Show that lines:

`vecr=hati+hatj+hatk+lambda(hati-hat+hatk)`

`vecr=4hatj+2hatk+mu(2hati-hatj+3hatk)` are coplanar

Also, find the equation of the plane containing these lines.

#### Solution

`vecr=hati+hatj+hatk+lambda(hati-hat+hatk) .......(i)`

Convert ing into cartesian form,

`(x-1)/1=(y-1)/-1=(z-1)/1`

(x_{1},y_{1},z_{1})=(1,1,1)

a_{1}=1,b_{1}=-1,c_{1}=1

`vecr=4hatj+2hatk+mu(2hati-hatj+3hatk)......(ii)`

Convert ing into cartesian form,

(x2,y2,z2)=(0,4,2)

a_{2}=2,b_{2}=-1,c_{2}=3

Condition for the lines to be coplanar is

`|[0-1,4-1,2-1],[1,-1,1],[2,-1,3]|=|[-1,3,1],[1,-1,1],[2,-1,3]|=0`

the lines are coplanar

Intersection of the two lines is

Let the equation be a(x-x_{1} )+b(y -y_{1} )+ c(z - z_{1} )= 0.....(iii)

Direction ratios of the plane is

a-b+c=0

2a-b+3c=0

Solving by cross-multiplication

`a/(-3+1)=b/(2-3)=c/(-1+2)`

Since the plane passes through (0,4,2) from line (ii)

a(x-0)+b(y-4)+c(z-2)=0

`=>-2lambdax-lambda(y-4)+lambda(z-2)=0`

`=>-2x-y+4+z-2=0`

`=>-2x-y+z=-2`

`=>2x+y-z=2`