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

Find An Equation Of The Plane. The Plane Through The Points

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 $.

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

Expert Answer

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

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