Alright no problem, it is a notation used in the Netherlands, so my apologies for that!
They are parallel vector equations, you have to read it as if it were parametric equations, so:
x = 2 + lambda
y = 1
z = 1 + lambda
and so on
The initial post of this problem did not mention that the system had no solution. This appeared in your second post.
Your coefficient matrix has either two sign errors or typos. Please double check the work to verify that.
Edit: The coefficient matrix is right after all. My apologies.
I believe that Plato had already mentioned the notation. d(l, V) is an extremely ambiguous statement. And I've never known the variable V to be referred to a plane. Maybe that is also a matter of difference of notation between here and the Netherlands as well. So now what would "d" mean? There are too many possibilities to consider.
Plato was trying to find out exactly what your problem statement meant. He was trying to help and you weren't being helpful. And you threw an insult at him. Not good for getting help in the future. And Plato is "ignorant?" Trust me: He's anything but ignorant.
Please be aware of these matters in the future.
-Dan
Oh please!
d is the standard symbol for a metric (d for distance),
V is the standard symbol for a vector space, and if it wasn't clear, the problem statement clarified that it was a plane,
// or || is the standard symbol for parallel, which indeed means that the system has no solution.
The OP did clarify in his second post without insulting.
After that things went downhill.
My only complaint is that the OP might have shown a little more effort.
Goeien avond!
to a)
If the straight line is parallel to the plane then it must be possible to construct the direction vector of the line by the 2 direction vectors of the plane:
$\displaystyle \begin{pmatrix}1\\0\\1 \end{pmatrix} = \mu \begin{pmatrix}0\\1\\2 \end{pmatrix} + \tau \begin{pmatrix}-1\\a\\1 \end{pmatrix}$
From the first line you see that $\displaystyle \tau = -1$
From the 2nd line you see that $\displaystyle a = 1$ if $\displaystyle \tau = -1$.
Use the 3rd line to check if these two values give a valid solution. Mar dat klopt!
to b)
Transform the equation of the plane into Hesse normal form. Substitute the coordinates of the stationary vector of the straight line into this equation to get the distance of the parallel to the plane.
Thank you all very much for your clear explanations. I would however like to note that the only analytical geometry for these types of problems that needed to be studied for this test, was to learn and use the distance equations for a point to a line or plane and the fact that the shared normal vector of two lines is used to calculate the distance. Do you guys think that is enough knowledge to solve a problem like this, especially question b? I did find a way to calculate problem a without the determinants (which are not yet known to a reader of solely the chapter that was tested), but is question b possible with only the information that I described in the second sentence of this comment?
This is the sort of clarification I was seeking in the first place. It may be that I am usually aware of this because of a project that I am now working on. Before me I have three pre-prints of pre-calculus texts (in North America) that includes what we call analytic geometry that I have been asked to review. Not one of them addresses a plane in parametric form. It is not even discussed in the classic Schaum's Outline series. So the disagreement here is simply a cultural different.
For anyone to claim that any notation is standard is simply naive.
On this side of the pond, $\displaystyle d(x,y)$ is the usual for distance (metric) between two points. Whereas, $\displaystyle D(A;x)$ or $\displaystyle D(A,B)$ denotes the distance between a set and a points or two sets.
It never occurred to me that "//" which is the LaTeX command for linefeed would be used for parallel.
To answer the first quoted sentence.
Lines look like: $\displaystyle \ell:~\frac{x-x_0}{d_x}=\frac{y-y_0}{d_y}=\frac{z-z_0}{d_z}$ where $\displaystyle d_x\cdot d_y\cdot d_z\ne 0$ (we adjust that otherwise).
A plane looks like $\displaystyle \Pi:~Ax+By+Cz+D=0$.
The line is parallel to the plane, $\displaystyle \ell\|\Pi$, iff $\displaystyle Ad_x+Bd_y+Cd_z=0$.
Now for distance: $\displaystyle D(\ell,\Pi)=\frac{|Ax_0+By_0+Cz_0|}{\sqrt{A^2+B^2+ C^2}}$
Just to make sure the point is hammered down the major problem I see here is that this post is in Pre-University, whereas the topic was apparently Linear Algebra. Not a topic you'd usually find in a High School level Geometry class. Even in the Netherlands I would think....$\displaystyle d(x,~y)$ is the usual for distance (metric) between two points. Whereas, $\displaystyle D(A;~x)$ or$\displaystyle D(A,~B)$ denotes the distance between a set and a points or two sets.
-Dan
So you need a normal vector to your plane.
That is, a vector normal to both $\displaystyle \begin{pmatrix}0 \\ 1 \\ 2\end{pmatrix}$ and $\displaystyle \begin{pmatrix}-1 \\ a \\ 1\end{pmatrix}$.
As earboth already showed, this last vector is actually $\displaystyle \begin{pmatrix}-1 \\ 1 \\ 1\end{pmatrix}$.
A normal vector must have a dot product of zero with the original vector.
So a vector normal to $\displaystyle \begin{pmatrix}0 \\ 1 \\ 2\end{pmatrix}$ must be of the form $\displaystyle \mathbf n = \begin{pmatrix}* \\ -2 \\ 1\end{pmatrix}$. (Why?)
To also be normal to $\displaystyle \begin{pmatrix}-1 \\ 1 \\ 1\end{pmatrix}$, it must be $\displaystyle \mathbf n = \begin{pmatrix}-1 \\ -2 \\ 1\end{pmatrix}$. (Why?)
Next we want to find the distance.
Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane.
The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector.
Such a vector is $\displaystyle \begin{pmatrix}2 \\ 2 \\ 2\end{pmatrix} - \begin{pmatrix}2 \\ 1 \\ 1\end{pmatrix} = \begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}$.
Take the projection on the normal unit vector that we will call $\displaystyle \mathbf{\hat n}$:
$\displaystyle d(l,V)=|\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix} \cdot \mathbf{\hat n}| = |\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix} \cdot {\mathbf{n} \over ||\mathbf{n}||}| = |\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix} \cdot {1 \over \sqrt 6}\begin{pmatrix}-1 \\ -2 \\ 1\end{pmatrix}| = {1 \over \sqrt 6}$.
If I can find the notation on wiki (in a general area), I consider it standard.
A metric d(x,y) is defined on any set X, which includes geometric objects.
Whatever the local usual usage is, that makes it unambiguous in this context, which is what I like to see in math, and what is sufficient.
This is a bit off-track, but the LaTeX command for linefeed is actually "\\".It never occurred to me that "//" which is the LaTeX command for linefeed would be used for parallel.
Seeing the level the OP has shown so far, it actually appears to be the case that the OP is pre-university, or otherwise just started.
Furthermore I believe we should not impose our own culture or usual usage of symbols on thread starters.
Rather we should adjust to their usage (if we can!).
Presumably we should be able to adjust to any symbol usage, whereas posters usually have difficulty enough with the problem at hand.
Yeah, I think that is a lack of the book that we use in class. This book introduces these notations before even covering Linear Algebra principles; it begins with vector geometry and analytical geometry using Linear Algebra even though basic matrix calculations and maps like lineair transformation are not explained until later chapters. Thank you guys for all the explanation; I have learned a lot as to how I can interprete the vector notation as Linear Algebra translations of subspaces!