Equation of a plane that is parallel to yz-plane

[Rawhem]

Homework Statement

Find the vector equation of a plane that contains the point P(2,-3,0) and is parallel to the yz-plane

Homework Equations

Vector equation is in the form... Pi: r = point + t(u) + s(v) s,t element of real numbers

The Attempt at a Solution

We know that the direction vectors (u and v) for a plane parallel to the yz-plane don't have x = 0 (no x component).

So the simplest answer to this would be...

Pi: r = P(2,-3,0) + t(0,1,0) + s(0,0,1) s,t element of R

However, another possible answer would be...

Pi: r = P(2,-3,0) + t(0,1,1) + s(0,1,2) s,t element of R

Although the second answer is a little more complex, it defines a plane that is parallel to the yz-plane.

However, my teacher insists that the plane I defined (second answer) isn't parallel to the yz-plane. Is there some way for me to prove that it is? (I am trying to explain my visualization to her but she is very insistent that my answer is wrong).

 

[Dick]

Homework Statement

Find the vector equation of a plane that contains the point P(2,-3,0) and is parallel to the yz-plane

Homework Equations

Vector equation is in the form... Pi: r = point + t(u) + s(v) s,t element of real numbers

The Attempt at a Solution

We know that the direction vectors (u and v) for a plane parallel to the yz-plane don't have x = 0 (no x component).

So the simplest answer to this would be...

Pi: r = P(2,-3,0) + t(0,1,0) + s(0,0,1) s,t element of R

However, another possible answer would be...

Pi: r = P(2,-3,0) + t(0,1,1) + s(0,1,2) s,t element of R

Although the second answer is a little more complex, it defines a plane that is parallel to the yz-plane.

However, my teacher insists that the plane I defined (second answer) isn't parallel to the yz-plane. Is there some way for me to prove that it is? (I am trying to explain my visualization to her but she is very insistent that my answer is wrong).

A plane is parallel to the yz plane if it's tangent vectors are normal to the x unit vector (1,0,0). Show that's true for dPi/ds and dPi/dt. And also show (0,1,1) and (0,1,2) are linearly independent, just to make sure you've got a plane and not a line.

 

[HallsofIvy]

Note, by the way. that, in two dimensions, a line that is "parallel to the x-axis", that is, parallel to the line y= 0, is of the form y= c for some number c. Similarly a line that is "parallel to the y-axis", that is, parallel to the line x= 0, is of the for x= c for some number c.

In three dimensions, the "xy-plane" is the plane z= 0. Any plane parallel to it has equation z= c for some number c.

 

[Rawhem]

Thanks for everyone's help!

The way I proved it to my teacher is by showing that the cross product of the direction vectors of each plane are a multiple of each other.

e.g. n1 = dir1 x dir2
n2 = dir2 x dir3

if n1 = kn2 then Pi: r1 = t(dir1) + s(dir2) is parallel to Pi: r2 = a(dir2) + b(dir3)

 

[haruspex]

To make it parallel to the yz plane you just need x constant. To pass through the given point the constant must be 2. So the generic form is (2, s, t).

 

[HallsofIvy]

Thanks for everyone's help!

The way I proved it to my teacher is by showing that the cross product of the direction vectors of each plane are a multiple of each other.

e.g. n1 = dir1 x dir2
n2 = dir2 x dir3

if n1 = kn2 then Pi: r1 = t(dir1) + s(dir2) is parallel to Pi: r2 = a(dir2) + b(dir3)

Actually, the simplest way to define the cross product is to define \vec{i}\times \vec{j}= \vec{k}, \vec{j}\times\vec{k}= \vec{i}, \vec{k}\times\vec{i}= \vec{j} (i.e. cyclically) and extend to all vectors by defining it to be both "linear" and "anti-commutative".