Find the point on line x+2/3=y+1/2=z-3/2 at a distance 5 units from the point P(1,3,3)....pllz help me out.

Answer 1

Let the new point be (x1, y1, z1)

Then, using the distance formula:

((x1-1)^2 + (y1-3)^2 + (z1-3)^2)^0.5 = 5

(x1-1)^2 + (y1-3)^2 + (z1-3)^2 = 25

Put all units in terms of one variable. I'm choosing x1 first:

x1+2/3 = y1+1/2; y1 = x1 + 2/3 - 1/2
= x1 + 1/6

x1 + 2/3 = z1-3/2; z1 = x1 + 2/3 + 3/2 = x1 + 4/6 + 9/6 = x1 + 13/6

(x1-1)^2 + (x1+1/6-3)^2 + (x1+13/6-3)^2 = 25

Multiply this out; solve for x1; then plug these back into the other equations to solve for y1 and z1