Answer:
27/77
Step-by-step explanation:
I hope this is correct
Have a good Day!
This involves quite a lot of arithmetic to do manually.
The first thing you do is to make the first number in row 2 = to 0.
This is done by R2 = -3/2 R1 + R2
so the matrix becomes
( 2 1 1) ( -3 )
( 0 -13/2 3/2) (1/2 )
(5 -1 2) (-2)
Next step is to make the 5 in row 5 = 0
then the -1 must become zero
You aim for the form
( 1 0 0) (x)
(0 1 0) (y)
(0 0 1) ( z)
x , y and z will be the required solutions.
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.
I pray to hod for yourgood day
The area of a circle is given by

We can solve this equation for r: divide both sides by pi:

Consider the (positive) square root of both sides:

So, in your case, we have
