To determine the coordinates of the foot of the perpendicular drawn from the origin to a given plane, we need to understand the relationship between the origin, the plane, and the line perpendicular to the plane. The problem asks us to find this specific point on the plane.

The given information is:

• The point from which the perpendicular is drawn: The origin &(\text{(0, 0, 0)})&.
• The equation of the plane: &(\text{4x - 2y + 3z - 6 = 0})&.

Plane Equation and Normal Vector

The general equation of a plane is given by &(\text{Ax + By + Cz + D = 0})&. From the given plane equation, &(\text{4x - 2y + 3z - 6 = 0})&, we can identify the coefficients that define its orientation in space.

• The coefficient of &(\text{x})& is &(\text{A = 4})&.
• The coefficient of &(\text{y})& is &(\text{B = -2})&.
• The coefficient of &(\text{z})& is &(\text{C = 3})&.

These coefficients &(\text{A, B, C})& form the components of the normal vector &(\vec{n})& to the plane. A normal vector is a vector that is perpendicular to the plane's surface. So, the normal vector for our given plane is &(\vec{n} = \langle 4, -2, 3 \rangle)&.

Line Perpendicular to the Plane

The line that passes through the origin &(\text{(0, 0, 0)})& and is perpendicular to the given plane will have its direction parallel to the normal vector of the plane. This means the direction ratios of the line will be the same as the components of the normal vector: &(\text{4, -2, 3})&.

The parametric equations of a line passing through a point &((x_1, y_1, z_1))& with direction ratios &((a, b, c))& are:

$x = x_1 + at$ $y = y_1 + bt$ $z = z_1 + ct$

For our problem:

• The starting point is the origin &((x_1, y_1, z_1) = (0, 0, 0))&.
• The direction ratios are &((a, b, c) = (4, -2, 3))&.

Substituting these values, the parametric equations of the line are:

• $x = 0 + 4t \implies x = 4t$
• $y = 0 - 2t \implies y = -2t$
• $z = 0 + 3t \implies z = 3t$

Here, &(\text{t})& is a scalar parameter that allows us to find any point on this line.

Foot of Perpendicular Coordinates Determination

The foot of the perpendicular is the specific point where the line we just defined intersects the given plane. To find the coordinates of this point, we substitute the parametric equations of the line into the equation of the plane.

The plane equation is: &(\text{4x - 2y + 3z - 6 = 0})&

Substitute &(\text{x = 4t})&, &(\text{y = -2t})&, and &(\text{z = 3t})& into the plane equation:

$4(4t) - 2(-2t) + 3(3t) - 6 = 0$

Now, we simplify the equation and solve for the parameter &(\text{t})&:

$16t + 4t + 9t - 6 = 0$ $29t - 6 = 0$ $29t = 6$ $t = \frac{6}{29}$

Finally, to find the exact coordinates of the foot of the perpendicular, we substitute this value of &(\text{t})& back into the parametric equations of the line:

• $x = 4t = 4 \left( \frac{6}{29} \right) = \frac{24}{29}$
• $y = -2t = -2 \left( \frac{6}{29} \right) = -\frac{12}{29}$
• $z = 3t = 3 \left( \frac{6}{29} \right) = \frac{18}{29}$

Thus, the coordinates of the foot of the perpendicular drawn from the origin to the plane &(\text{4x - 2y + 3z - 6 = 0})& are &(\left( \frac{24}{29}, -\frac{12}{29}, \frac{18}{29} \right))&.

Steps Summary for Foot of Perpendicular

Here is a concise summary of the steps taken to find the foot of the perpendicular coordinates:

1. Normal Vector Identification: Extract the normal vector &(\vec{n} = \langle A, B, C \rangle)& from the plane equation &(\text{Ax + By + Cz + D = 0})&.
2. Line Equation Formulation: Write the parametric equations of the line passing through the given point (in this case, the origin) using the normal vector as its direction vector.
3. Parameter Calculation: Substitute the line's parametric equations into the plane's equation and solve for the parameter &(\text{t})&.
4. Coordinates Determination: Substitute the calculated value of &(\text{t})& back into the line's parametric equations to obtain the &(\text{(x, y, z)})& coordinates of the foot of the perpendicular.