\\[0.3cm] z=-1+3t \end{array}\right.$

ANSWER

$\left\{\begin{array}{l} x=2+2t\\[0.3cm] y=t\\[0.3cm] z=-1+3t \end{array}\right.$

QUESRION 2

As we know, the parametric equation of the straight line has the form

$\left\{\begin{array}{l} x=x_0+at\\[0.3cm] y=y_0+bt\\[0.3cm] z=z_0+ct \end{array}\right.,\,\,\,\text{where} \left\{\begin{array}{l} p.M\left(x_0,y_0,z_0\right)-\text{any point on the line}\\[0.3cm] \overrightarrow{n}=\left(a,b,c\right)-\text{direction vector} \end{array}\right.$

In our case, since by the problem statement we know two points $p.A=\left(1, 2, - 3\right)$ and $p.B=\left(7, 2, - 4\right)$ through which the straight line passes, then

$\overrightarrow{n}=\overrightarrow{AB}=\left(7-1,2-2,-4-\left(-3\right)\right)\\[0.3cm] \boxed{\overrightarrow{n}=\left(6,0,-1\right)}$

As an "unknown" $p.M\left(x_0,y_0,z_0\right)$ , I propose to take $p.A=\left(1,2,-3\right)$ from the condition.

Then,

$\left\{\begin{array}{l} x=1+6t\\[0.3cm] y=2\\[0.3cm] z=-3-t \end{array}\right.$

ANSWER

$\left\{\begin{array}{l} x=1+6t\\[0.3cm] y=2\\[0.3cm] z=-3-t \end{array}\right.$

QUESTION 3

From the beginning, we find any point $p.M\left(x_0,y_0,z_0\right)$ on the desired straight line, for this we solve the system of equations

$\left\{\begin{array}{l} - 5x + y - 2z =3\to y=3+5x+2z\\[0.3cm] 2x - 3y + 5z =-7 \end{array}\right.\\[0.3cm] \left\{\begin{array}{l} y=3+5x+2z\\[0.3cm] 2x - 3\cdot\left(3+5x+2z\right) + 5z =-7 \end{array}\right.\\[0.3cm] \left\{\begin{array}{l} y=3+5x+2z\\[0.3cm] 2x - 9-15x-6z + 5z =-7 \end{array}\right.\\[0.3cm] \left\{\begin{array}{l} y=3+5x+2z\\[0.3cm] -13x-z =2\to z=-2-13x \end{array}\right.\\[0.3cm] \left\{\begin{array}{l} y=3+5x+2\cdot\left(-2-13x\right)\\[0.3cm] z=-2-13x \end{array}\right.\\[0.3cm] \left\{\begin{array}{l} y=-1-21x\\[0.3cm] z=-2-13x\\[0.3cm] x=Const \end{array}\right.\\[0.3cm]$

Let $x=0\longrightarrow \boxed{p.M\left(0,-1,-2\right)}$ .

As we know, from the canonical equation of the plane we can find the coordinates of the normal vector as follows

$l : ax+by+cz+d=0\to\overrightarrow{n_l}=\left(a,b,c\right)$

( more information : https://en.wikipedia.org/wiki/Normal_(geometry) )

In our case,

$\left\{\begin{array}{l} l_1 :-5x + y - 2z =3\to\overrightarrow{n_{l_1}}=\left(-5,1,-2\right)\\[0.3cm] l_2 :2x - 3y + 5z =-7\to\overrightarrow{n_{l_2}}=\left(2,-3,5\right) \end{array}\right.$

Then,

$\overrightarrow{n}=\overrightarrow{n_{l_1}}\times\overrightarrow{n_{l_2}}-\text{direction vector}\\[0.3cm]$

$\overrightarrow{n}=\left|\begin{array}{ccc} \overrightarrow{i}&\overrightarrow{j}&\overrightarrow{k}\\[0.3cm] -5&1&-2\\[0.3cm] 2&-3&5 \end{array}\right|=\\[0.3cm] =\overrightarrow{i}\cdot\left(5-6\right)-\overrightarrow{j}\cdot\left(-25+4\right)+\overrightarrow{k}\cdot\left(15-2\right)=\\[0.3cm] =-1\cdot\overrightarrow{i}+21\cdot\overrightarrow{j}+13\cdot\overrightarrow{k}\\[0.3cm] \boxed{\overrightarrow{n}=\left(-1,21,13\right)}\\[0.3cm] \left\{\begin{array}{l} p.M\left(0,-1,-2\right)\\[0.3cm] \overrightarrow{n}=\left(-1,21,13\right) \end{array}\right.\longrightarrow \left\{\begin{array}{l} x=-t\\[0.3cm] y=-1+21t\\[0.3cm] z=-2+13t \end{array}\right.$

ANSWER

$\left\{\begin{array}{l} x=-t\\[0.3cm] y=-1+21t\\[0.3cm] z=-2+13t \end{array}\right.$