Find the Equation of the Plane

[The Student]

Question: Find the equation of a plane which contains the line x = 1+t,y = 2t − 1,z = −2 + 4t and also the point P(1,1,0)
The equation has the form ax+by +cz = d. Substitute the line equations: t should disappear from the left side. Substitute the point equation. One gets three equations in 4 unknowns.

I can do this by finding three different points, but I really want to know what they are instructing; I don't understand the instruction that I put in bold. They seem to want me to create a matrix, but I have no idea what to use for the rows and thus how to find the a's, b's and c's.

 

[The Student]

They just want you to plug your expressions for x,y,z in terms of t into the general equation for the plane.

a(1+t) + b(2t-1) + c(-2+4t) = d

(a+2b+8c)t + (a-b-2c) = d

then clearly

(a+2b+8c) = 0 or t=0, and

(a-b-2c) = d

You can reduce that down a bit and then plug in your point P and reduce it further.

Play with it some more.

Ohhhh, thank-you so much!!!!

 

[pka]

Question: Find the equation of a plane which contains the line x = 1+t,y = 2t − 1,z = −2 + 4t and also the point P(1,1,0) The equation has the form ax+by +cz = d.

The point \(\displaystyle Q: (1,-1,-2)\) is on the given line. WHY?

The vector \(\displaystyle D=<1,2,4>\) is the direction vector of the given line. WHY?

The vector \(\displaystyle \overrightarrow {PQ} \times D\) is the normal of your plane. WHY?

All one needs is a point and a normal to have a plane. HOW?

 

[The Student]

The point \(\displaystyle Q: (1,-1,-2)\) is on the given line. WHY?

We know they are what the coordinates of the line are when t = 0.

The vector \(\displaystyle D=<1,2,4>\) is the direction vector of the given line. WHY?

I am not sure.

The vector \(\displaystyle \overrightarrow {PQ} \times D\) is the normal of your plane. WHY?

I am not sure, but I will assume that this is the result of multiplying vectors to get a perpendicular vector.

All one needs is a point and a normal to have a plane. HOW?

The point locates where the plane is along the normal line. But once again, I am not totally sure about this.

 

[pka]

We know they are what the coordinates of the line are when t = 0.
The vector [FONT=MathJax_Math]D[/FONT][FONT=MathJax_Main]=<[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]>[/FONT] is the direction vector of the given line. WHY? I am not sure.

Surely, you understand about the equation of a line?

 

[The Student]

Surely, you understand about the equation of a line?

I know that the direction and magnitude of a vector is described by its coefficients. But I don't know why it's a direction vector as opposed to a position vector. I don't really know the difference between the two other than how they are used in this area of linear algebra.

 

[The Student]

They just want you to plug your expressions for x,y,z in terms of t into the general equation for the plane.

a(1+t) + b(2t-1) + c(-2+4t) = d

(a+2b+8c)t + (a-b-2c) = d

then clearly

(a+2b+8c) = 0 or t=0, and

(a-b-2c) = d

You can reduce that down a bit and then plug in your point P and reduce it further.

Play with it some more.

I have got as far as the matrix,

|1-1-2 ? |
|1 2 4 ? |
|1 1 0 ? |

But I am struggling with how to know what the fourth unknown should be.

 

[The Student]

Never mind this thread; I get it now.