Question

Find the vector equation of the plane determined by the points $A(3,-1,2)$, $B(5,2,4)$ and $C(-1,-1,6)$. Also find the distance of point $P(6,5,9)$ from this plane.

Answer 1

Hint: We start solving the problem by checking whether the given three-point are non-collinear or not. If the points are not collinear, we find the equation of the plane using these three points. Once we find the equation of the plane, we find the distance of the point from the plane.

Complete step-by-step solution:
Given that we have three points $A(3,-1,2)$, $B(5,2,4)$ and $C(-1,-1,6)$. We need to find the vector equation of the plane and the distance of the point $P(6,5,9)$ from this plane.
Let us draw all the given information to get a better view.

To find the equation of the plane containing three points, we first need to make sure that the three points are not collinear.
We know if three points $(x_1,y_1,z_1)$, $(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$ are said to be collinear, then the condition to be satisfied is $$\left|\begin{array}{ccc}{x}_1& y_1& z_1\\ x_2& y_2& z_2\\ x_3& y_3& z_3\end{array}\right|=0$$.
Let us find the value of the determinant $$\left|\begin{array}{ccc}3& -1& 2\\ 5& 2& 4\\ -1& -1& 6\end{array}\right|$$.
We know that the determinant of a $3\times 3$ matrix is defined as $\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|=a\times \left|\begin{array}{cc}e& f\\ h& i\end{array}\right|-b\times \left|\begin{array}{cc}d& f\\ g& i\end{array}\right|+c\times \left|\begin{array}{cc}d& e\\ g& h\end{array}\right|$.
$\Rightarrow \left|\begin{array}{ccc}3& -1& 2\\ 5& 2& 4\\ -1& -1& 6\end{array}\right|=3\times \left|\begin{array}{cc}2& 4\\ -1& 6\end{array}\right|-(-1)\times \left|\begin{array}{cc}5& 4\\ -1& 6\end{array}\right|+2\times \left|\begin{array}{cc}5& 2\\ -1& -1\end{array}\right|$.
We know that determinant of $2\times 2$ matrix is defined as $\left|\begin{array}{cc}p& q\\ r& s\end{array}\right|=(p\times s)-(q\times r)$.
$\Rightarrow \left|\begin{array}{ccc}3& -1& 2\\ 5& 2& 4\\ -1& -1& 6\end{array}\right|=3\times ((2\times 6)-(-1\times 4))-(-1)\times ((5\times 6)-(-1\times 4))+2\times ((5\times -1)-(-1\times 2))$.
$\Rightarrow \left|\begin{array}{ccc}3& -1& 2\\ 5& 2& 4\\ -1& -1& 6\end{array}\right|=3\times (\left(12\right)-(-4))-(-1)\times (\left(30\right)-(-4))+2\times ((-5)-(-2))$.
$\Rightarrow \left|\begin{array}{ccc}3& -1& 2\\ 5& 2& 4\\ -1& -1& 6\end{array}\right|=3\times (12+4)-(-1)\times (30+4)+2\times (-5+2)$.
$\Rightarrow \left|\begin{array}{ccc}3& -1& 2\\ 5& 2& 4\\ -1& -1& 6\end{array}\right|=3\times \left(16\right)+1\times \left(34\right)+2\times (-3)$.
$\Rightarrow \left|\begin{array}{ccc}3& -1& 2\\ 5& 2& 4\\ -1& -1& 6\end{array}\right|=48+34-6$.
$\Rightarrow \left|\begin{array}{ccc}3& -1& 2\\ 5& 2& 4\\ -1& -1& 6\end{array}\right|=76$.
So, the points $A(3,-1,2)$, $B(5,2,4)$ and $C(-1,-1,6)$ are not collinear.
We know that the equation of the plane containing three non- collinear points $(x_1,y_1,z_1)$, $(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$ is $$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$$.
Now, we find the equation of the plane containing the points $A(3,-1,2)$, $B(5,2,4)$ and $C(-1,-1,6)$.
The equation of the required plane is $$\left|\begin{array}{ccc}x-3& y-(-1)& z-2\\ 5-3& 2-(-1)& 4-2\\ -1-3& -1-(-1)& 6-2\end{array}\right|=0$$.
$$\Rightarrow \left|\begin{array}{ccc}x-3& y+1& z-2\\ 2& 2+1& 2\\ -4& -1+1& 4\end{array}\right|=0$$.
$$\Rightarrow \left|\begin{array}{ccc}x-3& y+1& z-2\\ 2& 3& 2\\ -4& 0& 4\end{array}\right|=0$$
$$\Rightarrow (x-3)\times \left|\begin{array}{cc}3& 2\\ 0& 4\end{array}\right|-(y+1)\times \left|\begin{array}{cc}2& 2\\ -4& 4\end{array}\right|+(z-2)\times \left|\begin{array}{cc}2& 3\\ -4& 0\end{array}\right|=0$$.
$$\Rightarrow (x-3)\times ((3\times 4)-(0\times 2))-(y+1)\times ((2\times 4)-(2\times -4))+(z-2)\times ((2\times 0)-(-4\times 3))=0$$.
$$\Rightarrow (x-3)\times (\left(12\right)-\left(0\right))-(y+1)\times (\left(8\right)-(-8))+(z-2)\times (\left(0\right)-(-12))=0$$.
$$\Rightarrow (x-3)\times (12-0)-(y+1)\times (8+8)+(z-2)\times (0+12)=0$$.
$$\Rightarrow (x-3)\times \left(12\right)-(y+1)\times \left(16\right)+(z-2)\times \left(12\right)=0$$.
$$\Rightarrow 12x-36-16y-16+12z-24=0$$.
$$\Rightarrow 12x-16y+12z-76=0$$.
$$\Rightarrow 3x-4y+3z-19=0$$ ---(1).
Let us assume the vector in plane $$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$$.
We know that dot product of two vectors $(a\hat{i}+b\hat{j}+c\hat{k})$ and $(d\hat{i}+e\hat{j}+f\hat{k})$ is given as $(a\hat{i}+b\hat{j}+c\hat{k})\bullet (d\hat{i}+e\hat{j}+f\hat{k})=ad+be+cf$.
So, we get $$3x-4y+3z$$ as a dot product of two vectors $(3\hat{i}-4\hat{j}+3\hat{k})$ and $$(x\hat{i}+y\hat{j}+z\hat{k})$$.
So, $$3x-4y+3z=\overrightarrow{r}\bullet (3\hat{i}-4\hat{j}+3\hat{k})$$.
So, we get the vector equation of the plane $$3x-4y+3z-19=0$$ is given as $$\overrightarrow{r}\bullet (3\hat{i}-4\hat{j}+3\hat{k})-19=0$$.
We know that the perpendicular distance of the point $$(x_1,y_1,z_1)$$ to the plane $ax+by+cz+d=0$ is ${\displaystyle \frac{|a{x}_1+b{y}_1+c{z}_1+d|}{\sqrt{a^2+b^2+c^2}}}$.
We need to find the distance from point $P(6,5,9)$ to the plane $$3x-4y+3z-19=0$$. Let the perpendicular distance be ‘d’.
$\Rightarrow d={\displaystyle \frac{|3\left(6\right)-4\left(5\right)+3\left(9\right)-19|}{\sqrt{3^2+{(-4)}^2+3^2}}}$.
$\Rightarrow d={\displaystyle \frac{|18-20+27-19|}{\sqrt{9+16+9}}}$.
$\Rightarrow d={\displaystyle \frac{\left|6\right|}{\sqrt{34}}}$.
$\Rightarrow d={\displaystyle \frac{6}{\sqrt{34}}}$.
∴ We got the distance of the plane $$3x-4y+3z-19=0$$ from point $P(6,5,9)$ is ${\displaystyle \frac{6}{\sqrt{34}}}$.
The equation of the plane is $$3x-4y+3z-19=0$$ and the distance of plane $$3x-4y+3z-19=0$$ from point $P(6,5,9)$ is ${\displaystyle \frac{6}{\sqrt{34}}}$.

Note: We should not just start finding the equation of the plane without checking whether the given points are not collinear. To convert the cartesian form to vector form we always take the form of a general vector present in a plane $$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$$. If we get the perpendicular distance zero then the point lies in the plane.