Question

If the points (1, 1, p ) and (−3, 0, 1) be equidistant from the plane , then find the value of p .

Answer 1

The given points A=( 1,1,p ) and B=( −3,0,1 ) are equidistant from the plane r → ⋅( 3 i ^ +4 j ^ −12 k ^ )+13=0.

The position vector of a point ( x,y,z ) is given as,

r → =x i ^ +y j ^ +z k ^

Therefore, the position vector of the point ( 1,1,p ) is,

r → = i ^ + j ^ +p k ^

Similarly, the position vector of the point ( −3,0,1 ) is,

r → =−4 i ^ + k ^

We know that, the perpendicular distance between a point whose position vector is r → and the plane r → ⋅ N → =d is given by,

D=| r → ⋅ N → −d | N → | |(1)

Substitute, the values of r → , N → and din equation (1), to get the distance between the point ( 1, 1, p ) and the given plane,

D 1 =| ( i ^ + j ^ +p k ^ )⋅( 3 i ^ +4 j ^ −12 k ^ )+13 | 3 i ^ +4 j ^ −12 k ^ | | = | 3+4−12p+13 | ( 3 ) 2 + ( 4 ) 2 + ( −12 ) 2 = | 20−12p | 13

Substitute, the values of r → , N → and din equation (1), to get the distance between the point ( −3,0,1 ) and the given plane,

D 2 =| ( −3 i ^ + k ^ )⋅( 3 i ^ +4 j ^ −12 k ^ )+13 | 3 i ^ +4 j ^ −12 k ^ | | = | −9−12+13 | ( 3 ) 2 + ( 4 ) 2 + ( −12 ) 2 = 8 13

It is given that the distance D 1 and D 2 are equal.

So,

D 1 = D 2 | 20−12p | 13 = 8 13 | 20−12p |=8

Solve further.

20−12p=8 12p=12 p=1

or,

−( 20−12p )=8 −20+12p=8 12p=28 p= 7 3

Thus, the value of p for which the points A=( 1,1,p ) and B=( −3,0,1 ) are equidistant from the plane r → ⋅( 3 i ^ +4 j ^ −12 k ^ )+13=0 is either 1 or 7 3 .