Answer:
angle of intersection: 5.2°
Step-by-step explanation:
The direction vector normal to the plane is ...
n = (1, 1, 3)
The direction vector of the line is ...
m = (1, -3, 1)
Then the angle θ between them can be found from the dot product:
n•m = |n|·|m|·cos(θ)
(1·1 +1(-3) +3·1) = 1 -3 +3 = 1 = √(1²+1²+3²)·√(1²+(-3)²+1²)·cos(θ)
1 = 11·cos(θ)
θ = arccos(1/11) ≈ 84.8°
This is the angle between the line and the normal to the plane, so the angle between the line and the plane will be the complement of this. Since this angle is not 90°, <em>the line and plane must intersect</em>.
acute angle = 90° -84.8° = 5.2°
_____
The attached graph shows the line and plane meet at a shallow angle, consistent with the above answer.
The area of the entire sector of DEF = 60 / 360 * PI * radius^2
sector area = 1 / 6 * 3.14159265... * 20^2
sector area =
<span>
<span>
<span>
209.4395102393
</span>
</span>
</span>
segment area = sector area - triangle DEF Area
triangle DEF Area = (1/2) * 20 * sine 60 * 20
triangle DEF Area = (1/2) * 0.86603 * 400
triangle DEF Area = <span><span><span>(1/2) * 346.412
</span>
</span>
</span>
triangle DEF Area =
<span>
<span>
<span>
173.206
</span>
</span>
</span>
segment area = <span>
<span>
209.4395102393
</span>
-173.206
</span>
segment area =
<span>
<span>
<span>
36.2335102393
</span>
</span>
</span>
segment area =
36.23 m
Source:
http://www.1728.org/circsect.htm
Answer:I onesly dont no
Sorry I have no friends
Step-by-step explanation:
