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Suppose that the line ℓ is represented by r(t)=⟨18+4t,13+4t,10+2t⟩ and the plane P is represented by 3x−2y+6z=32. ------------------------------------------ 1. Find the intersection of the line ℓ and the plane P. Write your answer as a point (a,b,c) where a , b , and c are numbers. 2. Find the cosine of the angle θ between the line ℓ and the normal vector of the plane P .

Mathematics
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solve 3(18+4t) -2(13+4t) +6(10+3t) =32 for t
for 2, the normal to the plane is <+3 i -2 J +6k> and the vector along the line is <+4 i +4 j +2 k> use the definition of the dot product of two vectors to find the angle between them...
A dot B = |A||B| cos (theta)

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Other answers:

for 2, it worked but for 1 you are suppose to find a point.... how do i find a point with just one value (t)?
plug t into the parametric representation of the line
THANKS!!!!:)

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