Answer: -1
Step-by-step explanation:
The cross product of the normal vectors of two planes result in a vector parallel to the line of intersection of the two planes.
Corresponding normal vectors of the planes are
<5,-1,-6> and <1,1,1>
We calculate the cross product as a determinant of (i,j,k) and the normal products
i j k
5 -1 -6
1 1 1
=(-1*1-(-6)*1)i -(5*1-(-6)1)j+(5*1-(-1*1))k
=5i-11j+6k
=<5,-11,6>
Check orthogonality with normal vectors using scalar products
(should equal zero if orthogonal)
<5,-11,6>.<5,-1,-6>=25+11-36=0
<5,-11,6>.<1,1,1>=5-11+6=0
Therefore <5,-11,6> is a vector parallel to the line of intersection of the two given planes.
we know that
The measurement of the exterior angle is the semi-difference of the arcs which comprises
In this problem
∠FGH is the exterior angle
∠FGH=
∠FGH=
-----> equation A
--------> equation B
Substitute equation B in equation A
![100\°=(arc\ FEH-[360\°-arc\ FEH])](https://tex.z-dn.net/?f=100%5C%C2%B0%3D%28arc%5C%20FEH-%5B360%5C%C2%B0-arc%5C%20FEH%5D%29)
therefore
<u>The answer is</u>
The measure of arc FEH is equal to 
For a regular polygon with n sides, interior angle
= [(n-2) × 180°]/n
So, interior angle of this regular heptagon shape
= [(7 - 2) × 180°]/7
= (5 × 180°)/7
= 900°/7
= (900/7)°
= 128.571° [approximately]
