Answer to Question #103630 in Analytic Geometry for na

Question

Does there pass a plane through the lines (x+4)÷3 =y÷2 =(z−1)÷3 and x÷2 =(y−1)÷1 =(z+1)÷1?
Justify.

Answer 1

Solution

The plane passes through the lines if the lines are parallel or intersect. The line $(x+4)\div3=y\div2=(z-1)\div3$ has a direction vector$\overrightarrow{s}$ (3, 2, 3) and a point, that belongs to it, is A(-4, 0, 1).

The line $x\div2=(y-1)\div1=(z+1)\div1$ has a direction vector $\overrightarrow{n}$ (2, 1, 1) and a point, that belongs to it, is B(0, 1, -1).

If the lines are parallel, then the coordinates of their direction vectors are proportional:

$3\div2\not =2\div1\not =3\div1$ .

the lines are not parallel.

If the lines intersect, the vectors $\overrightarrow{s}$, $\overrightarrow{n}$, and $\overrightarrow{AB}$ are complanar,

$\overrightarrow{s}*\overrightarrow{n}*\overrightarrow{AB}=0$ ,

$\begin{vmatrix} 0+4 & 1-0&-1-1 \\ 3 & 2&3\\ 2&1&1 \end{vmatrix}=\begin{vmatrix} 4 & 1&-2 \\ 3 & 2&3\\ 2&1&1 \end{vmatrix}=\\=4*2*1+3*1*(-2)+1*3*2-(-2)*2*2-1*3*1-3*1*4=\\=8-6+6+8-3-12=1\not =0$

The lines do not intersect.

Answer:

the plane does not pass through the lines.