Question

Find the equation of the plane passing through the points (1,-1, 2) and (2,-2, 2) and which is perpendicular to the plane 6x-2y+2z=9.

Answer 1

Hint:For writing the equation of a plane with two points known and a plane that is perpendicular to this plane, we can follow the given procedure which is as follows
1.We will first write a vector that is perpendicular to the normal of the unknown plane.
2.Now, the normal of the plane that is perpendicular to this unknown plane is also perpendicular to the normal of the unknown plane, hence, we can write the equation of the normal by finding the cross product of the normal of the known plane and the vector that is formed from the two given points.
3.Now, we can represent the plane as follows
$a(x-{{x}_{1}})+b(y-{{y}_{1}})+c(z-{{z}_{1}})=d$
(Where a, b and c are the x-, y- and z- coefficients of the normal of this plane and $({{x}_{1}},{{y}_{1}},{{z}_{1}})$ is a point that lies on the plane)
Now, using this above mentioned procedure, one can get to the right answer.

Complete step-by-step answer:
As mentioned in the question, we have to find the equation of the required plane.
Now, following the procedure mentioned in the hint, on can write the following
The equation of the vector that is perpendicular to the normal of the unknown plane is as follows

& \Rightarrow \overrightarrow{v}=({{x}_{2}}-{{x}_{1}})\widehat{i}+({{y}_{2}}-{{y}_{1}})\widehat{j}+({{z}_{2}}-{{z}_{1}})\widehat{k} \\\ & \Rightarrow \overrightarrow{v}=(1-2)\widehat{i}+(-1-(-2))\widehat{j}+(2-z)\widehat{k} \\\ & \Rightarrow \overrightarrow{v}=-1\widehat{i}+1\widehat{j}+0\widehat{k} \\\ & \Rightarrow \overrightarrow{v}=-1\widehat{i}+1\widehat{j} \\\ \end{aligned}$$ (Using the formula to find the vector which is directed from point A $$\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$$ and point B $$\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$$ which is as follows $$\overrightarrow{v}=({{x}_{2}}-{{x}_{1}})\widehat{i}+({{y}_{2}}-{{y}_{1}})\widehat{j}+({{z}_{2}}-{{z}_{1}})\widehat{k}$$ ) Now, we can also write the equation of the normal of the given plane by taking the coefficients of x, y and z from the equation of the plane as follows $$\overrightarrow{u}=6\widehat{i}-2\widehat{j}+2\widehat{k}$$ (Using the formula to find the vector which represents the normal vector of a plane with equation $$ax+by+cz=d$$ which is as follows $$\overrightarrow{v}=a\widehat{i}+b\widehat{j}+c\widehat{k}$$ ) Now, on finding the cross product of these above equations, we can find the normal of the unknown plane as follows $$\begin{aligned} & \overrightarrow{v}\times \overrightarrow{u}=\left( -1\widehat{i}+1\widehat{j} \right)\times \left( 6\widehat{i}-2\widehat{j}+2\widehat{k} \right) \\\ & \overrightarrow{v}\times \overrightarrow{u}=\left| \left( \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\\ -1 & 1 & 0 \\\ 6 & -2 & 2 \\\ \end{matrix} \right) \right|=\left( 2-0 \right)\widehat{i}-\left( -2-0 \right)\widehat{j}+\left( 2-6 \right)\widehat{k} \\\ & \overrightarrow{v}\times \overrightarrow{u}=2\widehat{i}+2\widehat{j}-4\widehat{k} \\\ \end{aligned}$$ (Using the formula to find the cross product of two vectors $$\overrightarrow{a}=x\widehat{i}+y\widehat{j}+z\widehat{k}\ and\ \overrightarrow{b}=a\widehat{i}+b\widehat{j}+c\widehat{k}$$ as $$\overrightarrow{a}\times \overrightarrow{b}=\left| \left( \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\\ x & y & z \\\ a & b & c \\\ \end{matrix} \right) \right|$$ ) Hence, we get the equation of the normal of the unknown plane as follows $$\overrightarrow{z}=2\widehat{i}+2\widehat{j}-4\widehat{k}$$ Now, as mentioned in the hint, we can write the equation of the unknown plane as follows $$\begin{aligned} & \Rightarrow 2(x-1)+2(y-(-1))-4(z-2)=d \\\ & \Rightarrow 2(x-1)+2(y+1)-4(z-2)=d \\\ \end{aligned}$$ (Using the formula to write the equation of a plane when we know a point that lies on it (l, m, n)and the equation of normal $$\overrightarrow{v}=a\widehat{i}+b\widehat{j}+c\widehat{k}$$ as follows $$a(x-l)+b(y-m)+c(z-n)=d$$ Now, putting the other point that lies on the plane in this equation which is given in the hint, we get the following $$\begin{aligned} & \Rightarrow 2(2-1)+2(-2+1)-4(2-2)=d \\\ & \Rightarrow d=2+(-2)=0 \\\ & \Rightarrow d=0 \\\ \end{aligned}$$ Hence, the equation of the unknown plane is as follows $$\begin{aligned} & \Rightarrow 2(x-1)+2(y+1)-4(z-2)=0 \\\ & \Rightarrow 2x-2+2y+2-4z+8=0 \\\ & \Rightarrow 2x+2y-4z+8=0 \\\ & \Rightarrow x+y-2z=-4 \\\ \end{aligned}$$ Hence, this is the required equation of the unknown plane. Note: -Students must know the method to find the cross product and the method to find the vector which directs from point A to point B. These are two most important formulae that are used in the solution.