Posted by abhi on .
Find the point on line x+2/3=y+1/2=z3/2 at a distance 5 units from the point P(1,3,3)....pllz help me out.

maths 
Jennifer,
Let the new point be (x1, y1, z1)
Then, using the distance formula:
((x11)^2 + (y13)^2 + (z13)^2)^0.5 = 5
(x11)^2 + (y13)^2 + (z13)^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 = z13/2; z1 = x1 + 2/3 + 3/2 = x1 + 4/6 + 9/6 = x1 + 13/6
(x11)^2 + (x1+1/63)^2 + (x1+13/63)^2 = 25
Multiply this out; solve for x1; then plug these back into the other equations to solve for y1 and z1