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 prove that the line lies entirely in the plane, we need to show that every point on the line satisfies the equation of the plane.
Given the line {p: p = (1, 3, -1) + t(0, 3, 5)} and the plane {(x, y, z): 2x - 5y + 3z = -16}, we want to show that for any value of t, the point p = (1, 3, -1) + t(0, 3, 5) satisfies the equation 2x - 5y + 3z = -16.
Let's substitute the values of x, y, and z using the equation for the line:
x = 1 + 0t = 1
y = 3 + 3t
z = -1 + 5t
Now, let's substitute these values into the equation of the plane:
2(1) - 5(3 + 3t) + 3(-1 + 5t) = -16
2 - 15 - 15t - 3 + 15t = -16
-16 = -16
As we can see, the equation holds true for any value of t. Therefore, every point on the line lies in the plane. Hence, the line lies entirely in the plane {(x, y, z): 2x - 5y + 3z = -16}.
To prove that the line lies entirely in the plane, we need to show that all the points on the line satisfy the equation of the plane.
The equation of the line is given as:
p = (1, 3, -1) + t(0, 3, 5)
Let's substitute the coordinates of the line into the equation of the plane and see if it satisfies.
For any point on the line, let's consider the coordinates as (x, y, z). Then we have:
2x - 5y + 3z = -16
Substituting the coordinates of the line, we get:
2(1 + t * 0) - 5(3 + t * 3) + 3(-1 + t * 5) = -16
Simplifying this equation, we have:
2 - 15 - 15t - 3 + 15t = -16
Combine like terms:
-16 - 16 = -16
Therefore, the equation holds true for any value of 't'. This means that all the points on the line satisfy the equation of the plane.
Hence, the line {p:p = (1, 3, -1) + t(0, 3, 5)} lies entirely in the plane {(x, y, z): 2x – 5y + 3z = -16}.