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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)`
Solution
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.
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