In Exercises 1-8, write the equation of the plane with normal vector n passing through the given

step 1 Normal vector negative 1 to 1 through the point 319. So it's negative 1, x minus 3 plus 2y minus

In Exercises 1-8, write the equation of the plane with...

In Exercises $1-8,$ write the equation of the plane with normal vector $\mathrm{n}$ passing through the given point in the scalar form $a x+b y+c z=d$.
$$
\mathbf{n}=\langle-1,2,1\rangle, \quad(3,1,9)
$$

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Key Concepts

Scalar Form of the Plane Equation

Converting the point-normal form to the scalar form involves expanding the dot product and simplifying to achieve an equation in the form ax + by + cz = d. This scalar form is particularly useful for quickly assessing the geometric properties of the plane, and for solving problems related to the intersection of planes or distances from points to planes.

Equation of a Plane Using a Point and a Normal Vector

A primary method to define a plane in space is by using a known point on the plane and a normal vector perpendicular to the plane. This form, often called the point-normal form, is expressed as n • (r - r?) = 0, where n is the normal vector, r is a generic point (x, y, z) on the plane, and r? is the given point. This formulation is fundamental in understanding and deriving the plane's equation.

Normal Vector

The normal vector plays a central role in describing the plane's orientation since it is perpendicular to every line on the plane. Its components directly form the coefficients in the scalar equation of the plane, and they determine the direction in which the plane is tilted in three-dimensional space.

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