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:
0.25
Step-by-step explanation:
3.5 divided by 14
3.5 ÷ 14 = 0.25
Answer:
-0.4
Step-by-step explanation:
you just take 2 of them to find it
Answer:
90
Step-by-step explanation:
3600 / 40 = 90
so
90 x 40 = 3600
Answer:
-1/3
Step-by-step explanation:
Its opposite reciprocal, mean that 3/1 (which is 3) will have the numerator swapped with its denominator.
So basically 3/1 would turn into 1/3
Now the sign will be the opposite as well.
So since 3 is positive, the new slow should be a negative.
-1/3