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.
Here we go.
1) x/5 - 4h = 35
2) {(6-p)/3}+8
3) This is just 1.4k, since the distance you're trying to find is 1.4 times k feet.
You can include a diagram easily enough. Just draw a baseball diamond and two lines, one connecting first and third base and labeled 1.4k, while the other line, connecting home to third, is labeled k.
Hope that helped!
Answer:
n=4
Step-by-step explanation:
Attachment has my work.
Answer:
The student must pay 4247.5 - 4000 = 247.50 dollars in interest
Step-by-step explanation:
