Equation of a Plane Perpendicular to a Given Line and Passing Through a Specific Point

To find the equation of a plane that passes through a given point and is perpendicular to a specified line, follow these steps:

Identify the Given Information

Point through which the plane passes: (P(2, -3, 1)) Parametric equations of the line: x -1 - 12t, y 2 - t, z 3 - t

Extract the Direction Vector of the Line

From the parametric equations, extract the direction vector of the line:

(mathbf{d} [-12, -1, -1])

Use the Direction Vector as the Normal Vector of the Plane

Since the plane is perpendicular to the line, the direction vector of the line can be used as the normal vector of the plane. This normal vector is:

(mathbf{n} [-12, -1, -1])

For simplicity, we can normalize the direction vector to: (mathbf{n} [2, -1, -1])

Write the Equation of the Plane

The general form of the equation of a plane with a normal vector (A, B, C) that passes through the point ((x_0, y_0, z_0)) is:

(Ax - x_0 By - y_0 Cz - z_0 0)

Substituting (A 2), (B -1), (C -1), and the point ((2, -3, 1)) into the equation:

(2x - 2 - (-1)y - (-3) - (-1)z - 1 0)

Expanding this:

(2x - 2 - y 3 - z 1 0)

Combining like terms:

(2x - y - z 2 - 3 1 0)

Simplifying the equation:

(2x - y - z - 6 0)

Rearranging gives us the final equation of the plane:

(2x - y - z 6)

Therefore, the equation of the plane is:

(boxed{2x - y - z 6})

Verify Using a Different Approach

Verify the result using a different approach:

The normalised equation of the line is:

(frac{x 1}{2} frac{y - 2}{-1} frac{z - 3}{-1})

The equation of the plane passing through the point (2, -3, 1) and perpendicular to the line is:

(2(x - 2) - (y 3) - (z - 1) 0)

Simplifying:

(2x - 4 - y - 3 - z 1 0)

Combining like terms:

(2x - y - z - 6 0)

Therefore, the equation of the plane is:

(2x - y - z 6)

Conclusion

If t 0, we get the point ((-1, 2, 3)).

If t 1, we get the point ((1, 1, 2)).

The vector ((1, 1, 2) - (-1, 2, 3)) is in the direction of the line.

Hence, the plane has the equation (2x - y - z d).

Substituting the point ((2, -3, 1)) into the equation and finding d:

(4 - 3 - 1 d)

(d 6)

Thus, the equation of the plane is:

(2x - y - z 6)

The plane through ((2, -3, 1)) and perpendicular to the line (x -1 - 12t, y 2 - t, z 3 - t) indeed has the equation:

(2x - y - z 6)