Question

If the points $\left( 1,1,\lambda \right)$ and $\left( -3,0,1 \right)$ are equidistant from the plane $3x+4y-12x+13=0$, then $\lambda $ satisfies the equation:
(a) $3{{x}^{2}}+10x+7=0$
(b) $3{{x}^{2}}+7x-7=0$
(c) $3{{x}^{2}}-10x+7=0$
(d) $3{{x}^{2}}+10x-7=0$

Answer 1

Hint: Find the distance of $\left( -3,0,1 \right)$ from the given plane and equate this with the distance of points $\left( 1,1,\lambda \right)$ from the plane using the formula: distance of the point $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ from the plane $ax+by+cz+d=0$ is given as \[\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|\] . You will get an equation in $\lambda $. Solve to get the value of $\lambda $.
Complete step-by-step answer:
Given plane is $3x+4y-12z+13=0$ and the points whose distance are equal from this plane are $\left( 1,1,\lambda \right)$ and $\left( -3,0,1 \right)$.
Let us first calculate distance of point $\left( -3,0,1 \right)$ from plane $3x+4y-12z+13=0$-
We know that the distance of the point $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ from the plane ax + by + cz + d = 0 is given as \[\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|\] . So, the distance of $\left( -3,0,1 \right)$ from $3x+4y-12z+13=0$
\[=\left| \dfrac{3\left( -3 \right)+4\left( 0 \right)-12\left( 1 \right)+13}{\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 4 \right)}^{2}}+{{\left( -12 \right)}^{2}}}} \right|\]
\[=\left| \dfrac{-9-12+13}{\sqrt{9+16+144}} \right|=\left| \dfrac{-8}{\sqrt{169}} \right|=\dfrac{8}{13}\]
Now, let us calculate the distance of point $\left( 1,1,\lambda \right)$ from the plane in terms of $\lambda $.
Distance of $\left( 1,1,\lambda \right)$ from $3x+4y-12z+13=0$
$\begin{align}
  & =\left| \dfrac{3\left( 1 \right)+4\left( 1 \right)-12\left( \lambda \right)+13}{\sqrt{9+16+144}} \right| \\
 & =\left| \dfrac{3+4-12\lambda +13}{\sqrt{169}} \right|=\left| \dfrac{20-12\lambda }{13} \right| \\
 & =\left| \dfrac{20-12\lambda }{13} \right| \\
\end{align}$
We are given that both the points are equidistant from the plane.
\[\Rightarrow \left| \dfrac{20-12\lambda }{13} \right|=\dfrac{8}{13}\]
Multiplying both the sides of equation by 13, we will get-
$\left| 20-12\lambda \right|=8$
These are two possible cases: 1. $20-12\lambda \ge 0$ and
                                                       2. $20-12\lambda <0$
Case I: $20-12\lambda \ge 0$
So, modulus will open with +ve sign and we will get-
$20-12\lambda =8$
Taking constants to same side, we will get-
\[\Rightarrow -12\lambda =8-20\]
$\Rightarrow -12\lambda =-12$
Multiplying both sides of equations by “-1”, we will get-
$\Rightarrow 12\lambda =12$
Now dividing both sides of equation by 12, we will get-
$\Rightarrow \lambda =1$
Case II: $20-12\lambda <0$
So, modulus will open with negative sign and we will get-
$\begin{align}
  & -\left( 20-12\lambda \right)=8 \\
 & \Rightarrow -20+12\lambda =8 \\
\end{align}$
Taking constants to same side of equation, we will get-
$\Rightarrow 12\lambda =8+20$
$\Rightarrow 12\lambda =28$
$\Rightarrow \lambda =\dfrac{28}{12}$
Dividing side by 4, we will get-
$\Rightarrow \lambda =\dfrac{7}{3}$
So, we have got two values of $\lambda $: $\lambda =\dfrac{7}{3}\,$ and $\lambda =1$.
We have to form a quadratic equation whose roots are 1 and $\dfrac{7}{3}$.
We know, a quadratic equation can be written as: ${{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta =0$
Where $\alpha \,and\,\beta $ are two roots of the required equation.
So, the required equation will be:
$\begin{align}
  & {{x}^{2}}-\left( 1+\dfrac{7}{3} \right)x+\left( 1 \right)\left( \dfrac{7}{3} \right)=0 \\
 & \Rightarrow {{x}^{2}}-\left( \dfrac{10}{3} \right)x+\dfrac{7}{3}=0 \\
\end{align}$
On multiplying both sides of equation by 3, we will get-
$\begin{align}
  & \Rightarrow 3{{x}^{2}}-10x+7=0\times 3 \\
 & \Rightarrow 3{{x}^{2}}-10x+7=0 \\
\end{align}$
Hence $\lambda $ will satisfy the equation $3{{x}^{2}}-10x+7=0$ and option(c) is the correct answer.

Note: Students can do mistake by not considering the two cases while solving the equation$\left| 120-12\lambda \right|=8$. While solving an equation which involves modulus function, we should always consider two equations that the function inside modules may be greater than equal to zero or may be less than zero.