Question

If the lines `(x-1)/-3=(y-2)/(2k)=(z-3)/2 and (x-1)/(3k)=(y-1)/1=(z-6)/-7` are perpendicular  to each other, find the value of k and hence write the vector equation of a line perpendicular to these two lines and passing through the point (3, –4, 7).

Answer 1

The equation of the given lines are

`L_1: (x-1)/-3=(y-2)/(2k)=(z-3)/2`

`L_2: (x-1)/(3k)=(y-1)/1=(z-6)/-7`

Their Direction Ratio are (−3, 2k, 2) and (3k, 1, −7) when L₁ ⊥ L₂

a₁a₂ + b₁b₂ + c₁c₂ = 0

(−3)(+3k) + (2k)(1) + (2)(−7) = 0

−9k + 2k − 14 = 0

−7k − 14 = 0

k = −2

Now

`L_1: (x-1)/-3=(y-2)/(-4)=(z-3)/2`

`L_2: (x-1)/(-6)=(y-1)/1=(z-6)/-7`

We know that equation of any line through a given point (x₁, y₁, z₁) are 

`(x-x_1)/a=(y-y_1)/(b)=(z-z_1)/c`

∴ Equation of any line through (3, −4, −7)

`(x-3)/a=(y-4)/(b)=(z-7)/c`

If it is perpendicular to the lines then

−3a − 4b + 2c = 0

−6a + b − 7c = 0

`a/(28-2)=b/(-12-21)=c/(-3-24)`

`a/(+26) = b/-33 = c/-27`

Equation of line `(x-3)/26=(y-4)/(-33)=(z-7)/-27`