Consider the line through the points (3,2,5) and (1,1,1). Consider the plane A that is perpendicular to this line, and passing through the point (-1,0,2). Consider the plane B that passes through the points (1,-1,0), (0,2,0), and (0,5,2). Consider the intersection of plane A and plane B. Write the equation of the line of intersection.
A lot of considering taking place here.
plane A:
direction of given first line = (2,1,4)
so that becomes the normal to the plane and its
equation must be 2x + y + 4z = C
but it passes through (-1,0,2), so
-2 + 0 + 8 = C = 6
PLane A: 2x + y + 4z = 6
Plane B:
two directions are (1,-3,0) and (0,3,2)
the cross product is (6,2,-3) , which is the direction of the normal, so
6x + 2y - 3z = K
(1,-1,0) is on it, thus 6 - 2 + 0 = k = 4
Plane B : 6x + 2y -3z = 4
double equation of plane A
4x + 2y + 8z = 12
equation of plane B
6x + 2y - 3z = 4
subtract them:
2x - 11z = -8
x = (11z -8)/2
let z = 0, x = -4,
in plane A:
-8 + y + 0 = 6
y = 14 -----> point on line is (-4, 14, 0)
let z = 2, x = 7
in plane A:
14 + y +8 = 6
y = -16 -----> another point on the line is (7, -16, 2)
Now you have two points on your line.
Use whichever method you use to find the line
Thank you so much!
To find the equation of the line of intersection between plane A and plane B, we first need to determine the normal vectors of both planes.
1. Finding the normal vector of plane A:
Since plane A is perpendicular to the line passing through (3,2,5) and (1,1,1), we can use the direction vector of this line as the normal vector of plane A. The direction vector is obtained by subtracting the coordinates of the two points: (3-1, 2-1, 5-1) = (2, 1, 4). Therefore, the normal vector of plane A is (2, 1, 4).
2. Finding the normal vector of plane B:
To find the normal vector of plane B, we can use the cross product of two vectors lying on the plane. We can choose the vectors formed by the points (1,-1,0) and (0,2,0), and (1,-1,0) and (0,5,2) respectively.
Let's call the first vector u: u = (1,-1,0) - (0,2,0) = (1,-3,0)
And the second vector v: v = (1,-1,0) - (0,5,2) = (1,-6,-2)
Now, we can find the cross product of u and v to obtain the normal vector of plane B:
n = u x v = (6,2,3)
3. Finding the equation of the line of intersection:
The equation of a line can be written in parametric form as:
x = x0 + at
y = y0 + bt
z = z0 + ct
where (x0, y0, z0) is a point on the line, and (a, b, c) is the direction vector of the line.
Let's denote the coordinates of a point on the intersection line as (x, y, z). Since this point lies on both plane A and plane B, it will satisfy the equations of both planes.
For plane A:
The equation of plane A can be written as Ax + By + Cz + D = 0, where A, B, C are the coefficients of the normal vector (2, 1, 4), and D = -Ax0 - By0 - Cz0. Let's substitute the coordinates of the point (-1, 0, 2) into this equation:
2x + y + 4z + D = 0
2(-1) + 0 + 4(2) + D = 0
-2 + 8 + D = 0
D = -6
So, the equation of plane A is 2x + y + 4z - 6 = 0.
For plane B:
The equation of plane B can be written as Ex + Fy + Gz + H = 0, where E, F, G are the coefficients of the normal vector (6, 2, 3), and H = -Ex0 - Fy0 - Gz0. Let's substitute the coordinates of the point (1, -1, 0) into this equation:
6x + 2y + 3z + H = 0
6(1) + 2(-1) + 3(0) + H = 0
6 - 2 + H = 0
H = -4
So, the equation of plane B is 6x + 2y + 3z - 4 = 0.
By solving these two equations simultaneously, we can find the line of intersection between plane A and plane B.