Find the equation of the plane containing these lines

Question

So I was given these lines: $$\frac{x-1}{2} = \frac{y-2}{-2} = \frac{z}{-1}$$ $$ \frac{x}{-2} = y+\frac{5}{3} = z-\frac{4}{0} $$ And I was asked to find out if they are parallel or perpendicular. Now I found that they are not perpendicular since the determinant of the system is not equal to zero. But I was also asked to find the equation of plane containing these line but honestly don't know even where to start. Any Hints?

Answer 1

Since there's a $\frac{4}{0}$ in the second equation I'll only comment on the general procedure of finding out the solution but not the actual computation.

1)To check if the lines are perpendicular, do a dot product of vectors parallel to it(hint:its 0 if they were perpendicular and the 2 vectors would be scalar multiples if they were parallel). The equation of a line passing through a point $x_1,x_2,x_3$ and parallel to (a,b,c) in coordinate form is- $\frac{x-x_1}{a}=\frac{y-x_2}{b}=\frac{z-x_3}{c}=constant=\lambda (say)$ where x,y,z are any points on the line.

2) Finding out the point of intersection- Write the equation of the line in the parametric form, from above it'll give $(x,y,z)=(a\lambda+x_1,b\lambda +x_2,c\lambda+ x_3)$ and put this into the equation of the second line to find out the point of intersection.

3) Now having a point in the plane and 2 lines passing through it you would be able to write the equation of the plane containing the lines with the following procedure- given 2 lines are $\frac{x-x_1}{a_1}=\frac{y-x_2}{b_1}=\frac{z-x_3}{c_1}$ and $\frac{x-y_1}{a_2}=\frac{y-y_2}{b_2}=\frac{z-y_3}{c_2}$ and they pass through the point ($\alpha_1,\alpha_2,\alpha_3$), then the equation of all the points (x,y,z) in the plane would satisfy $(x-\alpha_1,y-\alpha_2,z-\alpha_3).\{(a_1,b_1,c_1)\times(a_2,b_2,c_2)\}=0$, i.e., this is the equaiton of the plane in coordinate form.(it can be interpreted as saying that the dot product of any vector inside the plane with the normal to the plane is always 0 since the vector is parallel to the plane and hence perpendicular to the normal.

Answer 2

Personally, I prefer to write lines in parametric form. $\frac{x- 1}{2}= \frac{y- 2}{-2}$ is the same as $1- x= y- 2$ or $y= 3- x$. $\frac{x- 1}{2}= \frac{z}{-1}$ is the same as $1- x= 2z$ or $x= 1- 2z$. Taking z as parameter, t, we have $x= 1- 2t$, $y= 3- x= 3- 1+ 2t= 2+ 2t$, $z= t$. For the second line, assuming you mean $\frac{x}{-2}= \frac{y- 5}{3}= \frac{z- 4}{0}$, from $\frac{x}{-2}= \frac{y- 5}{3}$ we get $3x= -2y+ 10$ or $2y= 10- 3x$. Taking parameter s such that $x= 2s$, $2y= 10- 6s$ so $y= 5- 3s$. The "$\frac{z- 4}{0}$" means that z is the constant, $z= 4$.

So the two lines are given by $x= 1- 2t$, $y= 2+ 2t$, $z= t$, and $x= 2s$, $y= 5- 3s$, $z= 4$. They are not parallel because the "direction vectors", -2i+ 2k+ z and 2i- 3s are not parallel- one is not a multiple of the other. The only other possibilities are that they intersect pr are skew. If they intersect then there exist two numbers, s and t, such that $x= 1- 2t= 2s$, $y= 2+ 2t= 5- 3s$ and $z= t= 4$. Setting t= 4 in the x-equation, $x= 1- 8= -7= 2s$ so $s= -7/2$. Setting t= 4 and s= -7/2 in the y-equation, $y= 2+ 8= 5+ 21/2$ which is false. There are no values of s and t that make all three equations true so the lines do not intersect.

The lines are neither parallel nor do they intersect. They are skew. There is no plane that contains both lines.