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# Find the equation of the plane passing through the points whose coordinates are (-1, 1, 1) and (1, -1, 1) and perpendicular to the plane x + 2y + 2z = 5

Answer Verified
Hint: Take the equation of plane as $a\left( x-{{x}_{1}} \right)+b\left( y-{{y}_{1}} \right)+c\left( z-{{z}_{1}} \right)=0$ where $a\widehat{i}+b\widehat{j}+c\widehat{k}$ is the normal vector to the plane and $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ is the point through which it passes. Since both planes are perpendicular, the dot produced of the normal vectors of both of them would be zero. Use this approach to find the equation of the plane.

Complete step-by-step answer:
Here we are given a plane passing through the points whose coordinates are (-1, 1, 1) and(1, -1, 1) and perpendicular to the plane x + 2y + 2z = 5.

We have to find the equation of this plane. We know that for any plane ax + by + cx = d, $a\widehat{i}+b\widehat{j}+c\widehat{k}$ is its normal vector.

Therefore, for the given plane x + 2y + 2z = 5, by comparing it to ax + by + cz = d, we get $\widehat{i}+2\widehat{j}+2\widehat{k}$ as its normal vector.

Let us consider the normal vector of this plane as
$\overrightarrow{P}=\widehat{i}+2\widehat{j}+2\widehat{k}....\left( i \right)$

We know that if any plane P passes through the point $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and the vector normal to it is $a\widehat{i}+b\widehat{j}+c\widehat{k}$, then its equation is given by,
$P:a\left( x-{{x}_{1}} \right)+b\left( y-{{y}_{1}} \right)+c\left( z-{{z}_{1}} \right)=0$

As we are given that plane to be focused passes through (-1, 1, 1) and the vector normal to it is $a\widehat{i}+b\widehat{j}+c\widehat{k}$ .
Then we get the equation of the plane as
$P:a\left[ x-\left( -1 \right) \right]+b\left( y-1 \right)+c\left( z-1 \right)=0.....\left( i \right)$

As we are given that this plane also passes through the point (1, -1, 1), therefore by substituting the value of x, y and z in the above equation, we get,
$a\left[ 1-\left( -1 \right) \right]+b\left( -1-1 \right)+c\left( 1-1 \right)=0$

By simplifying the above equation, we get,
$\Rightarrow a\left( 2 \right)+b\left( -2 \right)=0$
Or, $2a-2b=0$

Therefore, we get a = b
We know that both planes are perpendicular to each other, so the vectors normal to both of them would also be perpendicular.

We also know that, when two vectors are perpendicular to each other, their dot product is 0. Therefore, we get
(Normal vector of 1st plane) . (Normal Vector of 2nd plane) = 0
$\left( \widehat{i}+2\widehat{j}+2\widehat{k} \right).\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)=0$

As we know that
$\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right).\left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right)=ax+by+cz$
Therefore, we get,
$\left( 1.a \right)+\left( 2.b \right)+\left( 2.c \right)=0$

Since, we have found that a = b, therefore by substituting b = a in the above equation, we get
\begin{align} & a+2a+2c=0 \\ & \Rightarrow 2c+3a=0 \\ & \Rightarrow 2c=-3a \\ \end{align}

Therefore, we get $c=\dfrac{-3a}{2}$

As we have found that a = a, b = a and $c=\dfrac{-3a}{2}$, therefore by substituting the values of a, b and c in equation (i) in terms of a, we get,
$P:a\left( x+1 \right)+a\left( y-1 \right)+\left( \dfrac{-3a}{2} \right)\left( z-1 \right)=0$

By simplifying the equation and taking ‘a’ common, we get,
$P:a\left[ \left( x+1 \right)+\left( y-1 \right)-\dfrac{-3}{2}\left( z-1 \right) \right]=0$
$P:a\left[ \dfrac{2\left( x+1 \right)+2\left( y-1 \right)-3\left( z-1 \right)}{2} \right]=0$

By multiplying by $\dfrac{2}{a}$ on both sides and simplifying the equation, we get,
$P:\left( 2x+2+2y-2-3z+3 \right)=0$
Or, $P:2x+2y-3z+3=0$

Hence, we get the equation of plane as 2x + 2y – 3z + 3 = 0.

Note: Students must note that to find the equation of a plane, absolute values of a, b and c are not required but we must know their ratios like in the above solution, we have $a:b:c$ as $1:1:\dfrac{-3}{2}$. Also, students can cross-check their equation of plane by satisfying points (-1, 1, 1) and (1, -1, 1) in it.
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