Lines and planes (Scalar equation + pt. of int.)

[Lornt]

so I've been doing review and I got stuck on two questions, I was wondering if anyone could help me out.
Any help would be appericiated - whether it's a hint on what my first step should be or what's the final answer to confirm whether I'm right or wrong :)


1. Find the scalar equation of the plane that passes through the point (1, 1, 4) and is perpendicular to the line of intersection of the planes x + 2y + z = 1 and 2x + y + 3z = 3.


2. Determine the distance from the point P(2, 3, -1) to the plane 2x + y - 2z + 9 = 0. (I know I can plug the values into the equation, but that doesn't really help me. Could anyone show me a longer way to do this involving more steps? I don't really understand where the equation comes from).

 

[Hurkyl]

1. Find the scalar equation of the plane that passes through the point (1, 1, 4) and is perpendicular to the line of intersection of the planes x + 2y + z = 1 and 2x + y + 3z = 3.

Start simple, and build up a solution.

What is the scalar equation equation for a plane?
How do you tell if a plane passes through a point?
What is the scalar equation for a plane passing through that point?
What is the line of intersection of those two planes?
How do you tell if a plane is perpendicular to a line?
What is the scalar equation for a plane perpendicular to that line?
What is the scalar equation for a plane passing through that point and perpendicular to that line?

This is a basic problem solving technique: it's useful in just about anything you do.

2. Determine the distance from the point P(2, 3, -1) to the plane 2x + y - 2z + 9 = 0. (I know I can plug the values into the equation, but that doesn't really help me. Could anyone show me a longer way to do this involving more steps? I don't really understand where the equation comes from).

Starting with definitions is another basic problem solving technique.

What is the definition of the distance from a point to a plane?

 

[HallsofIvy]

so I've been doing review and I got stuck on two questions, I was wondering if anyone could help me out.
Any help would be appericiated - whether it's a hint on what my first step should be or what's the final answer to confirm whether I'm right or wrong :)


1. Find the scalar equation of the plane that passes through the point (1, 1, 4) and is perpendicular to the line of intersection of the planes x + 2y + z = 1 and 2x + y + 3z = 3.


2. Determine the distance from the point P(2, 3, -1) to the plane 2x + y - 2z + 9 = 0. (I know I can plug the values into the equation, but that doesn't really help me. Could anyone show me a longer way to do this involving more steps? I don't really understand where the equation comes from).

We could better "confirm whether I'm right or wrong" if you would show us what you have done.

 

[Kidphysics]

http://www.jtaylor1142001.net/calcjat/Solutions/VPlanes/VPPtLine.htm

little late but help for anyone who types this question into a search engine

 

[CellCoree]

1. To find the scalar equation of the plane, we first need to find the normal vector of the plane. We can do this by finding the cross product of the normal vectors of the two given planes. The normal vector of the first plane is <1, 2, 1> and the normal vector of the second plane is <2, 1, 3>. The cross product of these two vectors is <5, -1, -3>. This vector will be perpendicular to the line of intersection of the two planes.

Next, we can use the point (1, 1, 4) to find the scalar equation of the plane. The scalar equation of a plane is given by Ax + By + Cz = D, where A, B, C are the components of the normal vector and D is the constant term.

Substituting the values from our normal vector and the given point, we get the equation 5x - y - 3z = 5 as the scalar equation of the plane.

2. To find the distance from a point to a plane, we can use the formula d = |Ax + By + Cz - D| / √(A^2 + B^2 + C^2), where A, B, C are the components of the normal vector and D is the constant term of the plane.

In this case, the normal vector is <2, 1, -2> and the constant term is 9. Substituting the values of the point P(2, 3, -1), we get d = |2(2) + 1(3) - 2(-1) - 9| / √(2^2 + 1^2 + (-2)^2) = |4 + 3 + 2 - 9| / √(9) = 0.89 units.

Alternatively, we can also find the equation of the line perpendicular to the plane passing through the point P(2, 3, -1). This line will be parallel to the normal vector of the plane, and its equation can be written as x = 2 + 2t, y = 3 + t, z = -1 - 2t.

Now we can find the point of intersection between this line and the given plane 2x + y - 2z + 9 = 0. Sub