Please check my answers for me. They are two questions, both I have answered, but I am not sure if I am fully correct
2 answers:
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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°
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The attached graph shows the line and plane meet at a shallow angle, consistent with the above answer.
