Find an equation of the plane. The plane through the points (2, 1, 2), (3, −8, 6), and (−2, −3, 1)

This article aims to find the equation of the plane when points of the plane are given. The article uses the concept of vector multiplication. Cross Product – “vector product” is a binary operation on two vectors that results in another vector.

The cross product of two vectors in $3-space$ is defined as a vector perpendicular to the plane determined by two vectors whose magnitude is the product of magnitudes of two vectors and the sine of angle between the two vectors. Thus, if $\vec { n }$ is a unit vector perpendicular to the plane defined by vectors $A$ and $B$.

$A \times B = | A | \: | B | \: \sin \theta \vec { n }$

Let the given points be $P ( 2 , 1 , 2 ) , Q ( 3 , – 8 , 6 ) \: and \: R ( – 2 , – 3 , 1 )$.

$\vec { PQ } = \langle 3 – 2 , – 8 – 1, 6 – 2 \rangle = \langle 1, – 9 , 4 \rangle$

$\vec { PR } = \langle – 2 – 2 ,- 3 – 1 ,1 – 2 \rangle = \langle – 4 ,- 4 ,- 1 \rangle$

$\vec{PQ} \times \vec{PR} = \begin{vmatrix} i & j & k\\ 1 & -9 & 4\\ -4 & -4 & -1 \end{vmatrix} = ( 9 + 16 ) i + ( – 16 + 1 ) j + ( – 4 – 36 ) k$

$= 25i – 15j – 40k$

Therefore, the normal vector to the plane is:

$\vec { n } = \langle 25 , – 15 , -40 \rangle$

Since the plane passes through all three points, we can choose any point to find its equation. So the equation of the plane passing through the point $P(2,1,2)$ with the normal vector:

$\vec{n} = \langle 25,-15,-40\rangle$

$25 ( x – 2 ) – 15 ( y – 1 ) – 40 ( z – 2 ) = 0$

$\Rightarrow 25 x – 50 – 15 y + 15 – 40 z +80 = 0$

$\Rightarrow 25 x – 15 y – 40 z + 45 = 0$

The equation of the plane is $25 x – 15 y – 40 z + 45 = 0$.

Numerical Result

The equation of the plane is $25x-15y -40z+45=0$.

Example

Find the equation of the plane. The plane through the points $(6, 4, 2), (3, −8, 6) \:and \:(−2, −3, 1)$.

Solution

Let the given points be $P(6,4,2), Q(3,-8,6) \: and \:R(-2,-3,1)$.

$\vec{PQ}= \langle 6-3, -8-4, 6-2 \rangle= \langle 3,-12,4\rangle$

$\vec{PR} = \langle -2-2,-3-1,1-2\rangle = \langle -4,-4,-1\rangle$

$\vec{PQ} \times \vec{PR} = \begin{vmatrix} i & j & k\\ 3 & -12 & 4\\ -4 & -4 & -1 \end{vmatrix} = (12+16)i+(-3+16)j+(-12-48)k$

$= 28i – 13j – 60k$

Therefore, the normal vector to the plane is:

$\vec{n} = \langle 28,-13,-60\rangle$

Since the plane passes through all three points, we can choose any point to find its equation. So the equation of the plane passing through the point $P(6,4,2)$ with the normal vector:

$\vec{n} = \langle 28,-13,-60\rangle$

$28(x-6)-13(y-4)-60(z-2) = 0$

$\Rightarrow 28x-13y -60z+4=0$

The equation of the plane is $28x-13y -60z+4=0$.