14. If the points (1, 1, p) and (– 3, 0, 1) be equidistant from the plane \overrightarrow{r}.(3\widehat{i}+4\widehat{j}-12\widehat{k})+13 =0 then find the value of p.

Answers (1)
D Divya Prakash Singh

Given that the points A(1,1,p) and B(-3,0,1) are equidistant from the plane

\overrightarrow{r}.(3\widehat{i}+4\widehat{j}-12\widehat{k})+13 =0

So we can write the position vector through the point (1,1,p) is \vec{a_{1}} = \widehat{i}+\widehat{j}+p\widehat{k}

Similarly, the position vector through the point B(-3,0,1) is

\vec{a_{2}} = -4\widehat{i}+\widehat{k}

The equation of the given plane is \overrightarrow{r}.(3\widehat{i}+4\widehat{j}-12\widehat{k})+13 =0

and We know that the perpendicular distance between a point whose position vector is \vec{a}  and the plane, \vec{n} = 3\widehat{i}+4\widehat{j}-12\widehat{k}  and d =-13

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

D_{1} = \frac{\left | (\widehat{i}+\widehat{j}+p\widehat{k}).(3\widehat{i}+4\widehat{j}-12\widehat{k})+13 \right |}{3\widehat{i}+4\widehat{j}-12\widehat{k}}

D_{1} = \frac{\left | 3+4-12p+13 \right |}{\sqrt{3^2+4^2+(-12)^2}}

D_{1} = \frac{\left | 20-12p \right |}{13}                                 .........................(1)

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

D_{2} = \frac{\left | (-3\widehat{i}+\widehat{k}).(3\widehat{i}+4\widehat{j}-12\widehat{k})+13 \right |}{3\widehat{i}+4\widehat{j}-12\widehat{k}}

D_{2} = \frac{\left |-9-12+13 \right |}{\sqrt{3^2+4^2+(-12)^2}}

D_{2} = \frac{8}{13}                                                             .........................(2)

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

\therefore D_{1} =D_{2}

\implies \frac{\left | 20-12p \right |}{13} =\frac{8}{13}

therefore we have,

\implies 12p =12

or p =1  or  p = \frac{7}{3}