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.
Answer:
1 2/3 or 1 hour and 20 mins
Step-by-step explanation:
Answer:
Step-by-step explanation:
we know that
The area of the right triangle ABC is equal to
we have
substitute the values
The formula to solve a quadratic equation of the form
is equal to
in this problem we have
so
substitute in the formula
therefore
The solution is
7 is in the hundredths place.
There are 7 one hundredths.
Answer:
1. 50
2. 5
3.19
Step-by-step explanation:
