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}.
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}
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