The intersection point of the two lines is (2,1,0).
The respective direction vectors are
L1: <1,2,4>
L2: <4,1,15>
Since the normal vector of the required plane is perpendicular to both direction vectors, the normal vector of the plane is obtained by the cross product of L1 and L2.
i j k
1 2 4
4 1 15
=<30-4, 16-15, 1-8>
=<26, 1, -7>
We know that the plane must pass through (2,1,0), the equation of the plane is
26(x-2)+1(y-1)-7(z-0)=0
simplifying,
26x+y-7z=52+1+0=53
or
26x+y-7z=53
Check:
Put points on L1 in the plane
26(t+2)+(2t+1)-7(4t+0)=53 ok
For L2,
26(4t+2)+(t+1)-7(15t+0)=53 ok
(2^3)=2*2*2=8
(2^-7)=1/2*1/2*1/2*1/2*1/2*1/2*1/2=1/128
8*1/128=1/16=.0625
The 56th term in the sequence is 195
Answer:
4.67×10^6 is the correct answer
:)
