Question

Find the coordinates of the point where the line x-8/4=y-1/1=z-3/8 intersect the plane 2x+2y+z=3. also find the angle between the line and the plane.​

Answer 1

Answer:

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Answer 2

The coordinates of the point where the line x-8/4=y-1/1=z-3/8 intersect the plane 2x+2y+z=3 :

• Equation of line : x-8/4=y-1/1=z-3/8

• Equation of plane : 2x+2y+z=3

• Let, x-8/4 = y-1/1 = z-3/8 = k

x = 4k+8, y = k+1, z = 8k+3

• Substitute values of x, y and z in equation of plane

2(4k+8) + 2(k+1) + 8k + 3 = 3

18k = -18

• k = -1

x = -4 + 8 = 4 , y = -1 + 1 = 0, z = -8 + 3 = 5

• Point of intersection ( 4, 0, 5 )

• Angle between line and plane is given by formula ;

• Sin ɵ = | n . b / |n| |b| |

Here, n = ( 2, 2, 1 ) and b = ( 4, 1, 8 )

• Sin ɵ = | 2.4 + 2.1 + 1.8 / √( 4^2 +1^2 + 8^2) × √( 2^2 + 2^2 + 1^2) |

= | 18 / 3×√71 | = 6/√71

• ɵ = Sin^-1 { 6/√71 }