Answer:
The nonzero vector orthogonal to the plane is <-9,-8,2>.
Step-by-step explanation:
Consider the given points are P=(0,0,1), Q=(−2,3,4), R=(−2,2,0).
The nonzero vector orthogonal to the plane through the points P,Q, and R is
Expand along row 1.
Therefore, the nonzero vector orthogonal to the plane is <-9,-8,2>.
Answer:
see explanation
Step-by-step explanation:
(1)
Since FE = FG the triangle is isosceles with ∠ E = ∠ G, then
∠ E = =
= 37°
(2)
Since all 3 sides are congruent then triangle is equilateral with the 3 angles being congruent, 60° each , then
12y = 60 ( divide both sides by 12 )
y = 5
(3)
The 3 angles are congruent then triangle is equilateral with the 3 sides being congruent, then
KL = JL , that is
4t - 8 = 2t + 1 ( subtract 2t from both sides )
2t - 8 = 1 ( add 8 to both sides )
2t = 9 ( divide both sides by 2 )
t = 4.5
(4)
Given ∠ B = ∠ C then triangle is isosceles with 2 legs being congruent , that is
AB = AC
4x + 1 = 9 ( subtract 1 from both sides )
4x = 8 ( divide both sides by 4 )
x = 2
Then
perimeter = AB + BC + AC = 4x + 1 + 2x + 3 + 9
= 6x + 13
= 6(2) + 13
= 12 + 13
= 25