Posted by Shaila on Sunday, September 5, 2010 at 11:26pm.
As I understand it, the x-intercept of a line in 3D would be where y=0 and z=0
that is,
[x,0,0] = [-21,8,14] + t[-12,4,7]
this results in 2 different values of t, thus the line misses the x-axis.
The same would be true for the y-intercept of the 2nd line, it would miss the y-axis.
Writing the parametric equations of the given line one
x=-21-12t and since we want the x intercept the y=z=0 therefore
8+4t=0 and 14+7t=0 in both case the t value is '-2' repeating the similar procedure to find the y-intercept the value of s=-3
there after how do i find my parametric equations.
You are correct, I made an arithmetic error trying to do it in my head.
ok, so when t = -2
[x,y,z] = [3,0,0] , so the endpoint of that vector is(3,0,0,)
and in the second, when s = -3
[x,y,z] = [0, 7,0] so the endpoint of that vector is (0,7,0)
so the direction vector of the line joining the intercepts is (3,-7,0)
and the equation of that line is
[x,y,z] = (3,0,0) + k(3,-7,0)
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