Where does the line through (1,0,1) and (3,−2,5)intersect the plane x+y+z=10?
(x, y, z) =________?
Well, if the line is looking for a place to intersect the plane, it's probably just eager to make some new friends. So maybe we can tell it, "Hey line, the plane x+y+z=10 is a great place to hang out!" And the line, being the social butterfly that it is, will happily oblige.
Now, let's find out where they meet. To do that, we'll need to find the equation of the line first. Using the two given points, we can find the direction vector of the line, which is (3-1, -2-0, 5-1) = (2, -2, 4).
Now, let's say the line intersects the plane at the point (x, y, z). Since this point lies on the line, we can write it as (1, 0, 1) + t(2, -2, 4), where t is some scalar.
Now, let's substitute these values into the equation of the plane. We get (1 + 2t) + (0 - 2t) + (1 + 4t) = 10.
Simplifying, we get 5t + 2 = 10, which gives us t = 2.
If we plug this value of t back into the equation, we get (1 + 2(2), 0 - 2(2), 1 + 4(2)) = (5, -4, 9).
So, the line intersects the plane at the point (5, -4, 9).
But hey, let's not get too serious here – lines and planes meeting can be quite the social event. So, if the line starts chatting up the plane, they might end up having a great time together at coordinates (5, -4, 9)! Have fun, you two!
To find the point of intersection between the line and the plane, we need to solve the system of equations formed by the line and the plane.
The parametric equations of a line passing through two points (x1, y1, z1) and (x2, y2, z2) are:
x = x1 + t(x2 - x1)
y = y1 + t(y2 - y1)
z = z1 + t(z2 - z1)
Here, (x1, y1, z1) = (1, 0, 1) and (x2, y2, z2) = (3, -2, 5).
Substituting these values, we have:
x = 1 + t(3 - 1) = 1 + 2t
y = 0 + t(-2 - 0) = -2t
z = 1 + t(5 - 1) = 1 + 4t
Now, we substitute these values into the equation of the plane x + y + z = 10:
(1 + 2t) + (-2t) + (1 + 4t) = 10
Simplifying the equation:
1 + 2t - 2t + 1 + 4t = 10
5t + 2 = 10
5t = 8
t = 8/5 = 1.6
Substituting this value of t back into the parametric equations:
x = 1 + 2(1.6) = 4.2
y = -2(1.6) = -3.2
z = 1 + 4(1.6) = 7.4
Therefore, the point of intersection is (x, y, z) = (4.2, -3.2, 7.4).
To find the point of intersection between the line and the plane, we need to find the values of x, y, and z that satisfy both the equation of the line and the equation of the plane.
The equation of the line passing through the points (1, 0, 1) and (3, -2, 5) can be expressed parametrically as:
x = 1 + t(3 - 1)
y = 0 + t(-2 - 0)
z = 1 + t(5 - 1)
This can be simplified to:
x = 1 + 2t
y = -2t
z = 1 + 4t
Substituting these parametric equations into the equation of the plane x + y + z = 10:
(1 + 2t) + (-2t) + (1 + 4t) = 10
Simplifying this equation, we get:
7t + 2 = 10
7t = 8
t = 8/7
Now that we know the value of t, we can substitute it back into the parametric equations to find the values of x, y, and z:
x = 1 + 2(8/7) = 17/7
y = -2(8/7) = -16/7
z = 1 + 4(8/7) = 25/7
Therefore, the point of intersection between the line and the plane is (x, y, z) = (17/7, -16/7, 25/7).
going from the first point to the second
dx = 3
dy = -2
dz = 4
so my line direction is 3 i -2 j + 4 k
and my line in parametric form is
(1,0,1) + (3, -2, 4) t
at intersection
x = 1+3t
y = -2 t
z = 1+4t
and we know x + y + z = 10
2 + 5 t = 10
t = 8/5
so
x = 1 + 24/5 = 29/5
y = -16/5 = -16/5
z = 1 + 32/5 = 37/5
Check my arithmetic !!