Find the vector and cartesian equations of a plane

AI Thread Summary
The discussion revolves around solving vector and Cartesian equations related to planes and lines in three-dimensional space. To prove that a line is parallel to a plane, one must show that the direction vector of the line is perpendicular to the plane's normal vector. For finding the equation of a line perpendicular to a plane, the normal vector can be derived from the plane's coefficients. Additionally, the cosine of the angle between two lines can be determined by analyzing their direction vectors. Finally, any plane containing a specific line can be defined by finding a normal vector that is perpendicular to the line's direction vector.
~angel~
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These are just a few questions that a don't understand and any help would be great.

1. Prove that the line

(x-3)/2 = (y-4)/3 = (z-5)/4

is parallel to the plat 4x + 4y - 5z = 14

2. Find the equation of the line through (1,0,-2) and perpendicular to the plane
3x - 4y + z -6 = 0

I'm assuming you need to find the normal of the plane, but I'm not sure how to do that.

3. This is the 2nd part of a question: Find the cosine of the angle between the directions of the line in (a) and the line

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

The line in (a) is (x-1)i + (y-1)j + (z-2)k = t (3i + j + k), which becomes the cartesian equation:

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

4. Find the vector and cartesian equations of a plane containing the line

(x-4)/-2 = (y+3)/1 = (z-1)/3

I know all the basic things in vectors, but these are a few questions I just want to clear up due to my upcoming exam.

Any help would be greatly appreciated.
 
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For a plane

ax+by+cz = d

The normal vector can be denoted by

\vec{N} = <a,b,c>
 
~Angel~ said:
1. Prove that the line

(x-3)/2 = (y-4)/3 = (z-5)/4

is parallel to the plat 4x + 4y - 5z = 14
As whozum told you, a vector perpendicular to the plane is <4, 4,-5>.
A vector in the direction of the <2, 3, 4> (Set each of those fractions equal to the parameter t and solve for x, y, z). The line will be parallel to the plane if these two vectors are pependicular to one another.

2. Find the equation of the line through (1,0,-2) and perpendicular to the plane
3x - 4y + z -6 = 0

I'm assuming you need to find the normal of the plane, but I'm not sure how to do that.
Whozum's suggestion again. The components of the vector perpendicular to the plane are just the coefficients of the variables.

3. This is the 2nd part of a question: Find the cosine of the angle between the directions of the line in (a) and the line

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

The line in (a) is (x-1)i + (y-1)j + (z-2)k = t (3i + j + k), which becomes the cartesian equation:

(x-1)/3 = (y-1)/1 = (z-2)/1
So you know that a vector in the direction of the line in (a) is <3, 1, 1> and a vector in the direction of the other line is <3, 1, 1>. What do you think the angle between them is?

4. Find the vector and cartesian equations of a plane containing the line

(x-4)/-2 = (y+3)/1 = (z-1)/3

Notice "of a plane". There are an infinite number of planes containing any line- you are only asked to find one. Again, the vector <-2, 1, 3> is in the direction of the line. Any plane containing the line has normal vector perpendicular to that.
You just need to find one vector perpendicular to that and then be sure the plane contains that line.
 
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