Find the length of the perpendicular from the point (2,3,7) to the plane 3 x - y - z = 7 . Also , find the co-ordinates of the foot of the perpendicular .

Answer

To find the length of the perpendicular from the point \( P(2, 3, 7) \) to the plane given by the equation \( 3x - y - z = 7 \), we can follow these steps:

### Step 1: Write the equation of the plane in standard form
The equation of the plane can be rewritten as:
\[ 3x - y - z - 7 = 0 \]
Here, we identify \( A = 3 \), \( B = -1 \), \( C = -1 \), and \( D = -7 \).

### Step 2: Use the formula for the distance from a point to a plane
The formula for the distance \( D \) from a point \( (x_1, y_1, z_1) \) to a plane \( Ax + By + Cz + D = 0 \) is given by:
\[ D = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Substituting \( (x_1, y_1, z_1) = (2, 3, 7) \) into the formula, we have:
\[ D = \frac{|3(2) + (-1)(3) + (-1)(7) - 7|}{\sqrt{3^2 + (-1)^2 + (-1)^2}} \]

### Step 3: Calculate the numerator
Calculating the numerator:
\[ 3(2) + (-1)(3) + (-1)(7) - 7 = 6 - 3 - 7 - 7 = 6 - 3 - 14 = -11 \]
Taking the absolute value:
\[ |-11| = 11 \]

### Step 4: Calculate the denominator
Calculating the denominator:
\[ \sqrt{3^2 + (-1)^2 + (-1)^2} = \sqrt{9 + 1 + 1} = \sqrt{11} \]

### Step 5: Calculate the distance
Now substituting back into the distance formula:
\[ D = \frac{11}{\sqrt{11}} = \sqrt{11} \]

### Step 6: Find the coordinates of the foot of the perpendicular
To find the coordinates of the foot of the perpendicular \( Q(x, y, z) \), we use the direction ratios from the plane's normal vector \( (A, B, C) = (3, -1, -1) \) and the point \( P(2, 3, 7) \).

Using the parametric equations:
\[ \frac{x - 2}{3} = \frac{y - 3}{-1} = \frac{z - 7}{-1} = t \]
From this, we can express \( x, y, z \) in terms of \( t \):
\[ x = 2 + 3t, \quad y = 3 - t, \quad z = 7 - t \]

### Step 7: Substitute into the plane equation
Substituting these into the plane equation \( 3x - y - z - 7 = 0 \):
\[ 3(2 + 3t) - (3 - t) - (7 - t) - 7 = 0 \]
Expanding this:
\[ 6 + 9t - 3 + t - 7 + t - 7 = 0 \]
Simplifying:
\[ 9t + 3t - 11 = 0 \implies 11t - 11 = 0 \implies t = 1 \]

### Step 8: Find the coordinates of \( Q \)
Substituting \( t = 1 \) back into the equations for \( x, y, z \):
\[ x = 2 + 3(1) = 5, \quad y = 3 - 1 = 2, \quad z = 7 - 1 = 6 \]
Thus, the coordinates of the foot of the perpendicular \( Q \) are \( (5, 2, 6) \).

### Final Result
The length of the perpendicular from the point \( (2, 3, 7) \) to the plane is \( \sqrt{11} \) units, and the coordinates of the foot of the perpendicular are \( (5, 2, 6) \).

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