Question

If the perpendicular distance of a plane from the origin is 1 and d.c’s normal vector to the plane satisfies $\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{y}^{2}}}+\dfrac{1}{{{z}^{2}}}=k$, then find the minimum value of k.

Answer 1

Hint: Let us assume that equation of the plane is $ax+by+cz+d=0$.Find the perpendicular distance of the plane from the origin using the formula: $D=\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|$ and equate it to 1 as given in the question to get the value of $\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$.
Now, assume that direction cosines of the given plane are $\left( l,m,n \right)$. Since, we know that direction cosines of a plane are given as:
$\begin{align}
  & l=\cos \alpha =\dfrac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}} \\
 & m=\cos \beta =\dfrac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}} \\
 & n=\cos \gamma =\dfrac{z}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}} \\
\end{align}$
So, get the values of direction cosines of using the given formula. Since it is mentioned in the question that the direction cosines of the plane satisfy the equation of normal i.e. $\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{y}^{2}}}+\dfrac{1}{{{z}^{2}}}=k$. Therefore, substitute the values of d.c’s in the equation and get the value of k.

Complete step-by-step answer:
As we have assumed that the equation of plane is $ax+by+cz+d=0......(1)$
Now, find the perpendicular distance of the plane from origin using the distance formula: $D=\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|$
Since, at origin ${{x}_{1}}={{y}_{1}}={{z}_{1}}=0$, substituting the values in direction formula, we get:
 $\begin{align}
  & D=\left| \dfrac{a\times 0+b\times 0+c\times 0+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right| \\
 & D=\left| \dfrac{d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|......(2) \\
\end{align}$
As it is mentioned that the perpendicular distance of the plane from origin is 1, i.e. D = 1
So, we have:
$\begin{align}
  & \Rightarrow \left| \dfrac{d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|=1 \\
 & \Rightarrow d=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} \\
 & \Rightarrow {{d}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}......(3) \\
\end{align}$
Now, we need to find the values of direction cosines of the plane given. As we have assumed that the direction cosines of the plane are $\left( l,m,n \right)$
So, by using the formula for direction cosines:
$\begin{align}
  & l=\cos \alpha =\dfrac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}} \\
 & m=\cos \beta =\dfrac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}} \\
 & n=\cos \gamma =\dfrac{z}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}} \\
\end{align}$
We get:
$\begin{align}
  & l=\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \\
 & m=\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \\
 & n=\dfrac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \\
\end{align}$
As we know that $\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}=d$, we can write direction cosines as:
$\begin{align}
  & l=\dfrac{a}{d}......(4) \\
 & m=\dfrac{b}{d}......(5) \\
 & n=\dfrac{c}{d}......(6) \\
\end{align}$
Since it is mentioned that the direction cosines of the plane satisfy the equation of normal to the plane, substitute the values of $\left( l,m,n \right)$ in the equation of normal $\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{y}^{2}}}+\dfrac{1}{{{z}^{2}}}=k$, we get:
$\begin{align}
  & \Rightarrow \dfrac{1}{{{\left( \dfrac{a}{d} \right)}^{2}}}+\dfrac{1}{{{\left( \dfrac{b}{d} \right)}^{2}}}+\dfrac{1}{{{\left( \dfrac{c}{d} \right)}^{2}}}=k \\
 & \Rightarrow {{\left( \dfrac{d}{a} \right)}^{2}}+{{\left( \dfrac{d}{b} \right)}^{2}}+{{\left( \dfrac{d}{b} \right)}^{2}}=k \\
 & \Rightarrow {{d}^{2}}\left( \dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}} \right)=k \\
\end{align}$
Since we found that ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}={{d}^{2}}$
By substituting the vale of ${{d}^{2}}$, we get:
$\begin{align}
  & \Rightarrow \left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( \dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}} \right)=k \\
 & \Rightarrow \left( \left( {{b}^{2}}+{{c}^{2}} \right)+\left( {{a}^{2}}+{{c}^{2}} \right)+\left( {{a}^{2}}+{{b}^{2}} \right) \right)=k \\
 & \Rightarrow 2\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)=k \\
 & \Rightarrow 2{{d}^{2}}=k......(7) \\
\end{align}$
From equation (7), we conclude that k is a positive even number.
Since positive even numbers are 0, 2, 4, …. So, the minimum number that is positive and even is 0.
Therefore, the minimum value of k is 0.

Note: The direction cosine of a line is defined as the cosine of the angles between the positive directed lines and the coordinate axes. If$\alpha ,\beta $and $\gamma $ are the three angles between the directed line segment and the coordinate axes, then these three angles are considered as direction angles.