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: Order will be F,D,C and Fourth Option is correct and x = 10
Step-by-step explanation:
Since we have given that
We first transpose the square root to the right , so it becomes square of 8,i.e.
Now, transpose 4 to the right so it will get subtract from 64 i.e.
Since 6 is multiplied to x on tranposing it will get divided by 60 i.e.
Hence, on simplification, we get x=10.
Hence , the order is F,D,C.
Answer:
I think this is the correct solution
I'm not really sure because I have not done this in a while but I think True
Step-by-step explanation:
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Answer: dimond
Step-by-step explanation:
