Question

Let the plane $ax+by+cz+d=0$ bisect the line joining the points $\left(4,-3,1\right)$ and $\left(2,3,-5\right)$ at the right angles. If $a$, $b$, $c$, $d$ are integers, then the minimum value of $\left(a^2+b^2+c^2+d^2\right)$ is

Answer 1

Explanation for the correct option:

Step-1 Using mid point concept:

Given that the equation of the plane is $ax+by+cz+d=0$ and the line passed the plane having the coordinates as $P\left(4,-3,1\right)$ and $Q\left(2,3,-5\right)$.

Let $M\left(x,y,z\right)$ be the mid-point of the line passing through the plane.

Using the mid-point formula find the coordinate of $M\left(x,y,z\right)$.

$$\begin{gathered} x=\frac{x_1+x_2}{2} \\ \Rightarrow x=\frac{4+2}{2} \\ \Rightarrow x=\frac{6}{2} \\ \Rightarrow x=3 \end{gathered}$$

And

$$\begin{gathered} y=\frac{y_1+y_2}{2} \\ \Rightarrow y=\frac{-3+3}{2} \\ \Rightarrow y=\frac{0}{2} \\ \Rightarrow y=0 \end{gathered}$$

And

$$\begin{gathered} z=\frac{z_1+z_2}{2} \\ \Rightarrow z=\frac{1-5}{2} \\ \Rightarrow z=\frac{-4}{2} \\ \Rightarrow z=-2 \end{gathered}$$

Therefore, the mid-point $M\left(x,y,z\right)$ is obtained as $\left(3,0,-2\right)$.

Step-2 : Normal vector to the plane:

The normal of the plane $PQ$ is determined as,

$PQ=-2i+6j-6k\mathbf{}\left[\text{where coefficient of}i=2-4,\mathrm{coefficient}\mathrm{of}j=3-(-3),\mathrm{coefficient}\mathrm{of}k=-5-1\right]$

Then the value of $a$ is $-2$, $b$ is $6$ and $c$ is $-6$.

Now the equation of plane is $-2x+6y-6z+d=0$

Using the mid-point coordinate $\left(3,0,-2\right)$ finding the value of $d$.

$$\begin{gathered} -2\left(3\right)+6\left(0\right)-6\left(-2\right)+d=0 \\ \Rightarrow -6+0+12+d=0 \\ \Rightarrow 6+d=0 \\ \Rightarrow d=-6 \end{gathered}$$

Therefore, the equation of the plane is $-2x+6y-6z+-6=0$.

$\Rightarrow x-3y+3z+3=0\left[\text{equation is of the form}\mathrm{ax}+\mathrm{by}+\mathrm{cz}+d=0\right]$

Now substitute the values of $a$ as $1$, $b$ as $-3$, $c$ as $3$ and $d$ as $3$ in $\left(a^2+b^2+c^2+d^2\right)$.

$$\begin{gathered} a^2+b^2+c^2+d^2={\left(1\right)}^2+{\left(-3\right)}^2+{\left(3\right)}^2+{\left(3\right)}^2 \\ =1+9+9+9 \\ =28 \end{gathered}$$

Therefore, the minimum value of $\left(a^2+b^2+c^2+d^2\right)$ is $28$.