Vikas TU
Last Activity: 5 Years ago
Dear student
x + 2y + 3z = c
is a equation of a plane
the normal to the plane is
N= (1,2,3) = i + 2j + 3k
now we know the normal will point in the direction of motion
A is the angle between N the direction of light and the y axis
we know also
unit vector in x direction is
u = (1,0,0) = 1i
so we now can dot product
N dot u = |N| *|u| cos A
we want to solve for cos A so divide by the magnitudes
(N dot u)/|N| = cos A
I didnt write |u| because its a unit vector so that is 1 length so basically doesnt matter
cos A = (1i + 2j + 3k) dot (1i)/ sqrt(1^2 + 2^2 + 3^2)
cos A = 1 /sqrt(14)
thats the answer if you wish
