If the angle between the line x+1 1 = y-1 2 = z-2 2 and the plane 2 x- y + z +4=0 is such that = 1 3 , then +1=

Step-by-step solution

The angle between a line with direction vector b and a plane with normal vector n satisfies = | b n | | b | | n | . Given the line x+1 1 = y-1 2 = z-2 2 , its direction vector is b = 1, 2, 2 with magnitude | b | = 1^2 + 2^2 + 2^2 = 3 . The plane 2x - y + z + 4 = 0 has normal vector n = 2, -1, and magnitude | n | = 4 + 1 + = 5 + . The dot product evaluates to b n = 2 - 2 + 2 = 2 . Substituting = 1 3 into the formula yields: 1 3 = 2 3 5 + Multiplying both sides by 3 gives 1 = 2 5 + . Squaring both sides produces 1 =