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November 26, 2015
Posted by **Shaila** on Sunday, September 5, 2010 at 11:26pm.

[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.

- Math - equations of lines -
**Reiny**, Sunday, September 5, 2010 at 11:58pmAs 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.

- Math - equations of lines -
**Shaila**, Monday, September 6, 2010 at 12:47pmWriting 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.

- Math - equations of lines -
**Reiny**, Monday, September 6, 2010 at 1:47pmYou 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)