Question

Show that the lines `(x - 1)/2 = (y - 2)/3 = (z - 3)/4` and `(x - 4)/5 = (y - 1)/2` = z intersect. Also, find their point of intersection.

Answer 1

The given equations are `(x - 1)/2 = (y - 2)/3 = (z - 3)/4` and `(x - 4)/5 = (y - 1)/2` = z

Let `(x - 1)/2 = (y - 2)/3 = (z - 3)/4 = lambda`

∴ x = 2λ + 1, y = 3λ + 2 and z = 4λ + 3

And `(x - 4)/5 = (y - 1)/2 = z/1 = mu`

∴ x = `5mu + 4, y = 2mu + 1` and z = `mu`

If the two lines intersect each other at one point,

Then `2lambda + 1 = 5mu + 4`

⇒ `2lambda - 5mu` = 3  .....(i)

`3lambda + 2 = 2mu + 1`

⇒ `3lambda - 2mu = -1`  ......(ii)

And `4lambda + 3 = mu`

⇒ `4lambda - mu = - 3`  ......(iii)

Solving equations (i) and (ii) we get

`2lambda - 5mu` = 3      .......[Multiply by 3]

`3lambda - 2mu` = – 1    .......[Multiply by 2]

⇒ `6lambda - 15mu` =    9
    `6lambda - 14mu` = – 2
 (–)     (+)        (+)    
         `-11,u` =  11

∴ `mu = -1`

Putting the value of m in equation (i) we get,

2λ – 5(– 1) = 3

⇒ 2λ + 5 = 3

⇒ 2λ = – 2

∴ λ = – 1

Now putting the value of λ and m in equation (iii) then

4(– 1) – (– 1) = – 3

– 4 + 1 = – 3

– 3 = – 3  ....(Satisfied)

∴ Coordinates of the point of intersection are

x = 5 (– 1) + 4 = – 5 + 4 = – 1

y = 2(– 1) + 1 = – 2 + 1 = –

z = – 1

Hence, the given lines intersect each other at (– 1, – 1, – 1).

Alternately: If two lines intersect each other at a point, then the shortest distance between them is equal to 0.

For this we will use SD = `((vec"a"_2 - vec"a"_1)(vec"b"_1 xx vec"b"_2))/|vec"b"_1 xx vec"b"_2|` = 0.