Find the acute angle between the plane 5x - 4y + 7z - 13 = 0 and the y axis . Share with your friends Share 5 Manbar Singh answered this We know that the equation of y-axis is:x-00 = y-01 = 2-00The direction ratios of y-axis are: 0, 1 , 0∴ a1 = 0, b1 = 1 and c1 = 0The equation of given plane is 5x - 4y + 7z - 13 = 0The direction ratios of the normal to the plane are: 5, -4, 7∴ a2 = 5, b2 = -4 and c2 = 7Let θ be the angle between the given plane and y-axisthen, sin θ =a1 a2 + b1 b2 + c1 c2a12 + b12 + c12 a22 + b22 + c22⇒ sin θ = 0 × 5 + 1 × -4 + 0 × 702 + 12 + 02 52 + -42 + 72 ⇒ sin θ = 0 - 4 + 00 + 1 + 0 25 + 16 + 49⇒ sin θ = -41 90⇒ sin θ = -41 × 310⇒ sin θ = -4310⇒ θ = sin-1-4310∴ the acute angle between the given plane and the y-axis is:θ = sin-14310 9 View Full Answer
