Question

If the points (1, 1, p) and (– 3, 0, 1) be equidistant from the plane $\vec r.(3\hat i+4\hat j-12)+13=0$, then find the value of p.

Answer 1

Answer:

Step-by-step explanation:

Distance of the point(3,0,1) from the plane 3x+4y-12z+13=0 is=

|(3×3+4×0-12×1+13)/√(3^2+4^2+12^2)|=|10/13|

Distance of the point(1,1,p) from the plane =|(3×1+4×1-12p+13)/13|=10/13

|20-12p|=10

So, p=5/6 or 5/2

Hope I've not done any silly mistakes ^_^;