find the angle between the pair of lines `(x+3)/3=(y-1)/5=(z+3)/4` and `(x+1)/1=(y-4)/1=(z-5)/2`

Answer

`theta= 60^(@) , 2x+ (2pm sqrt2) y + (2 pm sqrt2) z + 4 =0, 2x+ (2pm sqrt2)y + (2 pm sqrt2)z - 20 =0`

Explanation

`ax+by+cz+1=0" "(i)`
It makes an `angle60^(@)` with the line x=y=z. So we get
`sin60^(@)=(a+b+c)/(sqrt3suma^(2))`
or `3sqrtsuma^(2)=2(a+b+c)" "(ii)`
Plane (i) makes an angle of `45^(@)` with the line
`x=y-z=0(or(x)/(0)=(y)/(1)=(z)/(1))`
`sin45^(@)=(b+c)/(sqrt2sqrt(suma^(2)))orsqrt(suma^(2))=a+c" "(iii)`
Plane (i) makes an `angletheta` with plane x=0. So we get `costheta=(a)/(sqrt(suma^(2)))" "(iv)`
From (ii) and (iii) we get
`(sqrtsuma^(2))=2a`
or `(a)/(sqrtsuma^(2))=(1)/(2)`
From (iv), `costheta=1//2ortheta=60^(@)`
Distance of plane (i) from the point (2,1,1) is 3 units.
`implies(2a+b+c+1)/(sqrt(suma^(2)))=+-3`
or `+-3sqrt(suma^(2))=2a+b+c+1`
Case I:
`+-3sqrt(suma^(2))=2a+b+c+1" "(v)`
From (ii) and (iv), we get `b+c-1=0" "(vi)`
and from (iii) and (iv), we get
`2a+b+c+1=3(b+c)" "(vii)`
From (vi) and (vii), we get `a=(1)/(2),b =(2pmsqrt2)/(4)andc=(2pmsqrt2)/(4)`
Hence, the set of such planes is
Case II:
`-3sqrtsuma^(2)=2a+b+c+1`
`a=(-1)/(10),b=(-(2pmsqrt2))/(20) andc=(-(2pmsqrt2))/(20)`
Hence, the other set of the planes is
`2x+(2pmsqrt2)y+(2pmsqrt2)z-20=0.`