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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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If the points $( 1, 1 , p)$ and $(– 3 , 0, 1)$ be equidistant from the plane $\hat{r} . (3\hat{i} +4\hat{j} -12\hat{k} +13) = 0$ then find the value of $p$.

$\begin{array}{1 1} (A)\; p=1\:\:or\:\:\large\frac{7}{3} \\(B)\; p=1 \\ (C)\; p=\large\frac{7}{3} \\ (D)\; p=-1 \end{array} $

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1 Answer

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Toolbox:
  • Perpendicular distance between a point whose position vector is $\overrightarrow a$ and the plane $\overrightarrow r.\overrightarrow N=d$ is given by $D=\begin{vmatrix}\large\frac{\overrightarrow a.\overrightarrow N-\overrightarrow d}{\mid\overrightarrow N\mid}\end{vmatrix}$
Step 1:
Let the position vector through the point $(1,1,p)$ is $\overrightarrow a_1=\hat i+\hat j+p\hat k$
Similarly let the position vector through the point $(-3,0,1)$ be $\overrightarrow a_2=-4\hat i+\hat k$
The equation of the given plane is $\overrightarrow r.(3\hat i+4\hat j-12\hat k)+13=0$
The perpendicular distance between a point whose position vector is $\overrightarrow a$ and the plane $\overrightarrow r.\overrightarrow N=d$,is given by $D=\begin{vmatrix}\large\frac{\overrightarrow a.\overrightarrow N-\overrightarrow d}{\mid\overrightarrow N\mid}\end{vmatrix}$
Here $\overrightarrow N=3\hat i+4\hat j-12\hat k$ and $d=-13$.
Step 2:
Now substituting this we get
$D_1=\large\frac{\mid(\hat i+\hat j+p\hat k).(3\hat i+4\hat j-12\hat k)+13\mid}{\mid 3\hat i+4\hat j-12\hat k\mid}$
We know that $\hat i.\hat i=\hat j.\hat j=\hat k.\hat k=1$
$D_1=\large\frac{\mid 3+4-12p+13\mid}{\sqrt{3^2+4^2-12^2}}$
$\quad\;\;=\large\frac{20-12p}{13}$
Step 3:
Similarly the distance between the point $(-3,0,1)$ and the given plane is
$D_2=\large\frac{\mid(-3\hat i+\hat k).(3\hat i+4\hat j-12\hat k)+13\mid}{\sqrt{ 3^2+4^2+(-12)^2}}$
$D_2=\large\frac{8}{13}$
Step 4:
But it is given that the distance between the required plane and the points $(1,1,p)$ and $(-3,0,1)$ equal.
Therefore $D_1=D_2$
Hence $\large\frac{20-12p}{13}=\large\frac{8}{13}$
Step 5:
On simplifying we get
$\mid 20-12p\mid=8$
(i.e) $20-12p=8$
Case (i)$\Rightarrow -12p=-12$
$p=1$
Case (ii)$\Rightarrow -20+12p=8$
$-20+12p=8$
$12p=28$
$p=\large\frac{28}{12}=\large\frac{7}{3}$
Therefore $p=1$ or $\large\frac{7}{3}$
answered Jun 5, 2013 by sreemathi.v
 

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