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.

$∴D_1=D_2$

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

⇒ $20-12p=8 \ or\ -(20-12p)=8$

⇒ $12p=12 \ or \ 12p=28$

⇒ $p=1\ or \ p=\frac 73$