Cfrgtkky

Find the algebraic equation of the plane that passes through the origin, is parallel to the line X = (-1, 0, 2) +t(2,-1,1), and contains the point (1,2,3).

I suspect you can do this with a cross-product, but our professor isn't letting us use that as he hasn't taught it to us yet. Is there a way to get another point on the plane from the parallel line, so you can make a system of equations to solve for an orthogonal vector?

The equation of a plane is usually of the form $ax+by+cz+d=0$. If the origin lies in the plane (as it does in the question), then $d=0$ and one may write the equation as $langle (a,b,c),(x,y,z)rangle=0$.

Now, we know $(1,2,3)$ lies in the plane. Moreover, for all $tin mathbb{R}$, $tcdot(2,-1,1)$ also lies in the plane. Hence, $(a,b,c)$ must satisfy

$$a+2b+3c=0\2a-b+c=0$$

You probably see this is a system of two equations with $3$ unknowns, so the solution is not unique. That's okay, because for any $alpha in mathbb{R}setminus{0}$ the equations $ax+by+cz=0$ and $alphacdot(ax+by+cz)=0$ are the same.

Follow these steps:

1. The origin $boldsymbol{r}_0$ and the point $ boldsymbol{r}_A $ define a line that belongs to the plane. This line has direction $(boldsymbol{r}_A - boldsymbol{r}_0)$. The normal to the plane must be perpendicular to all the vectors on this plane.




2. Define the normal $boldsymbol{n}$ as the vector that is perpendicular to the line $X$ and the line from the previous step.

$$ boldsymbol{n} = boldsymbol{e} times (boldsymbol{r}_A - boldsymbol{r}_0) $$

_{Here $boldsymbol{e}$ is the direction vector of line $X$. Note that a simplified scalar multiple of $boldsymbol{n}$ would do just fine.}

3. Find the distance $d$ of the plane to the origin. Use the equation of the plane $ boldsymbol{n} cdot boldsymbol{r} = d$ that must be true for all points on the plane. Since the origin is one of the points, it stands to reason that $$ d = boldsymbol{n} cdot boldsymbol{r}_0 = 0 $$




4. The equation of the plane is thus found by
   $$ boldsymbol{n} cdot pmatrix{x\y\z} = 0 $$

_{Here $cdot$ is the vector dot product, and $times$ the vector cross product.}

There are many different ways to solve this problem without cross product. The more straight could posssibly be the following.

Let's call $P=(1,2,3)$.

1) find the pencil of planes containg $O,P$

To do this, we just have to find two planes, not parallel to each other, and passing through those points. We can take for instance a plane parallel to the $z$ axis and passing through the projections of the points on $x,y$ plane: $pi_1; :; 2x-y=0$ . Same way, a plane parallel to $y$ axis: $ pi_2 ; : ; 3x-z=0$.

So the pencil is $0=mu pi_1 + lambda pi_2=(2mu +3lambda) x -mu y - lambda z$.

2)find a plane in the pencil, parallel to the given line

Just take two points on the line, the vector defined by the corresponding segment (in your case it is already given)and impose that it be normal to the normal of the pencil.
You get a simple homogeneous equation in $mu,lambda$. That is
$$
eqalign{
& 2left( {2mu + 3lambda } right) - ( - 1)mu - lambda = 5mu + 5lambda = 0quad Rightarrow cr
& Rightarrow quad mu = - lambda quad Rightarrow quad left{ matrix{
mu = 1 hfill cr
lambda = - 1 hfill cr} right. cr}
$$

Therefore the required plane is
$$
pi :quad - x - y + z = 0
$$
which in fact contains $(0,0,0)$ and $(1,2,3)$ and is parallel to the line because
$(2,-1,1) cdot (-1,-1,1)=0$.

This indicates another way to solve the problem, i.e. the system
$$
ax + by + cz = d;:quad left( {matrix{
0 & 0 & 0 cr
1 & 2 & 3 cr
2 & { - 1} & 1 cr
} } right)left( {matrix{
a cr
b cr
c cr
} } right) = left( {matrix{
d cr
d cr
0 cr
} } right)
$$