Question

Let `x-ysinalpha-zs inbeta=0,xs inalpha=zs ingamma-y=0a n dxsinbeta+ysingamma-z=0` be the equations of the planes such that `alpha+beta+gamma=pi//2(w h e r ealpha,betaa n dgamma!=0)dot` Then show that there is a common line of intersection of the three given planes.

Answer 1

`x-ysinalpha-zsinbeta=0" "(i)`
`xsinalpha+zsingamma-y=0" "(ii)`
`xsinbeta+ysingamma-z=0" "(iii)`
These planes pass through the origin. Let l, m and n be the direction cosines of the line of intersection of planes (i) and (ii). Then
`l.1-msinalpha-nsinbeta=0`
`lsinalpha-m.1+nsingamma=0`
`implies(l)/(-singammasinalpha-sinbeta)`
`=(m)/(-sinbetasinalpha-singamma)=(n)/(-1+sin^(2)alpha)" "(iv)`
If `alpha+beta+y=(pi)/(2)impliesbeta=(pi)/(2)-(alpha+gamma)`
`sinbeta=sin((pi)/(2)-(alpha+gamma))=cos(alpha+gamma)`
`sinbeta=cosalphacosgamma-sinalphasingamma`
`sinbeta+sinalphasingamma=cosalphacosbeta`
Similarly, `singamma+sinbetasinalpha=cosalphacosbeta`
From equation (iv), we get
`(l)/(cosalphacosgamma)=(m)/(cosalphacosbeta)=(n)/(cos^(2)alpha)`
`(l)/(cosgamma)=(m)/(cosbeta)=(n)/(cosalpha)" "(v)`
The line of intersection of plane (i) and (ii) also passes through the origin. Then the equation of the line is
`(x-0)/(l)=(y-0)/(m)=(z-0)/(n)`
`implies(x)/(cosgamma)=(y)/(cosbeta)=(z)/(cosalpha)" "(vi)`
If the line also lies on plane (iii), then the three planes will intersect on this straight line.
The angle between line and normal of plane (iii) should be `pi//2`. Therefore,
`cosgammasinbeta+cosbetasingamma+cosalpha(-1)`
`=sin(beta+gamma)-cosalpha`
`=sin((pi)/(2)-alpha)-cosalpha=0`
Hence, `(x)/(cosgamma)=(cosbeta)=(z)/(cosalpha)` is the common line of the intersection of the three given planes.