OA, OB and OC, with O as the origin, are three mutually perpendicular lines whose lines whose direction cosines are `l_(r),m_(r)andn_(r)(r=1,2 and 3)`. If the projections of OA and OB on the plane z=0 make angles `phi_(1)andphi_(2)`, respectively, with the x-axis, prove that `tan(phi_(1)-phi_(2))=+-n_(3)//n_(1)n_(2)`.

Explanation

Let `A_(1)andB_(1)` be the projections of A and B on the plane z=0. Let OA, OB and OC be of the unit length each so that the coordinates of A,B and C are `A(l_(1),m_(1),n_(1)),B(l_(2),m_(2),n_(2))andC(l_(3),m_(3),n_(3))`. The coordinates of `A_(1)andB_(1)`, therefore, are `A_(1)(l_(1),m_(1),0)andB_(1)(l_(2),m_(2),0)`. Since `OA_(1)andOB_(1)` make angles `phi_(1)andphi_(2),` respectively, with the x-axis, the angle between `OA_(1)andOB_(1)" is "phi_(1)~phi_(2)`. Hence,
`cso(phi_(1)~phi_(2))=(l_(1)l_(2)+m_(1)m_(2))/(sqrt(l_(1)^(2)+m_(1)^(2))sqrt(l_(2)^(2)+m_(2)^(2)))" "(i)`
Aslo OA, OB and OC are mutually perpendicular so that
`l_(1)l_(2)+m_(1)m_(2)+n_(1)n_(2)=0`
and `l_(1)^(2)+m_(1)^(2)+n_(1)^(2)=1`
Eq. (i), therefore, yields
`cos(phi_(1)-phi_(2))=(-n_(1)n_(2))/(sqrt(1-n_(1)^(2))sqrt(1-n_(2)^(2)))`
or `sec^(2)(phi_(1)-phi_(2))=(1-n_(1)^(2)-n_(2)^(2))/(n_(1)^(2)n_(2)^(2))=1+(n_(3)^(2))/(n_(1)^(2)n_(2)^(2))`
`=1+(1-n_(1)^(2)-n_(2)^(2))/(n_(1)^(2)n_(2)^(2))=1+(n_(3)^(2))/(n_(1)^(2)n_(2^(2)))`
or `tan^(2)(phi_(1)-phi_(2))=(n_(3)^(2))/(n_(1)^(2)n_(2)^(2))`
or `tan(phi_(1)-phi_(2))=pm(n_(3))/(n_(1)n_(2)`