is lines (x-1)/3=(y+1)/2=(z+4)/5 and (x-4)/3=y+3/-2=z-2/-1 are orthogonal, parallel or neither?

Question

how would I determine whether lines (x-1)/3=(y+1)/2=(z+4)/5 and (x-4)/3=y+3/-2=z-2/-1 are orthogonal, parallel or neither?

Answer 1

Given first line  (x-1)/3 = (y+1)/2 = (z+4)/5

We set the mutual value of those three equal to t.

(x-1)/3 = t

x-1 = 3t

x = 3t+1

(y+1)/2 = t

y+1 = 2t

y = 2t-1

(z+4)/5 = t

z+4 = 5t

z = 5t-4

These are parametic equations for the line and we can use them to write a vector in the direction of the line

(3t+1)i+(2t-1)j+(5t-4)k

When t = 0

i-j-4k is the vector from the origin to the point (1,-1,-4) on the line.

If we take t = 1

4i+j+k is the vector from the origin to the point (4,1,1) another point on the line.

And if we subtract those two vectors we get 3i+2j+5k a vector pointing in the direction of the line.

3i+2j+5k ---> (1)

Similarly  second line (x-4)/3 = (y+3)/-2 = (z-2)/-1

x = 3t+4,y = -2t-3,z = -t+2

(3t+4)i+(-2t-3)j+(-t+2)k

if t = 0

4i-3j+2k

t = 1

7i-5j+k

3i-2j-k ---->(2)

Angle between (1)&(2) CosѲ =a.b/|a||b|

a.b = 9-4-5 = 0

CosѲ = 0

Ѳ = 90

Given lines are orthogonal perpendicular lines.