NCERT Solutions
Class 12 Maths
Three Dimensional Geometry

Ex.Misc.Q.14
If the points (1, 1, p) and (−3, 0, 1) be equidistant from the plane r.(3i + 4j – 12k) + 13 = 0, then find the value of p.
The position vector through the point (1, 1, p) is a1 = i + j + pk
Similarly, the position vector through the point (−3, 0, 1) is a1 = -4i + k
The equation of the given plane is r.(3i + 4j – 12k) + 13 = 0
It is known that the perpendicular distance between a point whose position vector is a
and the plane r.N = d is given by D =
Here, N = 3i + 4j – 12k and d = -13
Therefore, the distance between the point (1, 1, p) and the given plane is
D1 =
D1 =
D1 =
D1 =
D1 = ……….1
Similarly, the distance between the point (−3, 0, 1) and the given plane is
D2 =
D2 =
D2 = ……….2
It is given that the distance between the required plane and the points, (1, 1, p) and (−3, 0, 1), is equal.
So, D1 = D2
=>
=> 20 – 12p = 8 or –(20 – 12p) = 8
=> 12p = 12 or 12p = 28
=> p = 1 or p =