Find and solve a recurrence relation for the number of different regions formed when n mutually intersecting planes are drawn in three-dimensional space such that no four planes intersect at a common point and no two planes have parallel intersection lines in a third plane

The number of region in which N non-parallel lines can divide a plane is equal to N×( N + 1 )/2 + 1.

Step-by-step explanation:

The attached figures show the maximum number of regions a line can divide a plane. One line can divide a plane into two regions, two non-parallel lines can divide a plane into 4 regions and three non-parallel lines can divide into 7 regions and so on.

When the nth line is added to a cluster of (n-1) lines then the maximum number of extra regions formed is equal to n.

Then the recurrence relation for the number of different regions formed when n mutually intersecting planes are

L(2) – L(1) = 2 … (i)

L(3) – L(2) = 3 … (ii)

L(4) – L(3) = 4 … (iii)

. . .

. . .

L(n) – L(n-1) = n ; … (n)

Adding all the above equation we get,

L(n) – L(1) = 2 + 3 + 4 + 5 + 6 + 7 + …… + n ;

L(n) = L(1) + 2 + 3 + 4 + 5 + 6 + 7 + …… + n ;

L(n) = 2 + 2 + 3 + 4 + 5 + 6 + 7 + …… + n ;

L(n) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + …… + n + 1 ;

L(n) = n ( n + 1 ) / 2 + 1 ;

Therefore, the number of region in which N non-parallel lines can divide a plane is equal to 

N × ( N + 1 )/2 + 1.

answer would be 8.64 or s = 11π/4 inches

step-by-step explanation:

s = r θ  

where s = arc length  

r = radius = 3 inches  

θ = central angle in radians = ( π /180)*165  

s = 3(π/180)*165  

= 8.64 inches