Multivarible calculus problem?

Question

Find the angle (in degrees) between the plane x+6y+z=3 and the line with symmetric equations (1/6)x+(1/2) = y-1 = z-1.

Answer 1

get the angle between the normal vector of the plane and the direction vector of the line.

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then get the complement of that angle.

Answer 2

The coeffs. of the variables defining the plane 1,6,1 are the components of a vector i +6j +k which is perp. to the plane.
The nos. in the denom for each of the parts of the eq of a line in the symmetric form 6, 1, 1 are the components of a vector parallel to the line. Call the first A and the second B.
A dot B =|A| |B| cos angle (between A and B)
A dot B = 13
|A||B|cos angle = sqrt38 sqrt 38 cos angle = 38 cos angle
cos angle = 13/38. The angle is approx. 70degs.
But that's the angle the line makes with the perp. to the plane. So the angle you want is it's complement. 20degs.
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Answer 3

Find a vector that points in the direction of the line.
Find a vector that is normal to the plane
Use the definition of the dot product to find the angle between the two.
Keep in mind that this is the angle between the line and the normal, not the plane.