The foot of the perpendicular from the point A(7,14,5) to the plane `2x+4y-z=2` is

Options

A. (3,1,8)
B. (1,2,8)
C. (3,-3,5)
D. (5,-3,-4)

Explanation

To find the foot of the perpendicular from the point A(7, 14, 5) to the plane given by the equation \(2x + 4y - z = 2\), we can use the formula for the foot of the perpendicular from a point to a plane.

### Step 1: Identify the components
The equation of the plane is given in the form \(Ax + By + Cz = D\), where:
- \(A = 2\)
- \(B = 4\)
- \(C = -1\)
- \(D = 2\)

The coordinates of point A are:
- \(x_1 = 7\)
- \(y_1 = 14\)
- \(z_1 = 5\)

### Step 2: Use the formula for the foot of the perpendicular
The foot of the perpendicular \(P(x, y, z)\) from the point \(A(x_1, y_1, z_1)\) to the plane \(Ax + By + Cz = D\) can be calculated using the following formula:

\[ \frac{x - x_1}{A} = \frac{y - y_1}{B} = \frac{z - z_1}{C} = -\frac{Ax_1 + By_1 + Cz_1 - D}{A^2 + B^2 + C^2} \]

### Step 3: Calculate the right-hand side
First, we need to compute \(Ax_1 + By_1 + Cz_1 - D\):

\[ Ax_1 = 2 \cdot 7 = 14 \]
\[ By_1 = 4 \cdot 14 = 56 \]
\[ Cz_1 = -1 \cdot 5 = -5 \]
\[ D = 2 \]

Now, substituting these values:

\[ Ax_1 + By_1 + Cz_1 - D = 14 + 56 - 5 - 2 = 63 \]

Next, we calculate \(A^2 + B^2 + C^2\):

\[ A^2 = 2^2 = 4 \]
\[ B^2 = 4^2 = 16 \]
\[ C^2 = (-1)^2 = 1 \]
\[ A^2 + B^2 + C^2 = 4 + 16 + 1 = 21 \]

Now, we can find the right-hand side:

\[ -\frac{63}{21} = -3 \]

### Step 4: Set up the equations
Now we can set up the equations using the value of \(-3\):

\[ \frac{x - 7}{2} = -3 \implies x - 7 = -6 \implies x = 1 \]
\[ \frac{y - 14}{4} = -3 \implies y - 14 = -12 \implies y = 2 \]
\[ \frac{z - 5}{-1} = -3 \implies z - 5 = 3 \implies z = 8 \]

### Step 5: Conclusion
Thus, the foot of the perpendicular from the point A(7, 14, 5) to the plane \(2x + 4y - z = 2\) is:

\[ P(1, 2, 8) \]