Find a point of intersection of the lines {p:p = (1, -5, 2) + t(-1, 1, 0)} and {p:p = (3, -3, 1) + t(4, 0, -1)}
Prove that the line {p:p = (1, 3, -1) + t(0, 3, 5)} lies entirely in the plane
{(x, y, z): 2x – 5y + 3z = -16}
To find the point of intersection of the two lines, we need to set their respective expressions equal to each other and solve for t:
(1, -5, 2) + t(-1, 1, 0) = (3, -3, 1) + t(4, 0, -1)
This gives us the following equations:
1 - t = 3 + 4t
-5 + t = -3
2 = 1 - t
Simplifying these equations, we get:
5t = -2
t = -2/5
Substituting this value of t into one of the line equations, we can find the point of intersection:
p = (3, -3, 1) + (-2/5)(4, 0, -1)
p = (3, -3, 1) + (-8/5, 0, 2/5)
p = (3 - 8/5, -3, 1 + 2/5)
p = (7/5, -3, 7/5)
So, the point of intersection of the two lines is (7/5, -3, 7/5).
To prove that the line {p: p = (1, 3, -1) + t(0, 3, 5)} lies entirely in the plane {(x, y, z): 2x - 5y + 3z = -16}, we need to show that every point on the line satisfies the equation of the plane.
Substituting the coordinates of a point on the line, (1, 3t + 3, -t + 5t), into the equation of the plane, we get:
2(1) - 5(3t + 3) + 3(-t + 5t) = -16
Simplifying this equation gives:
2 - 15t - 15 + 3t = -16
-12t - 13 = -16
-12t = -3
t = 1/4
So, for t = 1/4, the point (1, 3(1/4) + 3, -(1/4) + 5(1/4)) = (1, 3.75, 1.25) lies on the line and satisfies the equation of the plane:
2(1) - 5(3.75) + 3(1.25) = -16
2 - 18.75 + 3.75 = -16
-16 = -16
Since this is true, we can conclude that the line {p: p = (1, 3, -1) + t(0, 3, 5)} lies entirely in the plane {(x, y, z): 2x - 5y + 3z = -16}.
To find the point of intersection of the two lines, we can set their parametric equations equal to each other and solve for t:
(1, -5, 2) + t(-1, 1, 0) = (3, -3, 1) + t(4, 0, -1)
Simplifying each component, we get:
1 - t = 3 + 4t (for x-coordinate)
-5 + t = -3 (for y-coordinate)
2 = 1 - t (for z-coordinate)
Solving these equations, we find:
t = -1 (for x-coordinate)
t = 2 (for y-coordinate)
t = -1 (for z-coordinate)
Using these values of t, we can find the corresponding point of intersection:
p = (1, -5, 2) + (-1)(-1, 1, 0)
= (1, -5, 2) + (1, -1, 0)
= (2, -6, 2)
Therefore, the point of intersection of the two lines is (2, -6, 2).
To prove that the line {p: p = (1, 3, -1) + t(0, 3, 5)} lies entirely in the plane {(x, y, z): 2x – 5y + 3z = -16}, we can substitute the coordinates of any point on the line into the equation of the plane. If the equation holds true, then the line lies entirely in the plane.
Let's substitute the point (1, 3, -1) into the equation of the plane:
2(1) - 5(3) + 3(-1) = -16
Simplifying, we get:
2 - 15 - 3 = -16
-16 = -16
Since the equation holds true, we can conclude that the line {p: p = (1, 3, -1) + t(0, 3, 5)} lies entirely in the plane {(x, y, z): 2x – 5y + 3z = -16}.