Question

Find an equation of the plane passing through the point perpendicular to the given vector or line. $$ (3,2,2) \quad \mathbf{n}=2 \mathbf{i}+3 \mathbf{j}-\mathbf{k} $$

Short answer

The equation of the plane is \( 2x + 3y - z - 10 = 0 \).

Step-by-step solution

Identify given values

We have the point, \( P (3,2,2) \), and the normal vector \( \mathbf{n}=2 \mathbf{i}+3 \mathbf{j}-\mathbf{k} \), which gives us the coefficients \( a = 2, b = 3, c = -1 \). Thus, \( x_0 = 3, y_0 = 2, z_0 = 2 \).

Substitute values into the general equation

Substitute values into the general equation \( a(x – x_0) + b(y – y_0) + c(z – z_0) = 0 \). This gives us \( 2*(x - 3) + 3*(y - 2) - 1*(z - 2) = 0 \). Simplifying, we get \( 2x - 6 + 3y - 6 - z + 2 = 0 \).

Simplify the equation

Combine like terms to simplify the equation which results in: \( 2x + 3y - z - 10 = 0 \).