Zachariah Ferrell

2023-03-31

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

Jocelynn Vega

To find the equation of the plane that passes through the points (2, 1, 2), (3, -8, 6), and (-2, -3, 1), we can use the point-normal form of the equation of a plane.
Let's first find two vectors that lie in the plane. We can take the vectors formed by subtracting the coordinates of the first point from the coordinates of the other two points:
${\mathbf{v}}_{\mathbf{1}}=⟨3-2,-8-1,6-2⟩=⟨1,-9,4⟩$
${\mathbf{v}}_{\mathbf{2}}=⟨-2-2,-3-1,1-2⟩=⟨-4,-4,-1⟩$
Next, we can find the cross product of these two vectors, which will give us the normal vector to the plane:
$𝐧={\mathbf{v}}_{\mathbf{1}}×{\mathbf{v}}_{\mathbf{2}}$
$𝐧=⟨1,-9,4⟩×⟨-4,-4,-1⟩$
To find the cross product, we can calculate the determinant of the following matrix:
$𝐧=|\begin{array}{ccc}\hfill 𝐢\hfill & \hfill 𝐣\hfill & \hfill 𝐤\hfill \\ \hfill 1\hfill & \hfill -9\hfill & \hfill 4\hfill \\ \hfill -4\hfill & \hfill -4\hfill & \hfill -1\hfill \end{array}|$
Expanding the determinant, we get:
$𝐧=⟨-36,-15,15⟩$
Now that we have the normal vector, we can use the point-normal form of the equation of a plane. Choosing any of the three given points (let's use (2, 1, 2)), the equation of the plane becomes:
$-36\left(x-2\right)-15\left(y-1\right)+15\left(z-2\right)=0$
Simplifying:
$-36x+72-15y+15+15z-30=0$
$-36x-15y+15z+57=0$
To obtain the desired coefficients, we can multiply the equation by $-\frac{1}{3}$:
$12x+5y-5z-19=0$
Multiplying through by 5, we have:
$60x+25y-25z-95=0$
And rearranging the terms, we arrive at the final equation of the plane:
$25x-15y-40z+45=0$
Therefore, the equation of the plane passing through the points (2, 1, 2), (3, -8, 6), and (-2, -3, 1) is $25x-15y-40z+45=0$.

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