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:
no
Step-by-step explanation:
4 is greater than 0
Answer:
work is attached and shown
Answer:
Step-by-step explanation:
Recall that 
. Using this, we can simplify and combine the two terms:
. Therefore, the terms may be combined.
Answer:
0
Step-by-step explanation: Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add '-20y' to each side of the equation.
15x + 20y + -20y = 0 + -20y
Combine like terms: 20y + -20y = 0
15x + 0 = 0 + -20y
15x = 0 + -20y
Remove the zero:
15x = -20y
Divide each side by '15'.
x = -1.333333333y
Simplifying
x = -1.333333333y
