Planes. Find the equations of the planes in both cartesian and (vector) form.

[dodgedanpei]

Homework Statement

The plane that passes through the point (1, 6, 4) and contains the line
x = 1 + 2t; y = 2 - 3t; z = 3 - t where t is an element of R

Homework Equations

x = 1 + 2t; y = 2 - 3t; z = 3 - t

The Attempt at a Solution

Let L be the solution.
L = (1,6,4) - ?

t = (x -1)/ 2 = (2-y)/3 = 3-z

 

[Mark44]

Homework Statement

The plane that passes through the point (1, 6, 4) and contains the line
x = 1 + 2t; y = 2 - 3t; z = 3 - t where t is an element of R

Homework Equations

x = 1 + 2t; y = 2 - 3t; z = 3 - t

The Attempt at a Solution

Let L be the solution.
L = (1,6,4) - ?

t = (x -1)/ 2 = (2-y)/3 = 3-z

These equations just represent the line that you are given. Instead of just throwing up a bunch of equations, say something about your thought process in finding the equation of the plane.

 

[dodgedanpei]

Well I tried making 2 vectors by using the 3 equations.
I got
vector x = t(2,-3,-1) = (1,2,3)

But the two vectors are meant to be s(0,-4,7) + t(-8,0,25) , where s and t are real numbers.

 

[Mark44]

Well I tried making 2 vectors by using the 3 equations.
I got
vector x = t(2,-3,-1) = (1,2,3)

This doesn't make any sense at all. First off, <1, 2, 3> is a vector from the origin to the point (1, 2, 3) on the line. Second, the vector t<2, -3, -1> = <2t, -3t, -t> is a vector that has the same direction as the line.

There is no value of t for which t<2, -3, -1> = <1, 2, 3>. For every value of t, these vectors point in different directions.

But the two vectors are meant to be s(0,-4,7) + t(-8,0,25) , where s and t are real numbers.

Based on what you're showing is the answer, I don't believe that you have provided all of the information for this problem. If you know a point on a plane, and a line that goes through it, that is not enough information to determine the equation of a unique plane.

What is the complete statement of the problem?