Question

A line passing through the points A(1, 2, 3) and B(5, 8, 11) intersects the line `vecr=4hati+hatj+lambda(5hati+2hatj+hatk)`. Find the co-ordinates of the point of intersection. Hence, write the equation of a line passing through the point of intersection and perpendicular to both the lines.

Answer 1

Equation of a line passing through two points A and B.

`vecr_1=veca+t(vecb-veca)`

= `veci+2vecj+3veck+t[(5hati+8hatj+11hatk)-(hatl+2hatj+3hatk)]`

`vecr_1=hati+2hatj+3hatk+t(4hati+6hatj+8hatk)`

`vecr_2=4hati+hatj+λ(5hati+2hatj+hatk)`  ...(Given)

At a point of intersection co-ordinates of both the lines are equal.

1 + 4t = 4 + 5λ 

⇒ 4t – 5λ = 3   ...(1)

2 + 6t = 1 + 2λ

⇒ 6t – 2λ = –1   ...(2)

3 + 8t = λ    ...(3)

Put (3) in equation (2)

⇒ 6t – 2(3 + 8t) = –1

⇒ 6t – 6 – 16t = –1

⇒ –10t = 5

⇒ t = `(-1)/2`

Put t = `(-1)/2` in equation (3)

⇒ `3+3((-1)/2)=lambda`

⇒ λ = –1

Coordinates of the point of intersection:

`vecr_1=hati+2hatj+3hatk-1/2(4hati+6hatj+8hatk)`

= `hati+2hatj+3hatk-2hati-3hatj-4hatk`

`vecr_1=-hati-hatj-hatk`

Coordinates are (–1, –1, –1)

Direction of given lines are;

`vecd_1=4hati+6hatj+8hatk`

`vecd_2=5hati+2hatj+hatk`

`vecd_1xxvecd_2=|(hati,hatj,hatk),(4,6,8),(5,2,1)|`

= `(6-16)hati-(4-40)hatj+(8-30)hatk`

= `-10hati+36hatj-22hatk`

Hence, equation of line passing through point of intersect

`vecr=(-hati-hatj-hatk)+μ(-10hati+36hatj-22hatk)`

`vecr=veca+μvecb`

μ in any scalar.