Show that the line [x,y,z]=[10,5,16]+t[3,1,5] is contained in the plane x+2y-z-4=0
I tries writing the vector equation in parametric form and then substituting it into the respective x,y,z in the plane, after simplifying I get 0t=0 does this mean that the line is in the plane or did I do something wrong
Yes, that means that the line is in the given plane.
You can also take any two distinct points on the given line (i.e. two arbitrary values of t) and prove that each of the point lies in the plane.
To check if the line [x, y, z] = [10, 5, 16] + t[3, 1, 5] is contained in the plane x+2y-z-4=0, we need to substitute the parametric equation into the plane equation.
First, let's substitute the values of x, y, and z from the line equation into the plane equation:
x + 2y - z - 4 = 0
Substituting x = 10 + 3t, y = 5 + t, and z = 16 + 5t:
(10 + 3t) + 2(5 + t) - (16 + 5t) - 4 = 0
Simplifying the equation:
10 + 3t + 10 + 2t - 16 - 5t - 4 = 0
25t = 0
Therefore, we have 25t = 0.
This means that t = 0. When t = 0, we have the point [10, 5, 16], which is the initial point given in the line equation.
Thus, all values of t yield a point that satisfies the plane equation.
Therefore, the line [x, y, z] = [10, 5, 16] + t[3, 1, 5] is contained in the plane x+2y-z-4=0.
To show that the line [x, y, z] = [10, 5, 16] + t[3, 1, 5] is contained in the plane x + 2y - z - 4 = 0, you need to substitute the parametric equation of the line into the equation of the plane and check if it holds true for all values of t. Let's go through the steps:
1. Start with the equation of the plane: x + 2y - z - 4 = 0.
2. Substitute x = 10 + 3t, y = 5 + t, and z = 16 + 5t into the equation of the plane.
(10 + 3t) + 2(5 + t) - (16 + 5t) - 4 = 0
3. Simplify the equation:
10 + 3t + 10 + 2t - 16 - 5t - 4 = 0
(3t - 5t + 2t) + (10 + 10 - 16 - 4) = 0
0t - 0 = 0
0 = 0
4. The equation simplifies to 0 = 0, which is always true. This means that the line is contained in the plane x + 2y - z - 4 = 0.
Therefore, you have correctly shown that the line is contained in the plane.