Equation of a plane parallel to a line and perpendicaular to another plane

In summary: Yes, that's correct. Finding the cross product of the two vectors will give you the norm vector of the plane. Then you can use that vector and the point (2, 5, 7) to write the equation of the plane in scalar form.So, in summary, to find the equation of a plane parallel to a line and perpendicular to a given plane, you can use the direction vector of the line and the norm vector of the given plane to find the norm vector of the new plane, and then use a point on the new plane to write the equation in scalar form.
  • #1
popo902
60
0

Homework Statement


Find the equation of the plane through the point (2, 5, 7) that is parallel
to the line r = (3i + 2j - 2k) + t(i + 2j + 7k) and perpendicular to the plane 4x + 5y + 6z = 14
It should be written in scalar equation form


Homework Equations


scalar equation of a plane through a point: a(x - x0) + b(y - y0) + c(z - Z0) = 0
normal of a plane is (a, b, c)
cross product of two parallel vectors = 0 (v x u = 0)



The Attempt at a Solution



what I have so far is __(x-2) + __(y-5) + __(z-7)
but i can't figure out how I am supposed to find the normal of this plane...
 
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  • #2
The required normal passes through r = (3i + 2j - 2k) + t(i + 2j + 7k) and the normal of 4x + 5y + 6z = 14.

So find the equation of the normal. You will now have a point on the normal and a point on the line r=(3i + 2j - 2k) + t(i + 2j + 7k).

Just find the vector line joining these two points.
 
  • #3
k so the normal of the plane passes through line r and normal of plane "14"
the plane I am looking for is parallel to both of these and perpendicular to plane "14"
so basically look for the normal of the line r?
would the equation be (x-2) + 2(y-5) + 7(z-7) = 0?
 
  • #4
popo902 said:

Homework Statement


Find the equation of the plane through the point (2, 5, 7) that is parallel
to the line r = (3i + 2j - 2k) + t(i + 2j + 7k) and perpendicular to the plane 4x + 5y + 6z = 14
Incorrect, I misread the problem.

It should be written in scalar equation form


Homework Equations


scalar equation of a plane through a point: a(x - x0) + b(y - y0) + c(z - Z0) = 0
normal of a plane is (a, b, c)
cross product of two parallel vectors = 0 (v x u = 0)



The Attempt at a Solution



what I have so far is __(x-2) + __(y-5) + __(z-7)
but i can't figure out how I am supposed to find the normal of this plane...
 
Last edited by a moderator:
  • #5
popo902 said:
k so the normal of the plane passes through line r and normal of plane "14"
the plane I am looking for is parallel to both of these and perpendicular to plane "14"
so basically look for the normal of the line r?
would the equation be (x-2) + 2(y-5) + 7(z-7) = 0?

No, it's wrong.. When working with parallel, and perpendicular property like this, one should think about using direction, lying (on the plane), and norm vector. It goes like this:

Let [itex](\beta)[/itex] be our need-to-find plane, since its parallel to r, the vector (1, 2, 7) (the direction vector of r) must lie on it, right?

And since its perpendicular to the plane "4x + 5y + 6z = 14", another lying-on-[itex](\beta)[/itex] vector must be the norm vector of the 4x + 5y + 6z = 14 plane. Right?

Having 2 (non-parallel) vectors lying on [itex](\beta)[/itex], how can we find out the norm vector of it? Can you go from here? It's pretty simple. :)

HallsofIvy said:
This is impossible. The line r, itself, is not perpendicular to the plane so no line parallel to it can be. Are you sure you are not asked for two different lines?

Well, no HoI, it's a plane parallel to a line, not a line parallel to a line. So this can be possible. :)
 
  • #6
VietDao29 said:
No, it's wrong.. When working with parallel, and perpendicular property like this, one should think about using direction, lying (on the plane), and norm vector. It goes like this:

Let [itex](\beta)[/itex] be our need-to-find plane, since its parallel to r, the vector (1, 2, 7) (the direction vector of r) must lie on it, right?

And since its perpendicular to the plane "4x + 5y + 6z = 14", another lying-on-[itex](\beta)[/itex] vector must be the norm vector of the 4x + 5y + 6z = 14 plane. Right?

Having 2 (non-parallel) vectors lying on [itex](\beta)[/itex], how can we find out the norm vector of it? Can you go from here? It's pretty simple. :)



Well, no HoI, it's a plane parallel to a line, not a line parallel to a line. So this can be possible. :)
Yes, thanks. I misread the problem.
 
  • #7
VietDao29 said:
Let [itex](\beta)[/itex] be our need-to-find plane, since its parallel to r, the vector (1, 2, 7) (the direction vector of r) must lie on it, right?

And since its perpendicular to the plane "4x + 5y + 6z = 14", another lying-on-[itex](\beta)[/itex] vector must be the norm vector of the 4x + 5y + 6z = 14 plane. Right?

Having 2 (non-parallel) vectors lying on [itex](\beta)[/itex], how can we find out the norm vector of it? Can you go from here? It's pretty simple. :)

k sooo...
i find the cross product to those two vectors to find the one that's perpendicular to those two which are on the plane making the orthogonal vector i find also perpedicular to the plane itself...right?
so i got 23(x-2) - 22(y-5) - 3(z-7) = 0
if it's right, dude thanks, i don't know why i dint see this before :0
if it isn't, dude thanks, but there must be something wrong with me :eek:
 
  • #8
The equation of the plane is: Ax+By+Cz+D=0

The normal vector of the plane is <A,B,C>.

Since (2, 5, 7) lies on the plane substitute it in the equation so that:

2A+5B+7C+D=0

Because it is parallel of the line: r = (3i + 2j - 2k) + t(i + 2j + 7k)

the dot product of the directional vector of the line <1,2,7> and the normal vector of the plane <A,B,C> is zero, because they are perpendicular to each other.

<1,2,7>.<A,B,C>=0

A+2B+7C=0

Because it is perpendicular to the plane 4x + 5y + 6z = 14, the normal vectors of both planes are perpendicular, and their dot product is zero.

<A,B,C>.<4,5,6>=0

4A+5B+6C=0

Now just solve the system of equations:

2A+5B+7C+D=0

A+2B+7C=0

4A+5B+6C=0
 
  • #9
popo902 said:
k sooo...
i find the cross product to those two vectors to find the one that's perpendicular to those two which are on the plane making the orthogonal vector i find also perpedicular to the plane itself...right?
so i got 23(x-2) - 22(y-5) - 3(z-7) = 0
if it's right, dude thanks, i don't know why i dint see this before :0
if it isn't, dude thanks, but there must be something wrong with me :eek:

Well, you may have used the wrong formula for cross product, which should be:

Say, a = (a1, a2, a3)
b = (b1, b2, b3)

[tex]\mathbf{c} = \mathbf{a} \times \mathbf{b} = \left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{array} \right|[/tex]

(Be careful not to forget the (-1) factor, when finding the determinant of the above matrix)

After finding the cross product of a, and b, you can always check your result by using Dot Product: (Since the Cross Product of a and b is perpendicular to both of them)

Check to see if the 2 equations below hold:

  • ca = 0
  • cb = 0
 
Last edited:
  • #10
hihi...Vietdao..could u pls show me how to do this question?
The line L has equation r=j+k+s(i-2j+k).The plane p has equation x+2y+3z+5.
(a)Show that the line L lies in the plane p.
(b)A second plane is perpendicular to the plane p,parallel to the line L and contains the point with position vector 2i+j+4k.Find the equation of this plane,giving your answer in the form ax+by+cz=d.

Thank you very much..
 
  • #11
rockmylife89 said:
hihi...Vietdao..could u pls show me how to do this question?
The line L has equation r=j+k+s(i-2j+k).The plane p has equation x+2y+3z+5.
(a)Show that the line L lies in the plane p.
(b)A second plane is perpendicular to the plane p,parallel to the line L and contains the point with position vector 2i+j+4k.Find the equation of this plane,giving your answer in the form ax+by+cz=d.

Thank you very much..

You should start a new thread instead of just posting in an old one like this.

Well.. the forums' rules do not allow providing complete solutions though. So, may I ask what have you try, and how far have you got? So that I can know where you get stuck, and, hence, I can give you more accurate help. :)
 
  • #12
a)Line can be formed by two points. So show that two points of the line are also points on the plane.

b)If the plane ax+by+cz=d is perpendicular to p then the dot product of the normal vector (a,b,c) and the normal vector to the plane p: (1,2,3) is zero:

[tex](a,b,c) \cdot (1,2,3)=0[/tex]Try something and I will continue giving you hints.
 

What is the equation of a plane parallel to a line and perpendicular to another plane?

The equation of a plane parallel to a line and perpendicular to another plane can be written in the form Ax + By + Cz + D = 0, where A, B, and C are the coefficients of the variables x, y, and z, and D is a constant. This type of equation is known as the standard form of a plane.

How do you determine the coefficients A, B, and C in the equation of a plane?

The coefficients A, B, and C can be determined by finding the cross product of the two normal vectors of the given line and plane. The normal vector of a plane is the vector perpendicular to the plane, while the normal vector of a line is the vector perpendicular to the line. The cross product of these two vectors will result in the coefficients A, B, and C.

Can the equation of a plane parallel to a line and perpendicular to another plane be written in other forms?

Yes, the equation can also be written in the parametric form, which involves using two parameters (u and v) to represent the points on the plane. It can also be written in the symmetric form, which involves using the distance from the origin to the plane and the direction of the plane's normal vector.

What is the relationship between the normal vector of a line and the normal vector of a plane?

The normal vector of a line is always parallel to the normal vector of a plane, but they are in opposite directions. This means that the cross product of these two vectors will result in a zero vector, indicating that they are perpendicular to each other.

How can the equation of a plane parallel to a line and perpendicular to another plane be applied in real-world scenarios?

This type of equation is commonly used in engineering and architecture to design structures with specific orientations. For example, in construction, the equation can be used to determine the slope and angle of a roof or the orientation of walls in relation to each other. It can also be used in physics and astronomy to calculate the trajectory of objects or determine the orientation of a planet's orbit in relation to the sun.

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