Question

(1) Find the equation of the plane that passes through three points $(1,1,-1),(6,4,-5),(-4,-2,3)$

(2)Find the equation of the plane that passes through three points $(1,1,0),(1,2,1),(-2,2,-1)$

Answer 1

(1) The plane passes through the points

$A(1,1,-1),B(6,4,-5)$ and $C(-4,-2,3)$

Position vector of the points are :

$(1,1,-1),(6,4,-5),(-4,-2,3)$

$\overrightarrow{a}=\hat{i}+\hat{j}-\hat{k}$

$\overrightarrow{b}=6\hat{i}+4\hat{j}-5\hat{k}$

$\overrightarrow{c}=-4\hat{i}-2\hat{j}+3\hat{k}$

Now,

$(\overrightarrow{b}-\overrightarrow{a})=(6\hat{i}+4\hat{j}-5\hat{k})-(\hat{i}+\hat{j}-\hat{k})$

$\Rightarrow (\overrightarrow{b}-\overrightarrow{a})=5\hat{i}+3\hat{j}-4\hat{k}),(\overrightarrow{c}-\overrightarrow{a})=(-4\hat{i}-2\hat{j}+3\hat{k})-(\hat{i}+\hat{j}-\hat{k})$

$\Rightarrow (\overrightarrow{c}-\overrightarrow{a})=-5\hat{i}-3\hat{j}+4\hat{k}=-(5\hat{i}+3\hat{j}-4\hat{k})$

$(\overrightarrow{c}-\overrightarrow{a})$ and $(\overrightarrow{b}-\overrightarrow{a})$ are parallel.

As $\left(\overrightarrow{a}\right)$ is common to both the vectors, so the three points are collinear.

$\therefore$ There can be infinite planes passing through given three points.

(2) The plane passes through the points

$A(1,1,0),B(1,2,1)$ and $C(-2,2,-1)$

Position vector of the points are :

$\overrightarrow{a}=\hat{i}+\hat{j}$,

$\overrightarrow{b}=\hat{i}+2\hat{j}+\hat{k}$

$\overrightarrow{c}=-2\hat{i}-2\hat{j}-\hat{k}$

$(\overrightarrow{b}-\overrightarrow{a})=(\hat{i}+2\hat{j}+\hat{k})-(\hat{i}+\hat{j})$

$\Rightarrow (\overrightarrow{b}-\overrightarrow{a})=\hat{i}+\hat{k}$

$(\overrightarrow{c}-\overrightarrow{a})=(-2\hat{i}+2\hat{j}-\hat{k})-(\hat{i}+\hat{j})$

$\Rightarrow (\overrightarrow{c}-\overrightarrow{a})=-3\hat{i}+\hat{j}-\hat{k}$
Now,

$(\overrightarrow{b}-\overrightarrow{a})\times (\overrightarrow{c}-\overrightarrow{a})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 1& 1\\ -3& 1& -1\end{array}\right|$

$\Rightarrow (\overrightarrow{b}-\overrightarrow{a})\times (\overrightarrow{c}-\overrightarrow{a})=\hat{i}(-1-1)-\hat{j}(0+3)+\hat{k}(0+3)$

$\Rightarrow (\overrightarrow{b}-\overrightarrow{a})\times (\overrightarrow{c}-\overrightarrow{a})=-2\hat{i}-3\hat{j}+3\hat{k}$

Vector equation of a plane passing through given three points with position vectors
$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ is

$(\overrightarrow{r}-\overrightarrow{a}).\left[\right(\overrightarrow{b}-\overrightarrow{a})\times (\overrightarrow{c}-\overrightarrow{a}\left)\right]=0$

$\Rightarrow [\overrightarrow{r}-(\hat{i}+\hat{j}+0\hat{k}\left)\right].(-2\hat{i}-3\hat{j}+3\hat{k})=0$