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.
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
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