Question

14. In the following cases, find the distance of each of the given points from thecorresponding given plane.PointPlane(a)(0, 0, 0)3x- 4y + 12z 32xy + 2z3-0(d) -6, 0,0)

Answer 1

(a)

The given plane is 3x−4y+12z=3 and the point is ( 0,0,0 ).

Compare the equation of plane with the standard form of plane that is, Ax+By+Cz=d. Then,

A=3 B=−4 C=12 d=3

The distance of a point ( x 1 , y 1 , z 1 ) from the plane is,

D=| A x 1 +B y 1 +C z 1 −d A 2 + B 2 + C 2 | =| 3×0+( −4 )×0+12×0−3 3 2 + ( −4 ) 2 + 12 2 | = 3 9+16+144 = 3 169

Therefore, the distance between the point and the plane is 3 13  units.

(b)

The given plane is 2x−y+2z+3=0 and the point is ( 3,−2,1 ).

Compare the equation of plane with the standard form of plane that is, Ax+By+Cz=d. Then,

A=2 B=−1 C=2 d=−3

The distance of a point ( x 1 , y 1 , z 1 ) from the plane is,

D=| A x 1 +B y 1 +C z 1 −d A 2 + B 2 + C 2 | =| 2×3+( −1 )( −2 )+2×1−( −3 ) 2 2 + ( −1 ) 2 + 2 2 | = 6+2+2+3 4+1+4 = 13 9

Therefore, the distance between the point and the plane is 13 3  units.

(c)

The given plane is x+2y−2z=9 and the point is ( 2,3,−5 ).

Compare the equation of plane with the standard form of plane that is, Ax+By+Cz=d. Then,

A=1 B=2 C=−2 d=9

The distance of a point ( x 1 , y 1 , z 1 ) from the plane is,

D=| A x 1 +B y 1 +C z 1 −d A 2 + B 2 + C 2 | =| 1×2+2×3+( −2 )×( −5 )−9 1 2 + 2 2 + ( −2 ) 2 | = 2+6+10−9 1+4+4 = 9 9

Therefore, the distance between the point and the plane is 3 units.

(d)

The given plane is 2x−3y+6z−2=0 and the point is ( −6,0,0 ).

Compare the equation of plane with the standard form of plane that is, Ax+By+Cz=d. Then,

A=2 B=−3 C=6 d=2

The distance of a point ( x 1 , y 1 , z 1 ) from the plane is,

D=| A x 1 +B y 1 +C z 1 −d A 2 + B 2 + C 2 | =| 2×( −6 )+( −3 )×0+6×0−2 2 2 + ( −3 ) 2 + 6 2 | =| −12−2 4+9+36 | = 14 49

Simplify further,

D= 14 7 =2 units

Therefore, the distance between the point and the plane is 2 units.