Finding the Equation of a Plane

We are asked to find the equation of a plane that passes through a specific point and is perpendicular to a given vector. This perpendicular vector is also known as the normal vector to the plane.

Understanding the Key Concepts

• Point on the Plane: We are given a point \(P_0\) with position vector \(3\widehat i - \widehat j + \widehat k\). In Cartesian coordinates, this point is \((x_0, y_0, z_0) = (3, -1, 1)\).
• Normal Vector: The plane is perpendicular to the vector \(4\widehat i + 2\widehat j - \widehat k\). This is the normal vector to the plane, denoted by \(\vec{n}\). In Cartesian coordinates, \(\vec{n} = A\widehat i + B\widehat j + C\widehat k\), so we have \(A = 4\), \(B = 2\), and \(C = -1\).
• Equation of a Plane: The general equation of a plane passing through a point \((x_0, y_0, z_0)\) and having a normal vector \(\vec{n} = A\widehat i + B\widehat j + C\widehat k\) is given by:
  \(A(x - x_0) + B(y - y_0) + C(z - z_0) = 0\)
  This equation can also be written as \(Ax + By + Cz = D\), where \(D = Ax_0 + By_0 + Cz_0\).

Step-by-Step Derivation of the Plane Equation

We will use the formula \(A(x - x_0) + B(y - y_0) + C(z - z_0) = 0\).

Substitute the values of \(A, B, C\) and \(x_0, y_0, z_0\) into the formula:

\(A = 4\), \(B = 2\), \(C = -1\)

\(x_0 = 3\), \(y_0 = -1\), \(z_0 = 1\)

So the equation becomes:

\(4(x - 3) + 2(y - (-1)) + (-1)(z - 1) = 0\)

\(4(x - 3) + 2(y + 1) - 1(z - 1) = 0\)

Now, expand and simplify the equation:

\(4x - 12 + 2y + 2 - z + 1 = 0\)

Combine the constant terms:

\(4x + 2y - z + (-12 + 2 + 1) = 0\)

\(4x + 2y - z - 9 = 0\)

We can rewrite this equation by moving the constant term to the right side:

\(4x + 2y - z = 9\)

Verifying the Solution

The derived equation for the plane is \(4x + 2y - z = 9\). We can check if the given point \((3, -1, 1)\) lies on this plane by substituting the coordinates into the equation:

\(4(3) + 2(-1) - (1)\)

\(12 - 2 - 1\)

\(10 - 1\)

\(9\)

Since \(9 = 9\), the point \((3, -1, 1)\) lies on the plane \(4x + 2y - z = 9\).

The normal vector coefficients \((4, 2, -1)\) match the coefficients of \(x, y, z\) in the equation \(4x + 2y - z = 9\), confirming that the plane is perpendicular to the given vector \(4\widehat i + 2\widehat j - \widehat k\).

Comparing our derived equation \(4x + 2y - z = 9\) with the given options, we find that it matches one of the options.