At what point does the line r=(-2,3,1)+s(2,3,-1) intersect the y-axis?
x = -2 + 2 s
y = 3 + 3 s
z = 1 - s
on the y axis, x and z are 0
0 = -2 +2 s
0 = 1 - s
so it is where s = 1
then
y = 3+3 = 6
so
(0 , 6 , 0 )
Why did the line join the circus? Because it wanted to intersect with the y-axis and become a clown!
To find the point where the line intersects the y-axis, we need to substitute a value of 0 for the other two variables, x and z.
The given equation of the line is r=(-2,3,1)+s(2,3,-1).
Let's substitute x=0 and z=0 into the equation:
r=(0,3,0)+s(2,3,-1).
Simplifying the equation, we get:
r=(2s,3+3s,-s).
Since we want to find the point where the line intersects the y-axis, the x and z coordinates should be 0. Therefore, we need to solve for the value of s when x=0 and z=0.
From the equation, we have:
2s = 0 ----(1)
3 + 3s = 0 -----(2)
-s = 0 -----(3)
From equation (3), we get s = 0.
Substituting s=0 into equation (1), we get:
2(0) = 0
Therefore, the line intersects the y-axis at the point (0,3,0)
To find the point at which the line intersects the y-axis, we need to determine the value of the y-coordinate when the x and z coordinates are both equal to zero.
The given equation of the line is r = (-2,3,1) + s(2,3,-1).
Let's set x and z equal to zero in the parametric equation of the line:
x = -2 + 2s (equation 1)
y = 3 + 3s (equation 2)
z = 1 - s (equation 3)
Setting x = 0 and z = 0 in equation 1 and 3 respectively, we get:
0 = -2 + 2s => s = 1
0 = 1 - s => s = 1
Both equations give us s = 1.
Now, substituting s = 1 back into equation 2 to find the y-coordinate:
y = 3 + 3(1)
y = 3 + 3
y = 6
Therefore, the point at which the line intersects the y-axis is (0, 6, 0).
To summarize, the line r=(-2,3,1)+s(2,3,-1) intersects the y-axis at the point (0, 6, 0).