Find a plane parallel to a line and passing trough a given point

Question

I have to find a plane $\alpha$ parallel to the line: $$ \ell = \begin{cases} x+z=0 \\ x+3y=0 \end{cases} $$ and (the plane $\alpha$) passing trough $P=(1,1,1)$

Maybe I'm failing but I think the following method should be correct: If I take one of the equations of the line and generalize it in a form of a plane this's should be a good candidate since it i's parallel to the line. So this plane should look to somehting like this (I took the first equation forming $\ell$):

$\alpha: x+z+d=0$

and if I impose the condition of passing trough $P$ I can calulate the value of $d$:

$1+1+d=0, d = -2$

then $\alpha : x+z-2=0$ shoudl be a correct plane... (one of many to be coeherent with the selected answer).

Where did I fail? :-) [Ok I was not wrong after all but I needed a confirmation of my hypothesis]

Answer 1

The complaints against your solution seem to be considering the fact that there are many possible planes that solve the problem that your solution does not find.

But if the question was indeed to find a plane parallel to the given line $\ell$ and passing through the given point, your method is fine.

Any plane solving the problem will pass through $(1,1,1)$ and will be parallel to some plane that contains the line $\ell$. And vice versa, any plane through $(1,1,1)$ parallel to a plane containing the line $\ell$ is a solution to the problem.

The plane $x+z=0$ contains the line $\ell$. The plane $x+z-2=0$ contains the point $(1,1,1)$ and is parallel to $x+z=0,$ so it is parallel to $\ell$.

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For a more general solution, any linear combination of the two equations $x+z=0$ and $x+3y=0$ is the equation of a plane through line $\ell$. The linear combinations take two parameters but due to the equivalence of each equation with any multiple of itself, there is really only one degree of freedom. You can write $$ (x+z)\sin\theta + (x+3y)\cos\theta = 0 $$ for a constant $\theta$ in the range $[0,2\pi)$; every plane containing $\ell$ is described by one such equation. Now let a parallel plane be described by $$ (x+z)\sin\theta + (x+3y)\cos\theta +c = 0, $$ let $(x,y,z) = (1,1,1)$ and solve for $c.$

Answer 2

Your answer is correct, but you got lucky. You have no reason to assume that that plane is parallel to the given line. You only used the information that the line is contained in the plane $x+z=0$, but at no point you used the information that it is also contained in the plane $x+3y=0$.

Consider the points $(0,0,0)$ and $(-3,1,3)$, both of which belong to the line. A plane $ax+by+cz=d$ will contain the point $(1,1,1)$ and will be parallel to the line if and only if it also contains the point $(1,1,1)+(-3,1,3)-(0,0,0)=(-2,2,4)$. So, consider the system$$\left\{\begin{array}{l}a+b+c=d\\-2a+2b+4c=d\end{array}\right.$$One solution will be the one that you got: $a=1$, $b=0$, $c=1$, and $d=2$.