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.
It should be B If I am correct if I am not I am sorry
Answer:
-10
Step-by-step explanation:
-(x-5) ^2+25
-x+5^2+25
-5 -5
-x^2+20
-2x+20
---- ----
-2 -2
x=-10
// have a great day //
Try using an app called Symbolab it helps a lot !
Answer:
90
Step-by-step explanation:
180°-104°=76°( sum of angles A and B)
76°÷2=38°
x=38°
Bcuz it is an isosceles triangle ( a triangle with 2 equal sides). So, angle B and A are equal. thats why 76° must be divided by 2.
180°-104°=76°(angles on a straight line)
180°-76°=104° (sum of angle A and C)
104°÷2=52°
y=52°
it is also an isosceles triangle.
x+y=38+52=90°
