Let S be the plane defined by x−4y−3z = −18, and let T be the plane defined by −2x+8y+7z = 41. Find the vector equation for the line where S and T intersect.
x 0 0
y = 0 + t0
z 0 0
To find the vector equation for the line where S and T intersect, we need to solve the system of equations formed by the two plane equations.
First, let's write the equations of the planes in standard form:
For the plane S: x - 4y - 3z = -18 --> x - 4y - 3z + 18 = 0
For the plane T: -2x + 8y + 7z = 41 --> -2x + 8y + 7z - 41 = 0
Now, we can set up the system of equations:
x - 4y - 3z + 18 = 0
-2x + 8y + 7z - 41 = 0
To solve this system, we can use the method of elimination.
Multiply the first equation by 2 and the second equation by 1 to eliminate the x variable:
2(x - 4y - 3z + 18) = 0
-2x + 8y + 7z - 41 = 0
Expanding and simplifying, we get:
2x - 8y - 6z + 36 = 0
-2x + 8y + 7z - 41 = 0
Now, adding these two equations together, we can eliminate the x and y variables:
(2x - 8y - 6z + 36) + (-2x + 8y + 7z - 41) = 0
Simplifying:
-z - 5z - 5 = 0
Rearranging the terms:
-6z = 5
Dividing both sides by -6:
z = -5/6
Now, we can substitute this value of z back into one of the original equations to solve for the remaining variables. Let's use the equation for plane S:
x - 4y - 3z + 18 = 0
Substituting z = -5/6:
x - 4y - 3(-5/6) + 18 = 0
Simplifying:
x - 4y + 5/2 + 18 = 0
x - 4y + 41/2 = 0
Let's solve for x:
x = 4y - 41/2
Now, substituting the values for x and z into the equation for y:
y = 0 + t
Finally, combining the equations for x, y, and z, we get the vector equation for the line where S and T intersect:
r = <4y - 41/2, y, -5/6>, where y is a parameter.
To find the vector equation for the line where planes S and T intersect, we need to find a point on this line and a direction vector for the line.
First, let's find a point on the line of intersection. To do this, we need to find a solution that satisfies the equations of both planes S and T.
Let's solve the system of equations formed by the equations of planes S and T:
x - 4y - 3z = -18 (Equation of plane S)
-2x + 8y + 7z = 41 (Equation of plane T)
To solve this system, we can use the method of elimination. Let's eliminate x:
Multiply the equations by appropriate constants to make the coefficients of x in both equations equal:
2 * (x - 4y - 3z) = -2 * (-18)
-2 * (-2x + 8y + 7z) = -2 * 41
2x - 8y - 6z = 36
4x - 16y - 14z = -82
Now, add the two equations to eliminate x:
(2x - 8y - 6z) + (4x - 16y - 14z) = 36 + (-82)
6x - 24y - 20z = -46
Simplifying the equation:
6x - 24y - 20z = -46
Divide through by 2 to simplify further:
3x - 12y - 10z = -23
Now we have an equation in terms of x, y, and z.
Next, let's solve for x in terms of y and z:
3x = 12y + 10z - 23
x = (12y + 10z - 23) / 3
Now, let's let y = 0 and z = 0:
x = (12*0 + 10*0 - 23) / 3
x = -23/3
Therefore, a point on the line of intersection is (-23/3, 0, 0).
Now, let's find a direction vector for the line of intersection.
The direction vector of the line of intersection is orthogonal (perpendicular) to the normal vectors of planes S and T.
The normal vector of plane S is the coefficients of x, y, and z in the equation of plane S: (1, -4, -3).
Similarly, the normal vector of plane T is the coefficients of x, y, and z in the equation of plane T: (-2, 8, 7).
To find the direction vector, we can take the cross product of the normal vectors of the planes.
Direction vector = (1, -4, -3) × (-2, 8, 7)
Using the cross product formula:
i j k
1 -4 -3
-2 8 7
= (8*-3 - (-4*7), -2*7 - (1*8), 1*8 - (-4*-2))
= (-24 + 28, -14 - 8, 8 - 8)
= (4, -22, 0)
Therefore, the direction vector of the line of intersection is (4, -22, 0).
Now, we can write the vector equation for the line of intersection:
r = (-23/3, 0, 0) + t(4, -22, 0)
So, the vector equation for the line where planes S and T intersect is:
x = -23/3 + 4t
y = -22t
z = 0
1st one times 2
2x - 8y - 6z = -36
2nd
-2x + 8y + 7z= 41
add them
z = 5
sub into 1st x - 4y - 15 = -18
x - 4y = -3
x = 4y - 3
let y = 0 , x = -3, z = 5 ---> point (0,-3,5)
let y = 2 , x = 5, z = 5 ---> point (2,5,5)
so direction of line of intersection is (2, 8, 0)
using the point (2,5,5)
x = 2 + 2t
y = 5 + 8t
z = 5