Question

Find the equation of the plane through the points (3,2,2),(1,0,-1) and parallel to the line \[\dfrac{x-1}{1}=\dfrac{y-1}{-2}=\dfrac{z-2}{3}\]. Convert into vector form.

Answer 1

Hint: Find the vector equation of the normal to the plane by the cross product of the vectors which represent the given line parallel to the plane and the line joining the two points on the plane. Then, find the equation of the line joining any variable point on the plane to any of the given two points using the concept that the vector equation of a line passing through the points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is given as $\left( {{x}_{1}}-{{x}_{2}} \right)\widehat{i}+\left( {{y}_{1}}-{{y}_{2}} \right)\widehat{j}+\left( {{z}_{1}}-{{z}_{2}} \right)\widehat{k}$. Then, find the equation of the plane by equating the dot product of the normal and the equation of the line joining any variable point on the plane to any of the given two points to 0. Then, convert the resulting equation into vector equation of the plane.

Complete step by step solution:
In the question, it is given that the line \[\dfrac{x-1}{1}=\dfrac{y-1}{-2}=\dfrac{z-2}{3}\] is parallel to the plane and the points
A $=$ (3,2,2) and B= (1,0,-1) lie on the plane.
Now, let \[\left( x,y,z \right)\] be a variable point on the plane.
It is given that A (3,2,2) also lies on the plane.
Therefore, \[\left( x-3 \right)\widehat{i}+\left( y-2 \right)\widehat{j}+\left( z-2 \right)\widehat{k}\]vector lies on the plane\[....(1)\] \[\]
Now, we will find the vector normal to the plane.
It is given that the line \[\dfrac{x-1}{1}=\dfrac{y-1}{-2}=\dfrac{z-2}{3}\] is parallel to the plane. We know that the denominator terms will be taken as coefficient for the vector representation. So, this line will be
\[\widehat{i}-2\widehat{j}+3\widehat{k}\] \[....(2)\].
Since, A and B lies on plane then $\overrightarrow{AB}$ lies on the given plane.
Now, we know that the vector equation of a line passing through the points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is given as $\left( {{x}_{1}}-{{x}_{2}} \right)\widehat{i}+\left( {{y}_{1}}-{{y}_{2}} \right)\widehat{j}+\left( {{z}_{1}}-{{z}_{2}} \right)\widehat{k}$ .
So, the equation of the line passing through A (3, 2, 2) and B (1, 0, -1) is given as :\[\overrightarrow{AB}=\left( 3-1 \right)\widehat{i}+\left( 2-0 \right)\widehat{j}+\left( 2-\left( -1 \right) \right)\widehat{k}\]
\[\Rightarrow \overrightarrow{AB}=2\widehat{i}+2\widehat{j}+3\widehat{k}\]\[......\left( 3 \right)\]
Now, we know that the cross product of two vectors give a vector perpendicular to both the vectors. We also know that the vectors given by the equations (2) and (3) are parallel to the plane. So, any vector perpendicular to these vectors will also be perpendicular to the plane. Therefore, cross product of vectors 2 and 3 will give the normal to the plane.
Let the vector normal to the plane be $\overrightarrow{n}$ .
\[\Rightarrow \overrightarrow{n}=\left( \widehat{i}-2\widehat{j}+3\widehat{k} \right)\times \left( 2\widehat{i}+2\widehat{j}+3\widehat{k} \right)\]
$\Rightarrow \overrightarrow{n}=\left| \begin{matrix}
   \widehat{i} & \widehat{j} & \widehat{k} \\
   1 & -2 & 3 \\
   2 & 2 & 3 \\
\end{matrix} \right|$
$\Rightarrow \overrightarrow{n}=\widehat{i}\left( \left( -2\times 3 \right)-\left( 2\times 3 \right) \right)-\widehat{j}\left( \left( 3\times 1 \right)-\left( 3\times 2 \right) \right)+\widehat{k}\left( \left( 2\times 1 \right)-\left( -2\times 2 \right) \right)$
$\Rightarrow \overrightarrow{n}=\widehat{i}\left( -6-6 \right)-\widehat{j}\left( 3-6 \right)+\widehat{k}\left( 2+4 \right)$
\[\Rightarrow \overrightarrow{n}=-12\widehat{i}+3\widehat{j}+6\widehat{k}........(4)\]
Now, we know that the unit vector along normal to the plane is given as:
\[\widehat{n}=\dfrac{\overrightarrow{n}}{\left| \overrightarrow{n} \right|}\]
Also, $\left| \overrightarrow{n} \right|=\sqrt{{{\left( -12 \right)}^{2}}+{{\left( 3 \right)}^{2}}+{{\left( 6 \right)}^{2}}}$
$\Rightarrow \left| \overrightarrow{n} \right|=\sqrt{144+9+36}$
 $\Rightarrow \left| \overrightarrow{n} \right|=\sqrt{189}$
So, \[\widehat{n}=\text{ }\dfrac{-\left( 12\widehat{i}+3\widehat{j}+6\widehat{k} \right)}{\sqrt{189}}\]
\[\Rightarrow ~~\widehat{n}=\dfrac{3\left( -4\widehat{i}+3\widehat{j}+6\widehat{k} \right)}{3\sqrt{21}}\]
\[\Rightarrow ~~\widehat{n}=\dfrac{\left( -4\widehat{i}+3\widehat{j}+6\widehat{k} \right)}{\sqrt{21}}\]
Now, the vector given by equation (1) represents every possible vector on the plane. We know that the dot product of any two perpendicular vectors is 0. So, the dot product of the vectors given by equation (1) and (4) must be 0, as (4) represents the vector normal to the plane, and (1) represents a vector on the plane.
\[\Rightarrow vector\left( 1 \right).vector\left( 4 \right)=0\] \[\]
\[\Rightarrow \left[ \left( x-3 \right)\widehat{i}+\left( y-2 \right)\widehat{j}+\left( z-2 \right)\widehat{k} \right].\left[ -12\widehat{i}+3\widehat{j}+6\widehat{k} \right]=0\]
\[\Rightarrow -12x+36+3y-6+6z-12=0\]
\[\Rightarrow -12x+3y+6z+18=0\]
\[\Rightarrow -3\left( 4x-y-2z-6 \right)=0\]
\[\Rightarrow 4x-y-2z=6.\]
Therefore, equation of plane is 4x-y-2z=6.
Converting into vector form:
$4x-y-2z=6$ can be written as
\[\overrightarrow{n}=\left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right).\left( 4\widehat{i}-\widehat{j}-2\widehat{k} \right)=6\] ……………………………(5)
We know that $\left| \overrightarrow{n} \right|$ which we can find by using the formula:
, $\left| \overrightarrow{n} \right|=\sqrt{{{\left( 4 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( -2 \right)}^{2}}}$
$\Rightarrow \left| \overrightarrow{n} \right|=\sqrt{16+1+4}$
 $\Rightarrow \left| \overrightarrow{n} \right|=\sqrt{21}$
 so, dividing by \[\sqrt{21}\] on both sides of equation (5), we get
\[\dfrac{\left( x\hat{i}+y\hat{j}+z\hat{k} \right)\cdot \left( 4\hat{i}+\hat{j}-2\hat{k} \right)}{\sqrt{21}}=\dfrac{6}{\sqrt{21}}\]
Therefore, equation of the plane in vector form is \[~\left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right).\left( \dfrac{4\widehat{i}-\widehat{j}-2\widehat{k}}{\sqrt{21}} \right)\] \[=\dfrac{6}{\sqrt{21}}\]

Note: Vector form of the plane can also be evaluated by calculating \[\widehat{n}\]and d separately and substituting in the vector form of the plane equation, where d can be calculated by using the concept that if lx + my + nz = p is the equation of plane, then the perpendicular distance from the origin d can be calculated as \[\left| \dfrac{p}{\sqrt{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}} \right|\] . The equation of a plane in vector form is given as \[\overrightarrow{r}.\widehat{n}=d\] , where:
\[\vec{r}\]=position vector of a point in given plane,
\[\widehat{n}\]=unit vector perpendicular to the plane, and
d=perpendicular distance between origin and given plane.