Question

Find the vector equation of the line passing through the point (1, 2, − 4) and perpendicular to the two lines: 

`(x -8)/3 = (y+19)/(-16) = (z - 10)/7 and (x - 15)/3 = (y - 29)/8 = (z- 5)/(-5)`

Answer 1

Let us assume the required line

`vecr = hati + 2hatj - 4hatk + λ(b_1hati + b_2hatj + b_3hatk)`      .....(i)

The lines `(x - 8)/3 = (y + 19)/-16 = (z -10)/7` and `vecr = hati + 2hatj - 4hatk + λ(b_1hati + b_2hatj + b_3hatk)` are perpendicular to each other.

The direction ratios of these lines are 3, −16, 7 and b₁, b₂, b₃. These lines are perpendicular to each other if

3b₁ − 16b₂ + 7b₃ = 0               .....(ii) 

Similarly, the direction ratios of the lines `(x -15)/3 = (y - 29)/8 = (z - 5)/-5` and `vecr = hati + 2hatj - 4hatk + λ(b_1hati + b_2hatj + b_3hatk)` are 3, 8, −5 and b₁, b₂, b₃ are mutually perpendicular.

∴ 3b₁ + 8b₂ − 5b₃ = 0                   .....(iii)

From equations (ii) and (iii),

`b_1/(80 - 56) = b_2/(21 + 15) = b_3/(24 + 18)`

and `b_1/24 = b_2/36 = b_3/72`

`"b"_1/2 = "b"_2/3 = "b"_3/6`

Putting the proportional values ​​of b₁, b₂, b₃  (i) 

`vecr = hati + 2hatj - 4hatk + λ(2hati + 3hatj + 6hatk)`

This is the equation of the required line.