An angle between lines (r) and a plane (π) is usually equal to acute angle which forms between the direction of lines and the normal vector of the plane.
Determine the angle between the lines r ≡ 2x + 4y - z + 3 = 0, 4x - 2y - z - 1 = 0 and the plane π = 3x - 2y + 3z = 0
r ≡2x + 4y - z + 3 = 04x - 2y - z - 1 = 0 π = 3x - 2y + 3z = 0
Calculate the angle between lines and plane
u→ =2 , 4 , -1 i→ j→ k→ =4 , -2 , -1 =(-4-2)i - (-2+4)j + (-4-16)k =-6i - 2j - 20k =u(-6,-2,-20) π =3x - 2y + 3z = 0 =n(3,-2,3) sin α =|(-6x3) + (-2x-2) + (-20x3)| / √-62 + -22 + -202 x √32 + -22 + 32 = 74 / 20.976176963 x 4.69041576 α = arcsin (0.752131956) α =48.775 Therefore the angle is 48.775