Let be the line x = 1 + 2t, y = -1 + 3t, z = -5 + 7t. Find the two points on at a distance 3 units from the plane 2(x-1) + 2(y+3) –z = 0
To find the two points on the line x = 1 + 2t, y = -1 + 3t, z = -5 + 7t that are at a distance of 3 units from the plane 2(x-1) + 2(y+3) - z = 0, we will solve both equations simultaneously.
Step 1: Equation of the Plane
The equation of the plane is given as 2(x-1) + 2(y+3) - z = 0. We can simplify this equation by expanding the terms:
2x - 2 + 2y + 6 - z = 0
Simplifying further:
2x + 2y - z + 4 = 0
Step 2: Substituting into the Line Equation
We will substitute the expressions for x, y, and z from the line equation into the plane equation:
2(1 + 2t) + 2(-1 + 3t) - (-5 + 7t) + 4 = 0
Simplifying:
2 + 4t - 2 + 6t + 5 - 7t + 4 = 0
Combine like terms:
3t + 9 = 0
Subtracting 9 from both sides:
3t = -9
Dividing by 3:
t = -3
Step 3: Finding the Points
Now that we have the value of t, we can substitute it back into the line equation to find the corresponding points on the line:
For t = -3:
x = 1 + 2(-3) = 1 - 6 = -5
y = -1 + 3(-3) = -1 - 9 = -10
z = -5 + 7(-3) = -5 - 21 = -26
So, the first point is (-5, -10, -26).
For the second point, we will take the negative value of t:
For t = 3:
x = 1 + 2(3) = 1 + 6 = 7
y = -1 + 3(3) = -1 + 9 = 8
z = -5 + 7(3) = -5 + 21 = 16
So, the second point is (7, 8, 16).
Therefore, the two points on the line at a distance of 3 units from the plane 2(x-1) + 2(y+3) - z = 0 are (-5, -10, -26) and (7, 8, 16).
To find the points on the line at a distance 3 units from the plane, we can first find the intersection point of the line and the plane.
Given that the equation of the line is x = 1 + 2t, y = -1 + 3t, and z = -5 + 7t, we can substitute these values into the equation of the plane:
2(x-1) + 2(y+3) - z = 0
2((1+2t)-1) + 2((-1+3t)+3) - (-5+7t) = 0
2(2t) + 2(3t+3) - (-5+7t) = 0
4t + 6t + 6 + 5 - 7t = 0
3t + 11 = 0
t = -11/3
Substituting this value of t back into the equation of the line, we can find the point of intersection:
x = 1 + 2t
x = 1 + 2(-11/3)
x = 1 - 22/3
x = -19/3
y = -1 + 3t
y = -1 + 3(-11/3)
y = -1 - 11
y = -12
z = -5 + 7t
z = -5 + 7(-11/3)
z = -5 - 77/3
z = -92/3
So the intersection point of the line and the plane is (-19/3, -12, -92/3).
To find the two points on the line that are a distance 3 units from this intersection point, we can use the distance formula. Let the coordinates of one of the points be (x, y, z).
The distance between the two points is given by:
√((x - (-19/3))^2 + (y - (-12))^2 + (z - (-92/3))^2 ) = 3
Simplifying this equation gives:
√((3x + 19/3)^2 + (y + 12)^2 + (3z + 92/3)^2 ) = 9
Squaring both sides of the equation to eliminate the square root gives:
(3x + 19/3)^2 + (y + 12)^2 + (3z + 92/3)^2 = 9
Since we are looking for two points, we can consider two cases:
Case 1: Let x = 1 + 2t.
(3(1 + 2t) + 19/3)^2 + (y + 12)^2 + (3( -5 + 7t) + 92/3)^2 = 9
Expanding this equation gives:
(3 + 6t + 19/3)^2 + (y + 12)^2 + (-15 + 21t + 92/3)^2 = 9
(9 + 36t + 38 + 36/3 + 114t + 368/9) + (y + 12)^2 + (-225 + 315t + 1012/3) = 9
(24 + 150t + 368/9) + (y + 12)^2 + (787 + 315t) = 9
(24 + 150t + 368/9) + (y + 12)^2 + (787 + 315t) = 9
Combining like terms gives:
(150t + 800/9) + (y + 12)^2 + (315t + 811) = 9
Rearranging the equation gives:
(y + 12)^2 + (150t + 800/9) + (315t + 811) - 9 = 0
(y + 12)^2 + (150t + 800/9) + (315t + 811) - 9 = 0
Simplifying further gives:
(y + 12)^2 + (150t/3 + 800/9 + 315t/3 + 811) - 9 = 0
(y + 12)^2 + (50t + 800/9 + 105t + 811) - 9 = 0
(y + 12)^2 + (155t + 800/9 + 811) - 9 = 0
(y + 12)^2 + (155t + 800/9 + 811) - 9 = 0
Combining like terms gives:
(y + 12)^2 + (155t + 800/9 + 811) - 9 = 0
Expanding this equation gives:
(y^2 + 24y + 144) + (139t + 800/9 + 811) - 9 = 0
y^2 + 24y + 144 + 139t + 800/9 + 811 - 9 = 0
y^2 + 24y + 139t + 800/9 + 946 = 0
We can solve this equation for y in terms of t:
y^2 + 24y + 139t + 800/9 + 946 = 0
y^2 + 24y + 139t + 800/9 + 946 - 946 = 0 - 946
y^2 + 24y + 139t + 800/9 = -946
To find the other point, we can consider Case 2 where x = 1 + 2t + 3.
Similarly, we can substitute this value of x into the equation obtained from the distance formula, and solve for y in terms of t.
Once we have the equations for y in terms of t, we can substitute these equations back into the equation of the line to solve for t, and then find the corresponding values of x and z.
Note: The expressions for y in terms of t will most likely be quadratic equations, which means there may be multiple values of t that satisfy the equations. This would result in multiple pairs of points on the line that are a distance 3 units from the plane.