It would be 70r if you are using distributive property
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.
The first part of the second line, she left the -5 there. The correct work and solution should be this:
5(2x-1)-3x=5x+9
<span><span>7x</span>−5</span>=<span><span>5x</span>+<span>9
</span></span>2x-5=9
2x=14
x=7
Answer:
I need points!!!!!!!!
Step-by-step explanation:
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