Determine the line through which the planes in each pair intersect.
a) 3x+2y+5z=4
4x-3y+z=-1
http://www.jiskha.com/display.cgi?id=1283793096
1st times 3: --> 9x + 6y + 15z = 12
2nd times 2: --> 8x - 6y + 2z = -2
add them: 17x + 17z = 10 , we wanted to eliminate one of the variables
x+z = 10/17
Pick any value of z to get z
let z=0, then x = 10/17
in#2: 40/17 - 3y + 0 = -1
-3y = -57/17
y = 19/17 -------->a point is (10/17, 19/17, 0)
let x = 0 , then z = 10/17
in #2: 0 - 3y + 10/17 = -1
-3y = -27/17
y = 9/17 --------> a point (0, 9/17 , 10/17)
no we have two points on our line, not "nice" points, but hey ....
direction vector: (10/17 - 0 , 19/17 - 9/17 , 0 - 10/17)
= (10/17, 0 , -10/17)
or we could just use (10, 10, -10)
or even better:
(1, 1, -1)
so using the point (10/17, 19/17,0)
we have the parametric equation:
x = 10/17 + t
y = 19/17 + t
z = -t
be aware that this equation is not unique , but the direction vector must be a multiple of (1,1,-1)
To determine the line through which the planes in the given pair intersect, we need to find the direction vector of the line.
Step 1: Convert the given equations of the planes into their standard form Ax + By + Cz = D.
a) The first equation: 3x + 2y + 5z = 4 is already in standard form.
b) The second equation: 4x - 3y + z = -1 needs to be rearranged by moving the constant term to the other side: 4x - 3y + z + 1 = 0.
So, the standard equation for the second plane is 4x - 3y + z = -1.
Step 2: Determine the direction vector of the line of intersection by finding the cross product of the normal vectors of the planes.
The normal vector of the first plane is (3, 2, 5).
The normal vector of the second plane is (4, -3, 1).
To find the direction vector, take the cross product of these vectors:
Direction vector = (3, 2, 5) × (4, -3, 1)
= (2*1 - 5*(-3), 5*4 - 3*1, 3*(-3) - 2*4)
= (17, 17, -17).
Therefore, the line of intersection of the given pair of planes has a direction vector of (17, 17, -17).
To determine the line through which the planes intersect, we need to find the direction vector of this line.
First, let's express the equations of the planes in vector form by rearranging the equations:
Plane 1: 3x + 2y + 5z = 4 -> [3, 2, 5] · [x, y, z] = 4
Plane 2: 4x - 3y + z = -1 -> [4, -3, 1] · [x, y, z] = -1
In vector form, any point on the line of intersection can be represented as [x, y, z] = [x₀, y₀, z₀] + t[d₁, d₂, d₃], where [x₀, y₀, z₀] is a point on the line and [d₁, d₂, d₃] is the direction vector of the line.
Now, we can set up a system of equations to find the values for [x₀, y₀, z₀] and [d₁, d₂, d₃] that solve both planes simultaneously.
[3, 2, 5] · ([x₀, y₀, z₀] + t[d₁, d₂, d₃]) = 4
[4, -3, 1] · ([x₀, y₀, z₀] + t[d₁, d₂, d₃]) = -1
Expanding these equations, we get:
(3x₀ + 4t·d₁) + (2y₀ - 3t·d₂) + (5z₀ + t·d₃) = 4
(4x₀ + 4t·d₁) + (-3y₀ - 3t·d₂) + (z₀ + t·d₃) = -1
Simplifying, we can rewrite these equations as:
(3x₀ + 4t·d₁) + (2y₀ - 3t·d₂) + (5z₀ + t·d₃) - 4 = 0
(4x₀ + 4t·d₁) + (-3y₀ - 3t·d₂) + (z₀ + t·d₃) + 1 = 0
Now, equating the coefficients of each term to zero, we have:
3x₀ + 4t·d₁ + 2y₀ - 3t·d₂ + 5z₀ + t·d₃ = 4
4x₀ + 4t·d₁ - 3y₀ - 3t·d₂ + z₀ + t·d₃ = -1
Separating the variables, we get:
3x₀ + 2y₀ + 5z₀ + t[4d₁ - 3d₂ + d₃] = 4
4x₀ - 3y₀ + z₀ + t[4d₁ - 3d₂ + d₃] = -1
Since the coefficients of t need to be equal, we can write:
4d₁ - 3d₂ + d₃ = 0
Solving this equation will give us the direction vector of the line of intersection.
From the system of equations:
3x₀ + 2y₀ + 5z₀ = 4
4x₀ - 3y₀ + z₀ = -1
We can solve for [x₀, y₀, z₀] by any desired method, such as using elimination or substitution.
Once we have the values of [x₀, y₀, z₀], we can substitute them back into either of the original equations and solve for t.
After obtaining the values of t, we can determine the direction vector [d₁, d₂, d₃] by equating the coefficients of t in both equations.
Once we have the direction vector of the line of intersection, we can express it as two equations using parameter t, such as:
x = x₀ + t·d₁
y = y₀ + t·d₂
z = z₀ + t·d₃
Finally, the line of intersection of the two planes will be described by these equations.