I think from the information given, it can be inferred that l1 and l2 already intersect at the origin since the position vector of the given point isn't shown in both equations, which implies that it is (0,0,0)

Hi ! Thanks J! I still cant solve it with that info though

If l1 and l2 are reflections of each other about l3, then they must be mirror images of each other.

It also means they are equidistant from l3.

Any point on l1 (call it A for easier reference) and its corresponding reflected point on l2 (call it A'), will be equidistant from l3, and vice versa.


Similarly, as both lines (in fact all three lines) intersect at the origin), the length of OA must be equal to OA'

Now we know that P(18,27,54) lies on l2. The idea here is to find a point P' on l1 such that OP = OP'.

P' would be the reflection of P.

We can find length of OP first. Then find the coordinates of point P' such that the vector OP' is a scalar multiple of the direction vector of l1(1,2,-2), and the length of OP' = OP

OP = √(18² + 27² + 54²) = 63
Surprisingly, a very nice whole number.

Now let P' = (k, 2k, -2k) for some k ∈ R

Vector OP' = k(1, 2, -2) for some k∈ R
(Here the bracket is the column vector as it's not possible to do a column brackets on text)

OP' = |k|√(1² + 2² + (-2)²) = |k|√9 = 3|k|

(The modulus sign or absolute value sign is placed over k as k could be negative or positive, but length of a line /magnitude of a vector is always ≥ 0)

Now let OP' = OP
3|k| = 63
|k| = 21

k = 21 or -21


There are actually two possible l3's since every pair of intersecting lines has two lines of symmetry (this is similar to vertically opposite angles, where you can have angle bisectors for both pairs of opposite angles)

So we can choose k = 21 or k = -21

Next we just need to find a point on l3 to get its possible equation.

Now because l2 and l1 are equidistant from l3, l3 is like an 'angle bisector' and so we can just find the midpoint of P and P'. It would lie on l3.

If we choose k = 21, then P' = (21, 42, -42)


Midpoint of P and P' (call it M) would be :

( (18 + 21)/2 , (27 + 42)/2 , (54 - 42)/2 )

= (19.5, 34.5, 6)


A direction vector for l3 would be vector OM

Possible equation of l3 would be :

r = β(19.5, 34.5, 6) , β ∈ R

Alternatively, if we choose k = -21, then P' = (-21, -42, 42)


Midpoint of P and P' (call it M) would be :

( (18 - 21)/2 , (27 - 42)/2 , (54 + 42)/2 )

= (-1.5, -7.5, 48)


A direction vector for l3 would be vector OM

Possible equation of l3 would be :

r = β(-1.5, 7.5, 48) , β ∈ R


Notice that OM is actually a scalar multiple of the direction vector given in the answer, since (-1.5,7.5,48) = -1.5(1,5,32)

It's has 1.5 times the magnitude but goes in the opposite direction.

beautiful solution and explanation, thanks J !!