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## Homework Statement

An arrow is shot in a straight line through the points (3, 4, -1) and (-2, 5, 2) and hits a plane containing the points (2, -1, 3) and (3, -3, 5) and (4, 0, 2). What is the angle between the arrow and the plane?

## The Attempt at a Solution

I've done it like this, I first calculated the equation for the plane to be 0x + 5y + 5z - 10 = 0 and the direction of the arrow to (-5, 1, 3).

The arrow hits the plane in (17/4, 15/4, -7/4) and together with the arrow vector and the normal to the plane (0, 5, 5) a new plane can be calculated with help of the determinant. Its equation is 2x - 5y + 5z + 19 = 0.

I then calculated the line which the two planes have in common: x = -9/2-5t, y = 2-t, z = t.

And the last thing I did was to use the scalar product with (-5, -1, 1) and (-5, 1, 3), which gave me the answer that the angle equals about 29 degrees (28,56).

Is this correct?