Question

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

Answer 1

The position vector through the point (1, 1, p) is
\(\vec a_1 =\hat i+\hat j+p\hat k\)

Similarly, the position vector through the point (-3, 0, 1)is
\(\vec a_2 =-4\hat i+\hat k\)

The equation of the given plane is
\(\vec r.(3\hat i+4\hat j-12\hat k)+13=0\)

It is known that the perpendicular distance between a point whose position vector is \(\vec a\) and the plane,\(\vec r.\vec N =d\), is given by,\(D=\frac {|\vec a.\vec N-d|}{|\vec N|}\)

Here, \(\vec N=3\hat i+4\hat j-12\hat k\) and \(d=-13\)

Therefore,the distance between the point(1,1,p)and the given plane is

\(D_1=\frac {|(\hat i+\hat j+p\hat k).(3\hat i+4\hat j-12\hat k)+13|}{|3\hat i+4\hat j-12\hat k|}\)

⇒\(D_1=\frac {|3+4-12p+13|}{\sqrt {3^2+4^2+(-12)^2}}\)

⇒\(D_1=\frac {|20-12p|}{13}\)        ...(1)

Similarly,the distance between the point(-3,0,1)and the given plane is

\(D_2=\frac {|(-3\hat i+\hat k).(3\hat i+4\hat j-12\hat k)+13|}{|3\hat i+4\hat j-12\hat k|}\)

⇒\(D_2=\frac {|-9-12+13|}{\sqrt {3^2+4^2+(-12)^2}}\)]

⇒\(D_2=\frac {8}{13}\)        ...(2)

It is given that the distance between the required plane and the points (1, 1, p) and (-3, 0, 1) is equal.

\(∴D_1=D_2\)

⇒ \(\frac {|20-12p|}{13} =\frac {8}{13}\)

⇒ \(20-12p=8 \ or\ -(20-12p)=8\)

⇒ \(12p=12 \ or \ 12p=28\)

⇒ \(p=1\  or \ p=\frac 73\)