Let S be the plane defined by 5x+5y−5z = 45, and let T be the plane defined by 2x+3y+z = 22. 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 a point on the line and a direction vector for the line.
First, we need to find a point that lies on both planes. To do this, we set the equations for S and T equal to each other and solve for x, y, and z.
5x + 5y - 5z = 45 (equation for S)
2x + 3y + z = 22 (equation for T)
Setting these equations equal to each other, we have:
5x + 5y - 5z = 2x + 3y + z
3x + 2y - 6z = 0
Now, we can solve this system of equations. We can start by solving for one variable in terms of the other two:
y = (3x + 6z) / 2
x = (2y - 6z) / 3
Now, substitute these expressions for x and y into one of the original equations to solve for z:
5 x + 5 y - 5 z = 45
5((2 y - 6 z) / 3) + 5y - 5z = 45
10y - 30z + 15y - 15z = 135
25y - 45z = 135
Simplifying further:
5y - 9z = 27
Now, we can solve for y in terms of z:
y = (27 + 9z) / 5
Substituting this back into the expression for x:
x = (2y - 6z) / 3
x = (2((27 + 9z) / 5) - 6z) / 3
x = (54 + 18z - 30z) / 15
x = (54 - 12z) / 15
x = (9 - 2z) / 5
Now, we have expressions for x, y, and z in terms of z. We can choose any value to substitute in for z to get the corresponding values for x, y, and z. Let's take z = 1:
x = (9 - 2(1)) / 5 = 7/5
y = (27 + 9(1)) / 5 = 6
z = 1
So, the point (7/5, 6, 1) lies on both planes S and T.
Next, we need to find a direction vector for the line. We can do this by taking the cross product of the normal vectors of planes S and T.
The normal vector of plane S is <5, 5, -5> and the normal vector of plane T is <2, 3, 1>.
Taking the cross product of these vectors gives us:
<5, 5, -5> x <2, 3, 1> = (5(1) - 5(3))i - (5(2) - 5(1))j + (5(3) - 5(2))k
= -10i - 5j + 5k
So, the direction vector of the line is <-10, -5, 5>.
Therefore, the vector equation for the line where S and T intersect is:
R = <7/5, 6, 1> + t<-10, -5, 5>, where t is a parameter.