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Question: Find the equation of a plane passing through the point P = \((-3,-3,1)\) and perpendicular to the li...

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

Explanation

Solution

We know that the equation of a plane in cartesian form is a(x−x0)+b(y−y0)+c(z−z0)=0a(x-{{x}_{0}})+b(y-{{y}_{0}})+c(z-{{z}_{0}})=0 and whose directional ratios are a=x2−x1,b=y2−y1,c=z2−z1a={{x}_{2}}-{{x}_{1}},b={{y}_{2}}-{{y}_{1}},c={{z}_{2}}-{{z}_{1}} and plane passing through a point P =(x0,y0,z0)({{x}_{0}},{{y}_{0}},{{z}_{0}}).Also the line equation passing through two points(x1,y1,z1),(x2,y2,z2)({{x}_{1}},{{y}_{1}},{{z}_{1}}),({{x}_{2}},{{y}_{2}},{{z}_{2}}) in cartesian form is x−x1x2−x1=y−y1y2−y1=z−z1z2−z1\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 (x0,y0,z0)({{x}_{0}},{{y}_{0}},{{z}_{0}})=(−3,−3,1)(-3,-3,1)
Also given a line AB whose joining points as A=(x1,y1,z1)=(2,6,1);B=(x2,y2,z2)=(1,3,0);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 x−x1x2−x1=y−y1y2−y1=z−z1z2−z1→equation(1)\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 (x1,y1,z1),(x2,y2,z2)({{x}_{1}},{{y}_{1}},{{z}_{1}}),({{x}_{2}},{{y}_{2}},{{z}_{2}}) values in equation(1) , then we get as follows
⇒x−x1x2−x1=y−y1y2−y1=z−z1z2−z1\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}}}
⇒x−21−2=y−63−6=z−10−1\Rightarrow \dfrac{x-2}{1-2}=\dfrac{y-6}{3-6}=\dfrac{z-1}{0-1}
On simplifying above equation we get as follows
⇒x−2−1=y−6−3=z−1−1\Rightarrow \dfrac{x-2}{-1}=\dfrac{y-6}{-3}=\dfrac{z-1}{-1}
Therefore, the required directional ratios are a=x2−x1=−1;b=y2−y1=−3;c=z2−z1=−1;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−x0)+b(y−y0)+c(z−z0)=0→equation(2)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 (x0,y0,z0)({{x}_{0}},{{y}_{0}},{{z}_{0}}) in equation(2) we get the plane equation as
a(x−x0)+b(y−y0)+c(z−z0)=0a(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(-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(-1)(x-(-3))+(-3)(y-(-3))+(-1)(z-(1))=0
−x−3−3y−9−z+1=0-x-3-3y-9-z+1=0
Add constant terms i.e., −3,−9,+1-3,-9,+1 then we get −11-11 as result
−x−3y−z−11=0-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=−11x+3y+z=-11

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

Note: We can also find the equation of a plane in vector form by using vector equation of a plane
(r⃗−a⃗).n⃗=0\left( \vec{r}-\vec{a} \right).\vec{n}=0 this can also be written as r⃗.n⃗=d,\vec{r}.\vec{n}=d,where d=a⃗.n⃗d=\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=(x0,y0,z0)({{x}_{0}},{{y}_{0}},{{z}_{0}})= (−3,−3,1)(-3,-3,1) and
plane perpendicular to the line joining two point are
A=(x1,y1,z1)=(2,6,1);B=(x2,y2,z2)=(1,3,0);A=({{x}_{1}},{{y}_{1}},{{z}_{1}})=(2,6,1);B=({{x}_{2}},{{y}_{2}},{{z}_{2}})=(1,3,0);
Here n⃗=Position\vec{n}=Position B⃗\vec{B} −Position-Position A⃗\vec{A}
n⃗=((1)i^+(3)j^+(0)k^)−((2)i^+(6)j^+(1)k^)\vec{n}=((1)\hat{i}+(3)\hat{j}+(0)\hat{k})-((2)\hat{i}+(6)\hat{j}+(1)\hat{k})
n⃗=(−i^−3j^−k^)\vec{n}=(-\hat{i}-3\hat{j}-\hat{k})
Also a⃗=−3i^−3j^+1k^\vec{a}=-3\hat{i}-3\hat{j}+1\hat{k}
r⃗=xi^+yj^+zk^\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}
Therefore,now the required plane equation is
(r⃗−a⃗).n⃗=0⇒((xi^+yj^+zk^)−(−3i^−3j^+1k^)).(−i^−3j^−k^)=0(\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