I really can't understand what is going on in this example.
I understand the vector and parametric equations of a plane in R3, and I understand the examples below and above this one where they give a point on the plane, and then 2 vectors parallel to the plane.
I thought that the parametric equation is in terms of vectors if in R3 (e.g. x = some point + t1v1 + t2v2. But this equations seems to use x, y and z as individual components of the vector in R3 (or as x y and z coordinates), as apposed to y = v1 and z = v2.
I also do not understand how the coeffecient of y (1) is equated to the coordinate (1,1,0) and how the coeffecient of z is equated to the coordinate (-2,0,1).
Could you please shed some light on this example. It's taken from page 166 of Elementary Linear Algebra 11th ed by Anton (it's in the 10th edition as well).
EXAMPLE 2: Vector and Parametric Equations of a Plane in R3
Find vector and parametric equations of the plane x − y + 2z = 5. Solution: We will find the parametric equations first. We can do this by solving the equation for any one of the variables in terms of the other two and then using those two variables as parameters. For example, solving for x in terms of y and z yields x = 5 + y − 2z (8) and then using y and z as parameters t1 and t2, respectively, yields the parametric equations x = 5 + t1 − 2t2, y= t1, z= t2 To obtain a vector equation of the plane we rewrite these parametric equations as
(x, y, z) = (5 + t1 − 2t2, t1, t2) or, equivalently, as (x, y, z) = (5, 0, 0) + t1(1, 1, 0) + t2(−2, 0, 1)
PLEASE HELP!!!???
Thanks in advance.
To address your first point, you are right: parametric equations are of the form
$$ \vec{r} = \vec{r}_0 + t_1 \vec{v}_1 + t_2 \vec{v}_2 $$
Also, $x,y,$ and $z$ are numbers, not vectors, so it makes perfect sense that $x,y,z$ are the components/coordinates of a vector. It does not make sense to say $y=\vec{v_1}$ or $z=\vec{v}_2$, again because $y$ and $z$ are scalars/numbers, and NOT vectors. In the notation I used above, $\vec{r} = (x,y,z)$.
To address your second point, $y$ and $z$ are not being "equated" with the vectors $(1,1,0)$ and $(-2,0,1)$. What is happening is that $y$ and $z$ are being equated with the parameters $t_1$ and $t_2$ (again as above, this makes more sense, because $y$ and $z$ are numbers, not vectors). The author is simply saying that if you solve the equation for $x$, you get
$$ x = 5 + y - 2z $$
Then they are saying let $y=t_1$ and $z=t_2$. Then $x = 5 + t_1 - 2t_2$. Making these three substitutions gives
$$ (x,y,z) = (5+t_1-2t_2, \, t_1, \, t_2) $$
The last version of the equation (where it says "or equivalently" at the end), is simply obtained by splitting this vector apart and factoring out the scalars (using regular vector addition and scalar multiplication). That is, write it as the sum of three vectors, where the first only contains the constant terms of each slot, the second only contains the $t_1$ terms, and the third only contains the $t_2$ terms:
$$ \begin {eqnarray} (x,y,z) &=& (5+t_1-2t_2, \, t_1, \, t_2) \\ &=& (5,0,0) + (t_1, \, t_1, \, 0) + (-2t_2, \, 0, \, t_2) \\ &=& (5,0,0) + t_1(1,1,0) + t_2(-2,0,1) \end {eqnarray} $$