A line has the same x-intercept as
[x,y,z]=[-21,8,14]+t[-12,4,7] and the same y-intercept as
[x,y,z]=[6,-8,12]+s[2,-5,4]. Write the parametric equation of the line.
To write the parametric equation of the line, we need to find the direction vector of the line and a point on the line.
Let's start by finding the x-intercept of the line. The x-intercept occurs when the y and z coordinates are both zero. So, we can set y = 0 and z = 0 in the equation [x, y, z] = [-21, 8, 14] + t[-12, 4, 7] and solve for t.
0 = 8 + 4t (Setting y = 0)
0 = 14 + 7t (Setting z = 0)
Solving these equations, we get t = -2 for both.
Now, let's find the y-intercept of the line. The y-intercept occurs when the x and z coordinates are both zero. So, we can set x = 0 and z = 0 in the equation [x, y, z] = [6, -8, 12] + s[2, -5, 4] and solve for s.
0 = 6 + 2s (Setting x = 0)
0 = 12 + 4s (Setting z = 0)
Solving these equations, we get s = -3 for both.
Now that we have the values of t and s, we can substitute them back into the original equations to find the corresponding points.
For the x-intercept:
When t = -2, substituting into [x, y, z] = [-21, 8, 14] + t[-12, 4, 7], we get:
[x, y, z] = [-21, 8, 14] + (-2)[-12, 4, 7]
= [-21 + 24, 8 - 8, 14 - 14]
= [3, 0, 0]
For the y-intercept:
When s = -3, substituting into [x, y, z] = [6, -8, 12] + s[2, -5, 4], we get:
[x, y, z] = [6, -8, 12] + (-3)[2, -5, 4]
= [6 - 6, -8 + 15, 12 - 12]
= [0, 7, 0]
Now, we have a point on the line: (3, 0, 0), and the direction vector of the line: [-12, 4, 7].
Therefore, the parametric equation of the line is:
x = 3 - 12t
y = 4t
z = 7t