Determine the values of a and c on a line.

[ND3G]

A line is defined by the equations x = 3-4t, y = 8t, and z = 6+t

a) Determine the vector and symmetric equations of the line

b) Determine the values of a and c if the point (a, 2, c) lies on the line.

I have part a) solved: (x,y,z) = (3,0,6) +t(-4,8,1)

(3-x)/4 = y/8 = z-6/1

I am a little unclear has to how I would solve for a and c though in part b)

y-2/8 = a-x/4 = z-c/1 but I am not sure where to go from there...

 

[NateTG]

At the point (a,2,c), what is the value of y?

 

[ND3G]

y = 2, so 2/8 = 1/4

therefore, 3-x/4 = z-6/1 = 1/4

(2, 2, 25/4) I think I got it. Thanks

 

[VietDao29]

y = 2, so 2/8 = 1/4

therefore, 3-x/4 = z-6/1 = 1/4

(2, 2, 25/4) I think I got it. Thanks

Yup. Correct. Or you can also use its parametric equation to solve the problem.
When y = 2, that means 8t = 2 ~~> t = 1 / 4
Sub t = 1 / 4 in x, and z, we have:
x = 3 - 4t = 3 - 4 \times \frac{1}{4} = 2
z = 6 + t = 6 + \frac{1}{4} = \frac{25}{4}, so the coordinate of the point is:
\left( 2; \ 2 ; \ \frac{25}{4} \right), so a = 2, and c = 25/4.