Question

Find the equation of a plane passing through the point P = $(-3,-3,1)$ and perpendicular to the line AB whose joining points are A=$(2,6,1)$ and B=$(1,3,0)$.

Answer 1

Hint: We know that the equation of a plane in cartesian form is $a(x-{{x}_{0}})+b(y-{{y}_{0}})+c(z-{{z}_{0}})=0$ and whose directional ratios are $a={{x}_{2}}-{{x}_{1}},b={{y}_{2}}-{{y}_{1}},c={{z}_{2}}-{{z}_{1}}$ and plane passing through a point P =$({{x}_{0}},{{y}_{0}},{{z}_{0}})$.Also the line equation passing through two points$({{x}_{1}},{{y}_{1}},{{z}_{1}}),({{x}_{2}},{{y}_{2}},{{z}_{2}})$ in cartesian form is $\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}$.

Complete step by step answer:
From the problem,
 we have given that the plane passing through the point P is $({{x}_{0}},{{y}_{0}},{{z}_{0}})$=$(-3,-3,1)$
Also given a line AB whose joining points as $A=({{x}_{1}},{{y}_{1}},{{z}_{1}})=(2,6,1);B=({{x}_{2}},{{y}_{2}},{{z}_{2}})=(1,3,0);$
Now the required line equation in cartesian form is $\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}\to equation(1)$.
On substituting $({{x}_{1}},{{y}_{1}},{{z}_{1}}),({{x}_{2}},{{y}_{2}},{{z}_{2}})$ values in equation(1) , then we get as follows
$\Rightarrow \dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}$
$\Rightarrow \dfrac{x-2}{1-2}=\dfrac{y-6}{3-6}=\dfrac{z-1}{0-1}$
On simplifying above equation we get as follows
$\Rightarrow \dfrac{x-2}{-1}=\dfrac{y-6}{-3}=\dfrac{z-1}{-1}$
Therefore, the required directional ratios are $a={{x}_{2}}-{{x}_{1}}=-1;b={{y}_{2}}-{{y}_{1}}=-3;c={{z}_{2}}-{{z}_{1}}=-1;$
We know that the equation of a plane in cartesian form is as follows
$a(x-{{x}_{0}})+b(y-{{y}_{0}})+c(z-{{z}_{0}})=0\to equation(2)$
And now let us substitute the a,b,c and $({{x}_{0}},{{y}_{0}},{{z}_{0}})$ in equation(2) we get the plane equation as
$a(x-{{x}_{0}})+b(y-{{y}_{0}})+c(z-{{z}_{0}})=0$
Simplify carefully by placing all the values in respective places
$\Rightarrow $$(-1)(x-(-3))+(-3)(y-(-3))+(-1)(z-(1))=0$
On simplifying we get as shown below
\[(-1)(x-(-3))+(-3)(y-(-3))+(-1)(z-(1))=0\]
\[-x-3-3y-9-z+1=0\]
Add constant terms i.e., $-3,-9,+1$ then we get $-11$ as result
\[-x-3y-z-11=0\]
On shifting the variable terms to L.H.S and constant terms to R.H.S we get
 \[x+3y+z=-11\]

The required plane equation in cartesian form is “\[x+3y+z=-11\]”.

Note: We can also find the equation of a plane in vector form by using vector equation of a plane
$\left( \vec{r}-\vec{a} \right).\vec{n}=0$ this can also be written as $\vec{r}.\vec{n}=d,$where $d=\vec{a}.\vec{n}$ known as scalar product of a plane.
Alternate Method:
From the given problem the plane passing through the point P= a=$({{x}_{0}},{{y}_{0}},{{z}_{0}})$= $(-3,-3,1)$ and
plane perpendicular to the line joining two point are
$A=({{x}_{1}},{{y}_{1}},{{z}_{1}})=(2,6,1);B=({{x}_{2}},{{y}_{2}},{{z}_{2}})=(1,3,0);$
Here \[\vec{n}=Position\] $\vec{B}$ \[-Position\] $\vec{A}$
$\vec{n}=((1)\hat{i}+(3)\hat{j}+(0)\hat{k})-((2)\hat{i}+(6)\hat{j}+(1)\hat{k})$
\[\vec{n}=(-\hat{i}-3\hat{j}-\hat{k})\]
Also $\vec{a}=-3\hat{i}-3\hat{j}+1\hat{k}$
$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$
Therefore,now the required plane equation is
$(\vec{r}-\vec{a}).\vec{n}=0\Rightarrow ((x\hat{i}+y\hat{j}+z\hat{k})-(-3\hat{i}-3\hat{j}+1\hat{k})).(-\hat{i}-3\hat{j}-\hat{k})=0$