Let S be the plane defined by 2x+4y+2z = 4, and let T be the plane defined by x+3y+3z = 2. Find the vector equation for the line where S and T intersect.
To find the vector equation for the line where planes S and T intersect, we need to find the point of intersection and the direction vector of the line.
First, let's find the point of intersection.
To find the intersection, we need to solve the system of equations formed by the two plane equations:
2x + 4y + 2z = 4 (Equation 1)
x + 3y + 3z = 2 (Equation 2)
To solve the system, we can use any method (e.g., substitution, elimination, etc.). Here, let's use the elimination method.
Multiply Equation 2 by 2 to eliminate the x term:
2(x + 3y + 3z) = 2(2)
2x + 6y + 6z = 4 (Equation 3)
Now, subtract Equation 1 from Equation 3 to eliminate the x term:
(2x + 6y + 6z) - (2x + 4y + 2z) = 4 - 4
2y + 4z = 0
Simplify the equation further:
y + 2z = 0 (Equation 4)
Let's solve the system formed by Equations 1 and 4:
2x + 4y + 2z = 4 (Equation 1)
y + 2z = 0 (Equation 4)
We can solve this system using the substitution method.
From Equation 4, we have:
y = -2z
Substitute y = -2z into Equation 1:
2x + 4(-2z) + 2z = 4
2x - 8z + 2z = 4
2x - 6z = 4
x - 3z = 2
Now, let's express x and z in terms of a parameter t:
x = 2 + 3z
z = t (let t be a parameter)
Substitute z = t into x = 2 + 3z:
x = 2 + 3t
Now that we have expressions for x, y, and z in terms of a parameter t, let's rewrite the point of intersection in vector form.
The point of intersection lies on both planes S and T, so we can use the coordinates (x, y, z) as the position vector of this point.
Therefore, the position vector of the point of intersection is given by:
r = <x, y, z> = <2 + 3t, -2t, t>
Now, let's find the direction vector of the line of intersection.
To obtain the direction vector, we will consider the normal vectors of the planes S and T.
The normal vector of plane S is given by the coefficients of x, y, and z in its equation:
Vector n_s = <2, 4, 2>
The normal vector of plane T is given by the coefficients of x, y, and z in its equation:
Vector n_t = <1, 3, 3>
Now, since the line lies in both planes, its direction vector will be perpendicular to both normal vectors n_s and n_t.
We can find the direction vector by taking the cross product of the normal vectors n_s and n_t:
Vector v = n_s x n_t
v = <2, 4, 2> x <1, 3, 3>
= <(-1)(3) - (4)(3), (2)(3) - (2)(1), (2)(3) - (2)(1)>
= <-15, 4, 4>
Therefore, the direction vector of the line of intersection is v = <-15, 4, 4>.
In summary, the vector equation for the line where planes S and T intersect is:
r = <2 + 3t, -2t, t> + t<-15, 4, 4>, where t is a parameter.
To find the vector equation for the line where planes S and T intersect, we first need to find a point on this line and a direction vector for the line.
Step 1: Find a point on the line of intersection.
To find a point on the line of intersection, we need to solve the system of equations formed by the equations of planes S and T. Let's solve this system of equations.
2x + 4y + 2z = 4 (Equation 1)
x + 3y + 3z = 2 (Equation 2)
To solve this system, we can use the method of elimination or substitution. In this case, let's use the method of elimination.
First, let's multiply Equation 2 by 2 to eliminate the x term:
2(x + 3y + 3z) = 2(2)
2x + 6y + 6z = 4 (Equation 3)
Next, we can subtract Equation 3 from Equation 1 to eliminate the x term:
(2x + 4y + 2z) - (2x + 6y + 6z) = 4 - 4
-2y - 4z = 0
-2(y + 2z) = 0
y + 2z = 0 (Equation 4)
Now, we have two equations:
y + 2z = 0 (Equation 4)
x + 3y + 3z = 2 (Equation 2)
Let's solve this system of equations.
We can solve Equation 4 for y:
y = -2z
Now we can substitute this expression for y in Equation 2:
x + 3(-2z) + 3z = 2
x - 3z = 2
We can rewrite this equation as:
x = 3z + 2
So, now we have:
x = 3z + 2
y = -2z
Now, let's choose a value for z, substitute it into the equations for x and y, and find the corresponding values for x and y.
Let z = 0:
x = 3(0) + 2 = 2
y = -2(0) = 0
So, when z = 0, we have x = 2 and y = 0. Therefore, a point on the line of intersection is (2, 0, 0).
Step 2: Find the direction vector of the line of intersection.
The direction vector of the line of intersection is perpendicular to both planes S and T. It can be found by taking the cross product of the normal vectors of the two planes.
The normal vector of plane S is (2, 4, 2).
The normal vector of plane T is (1, 3, 3).
Now, let's find the cross product of these two vectors:
(2, 4, 2) × (1, 3, 3) = (6 - 12, 6 - 2, 2 - 12)
= (-6, 4, -10)
So, the direction vector of the line of intersection is (-6, 4, -10).
Step 3: Write the vector equation for the line of intersection.
Now that we have a point on the line of intersection, (2, 0, 0), and a direction vector for the line, (-6, 4, -10), we can write the vector equation for the line of intersection.
The vector equation for a line passing through a point (a, b, c) with direction vector (d, e, f) is given by:
r = <a, b, c> + t<d, e, f>
In our case, the point on the line is (2, 0, 0), and the direction vector is (-6, 4, -10). So, the vector equation for the line of intersection is:
r = <2, 0, 0> + t<-6, 4, -10>
Therefore, the vector equation for the line where planes S and T intersect is:
r = <2 - 6t, 4t, -10t>