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Question
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.
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Solution
`vecr=hati+hatj+hatk+lambda(hati-hat+hatk) .......(i)`
Convert ing into cartesian form,
`(x-1)/1=(y-1)/-1=(z-1)/1`
(x1,y1,z1)=(1,1,1)
a1=1,b1=-1,c1=1
`vecr=4hatj+2hatk+mu(2hati-hatj+3hatk)......(ii)`
Convert ing into cartesian form,
(x2,y2,z2)=(0,4,2)
a2=2,b2=-1,c2=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-x1 )+b(y -y1 )+ c(z - z1 )= 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`
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