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

Answer

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

### Step 1: Identify the normal vector of the plane
The equation of the plane can be rewritten in the standard form \( Ax + By + Cz + D = 0 \). Here, we can rearrange the equation as:
\[ 3x - y - z - 7 = 0 \]
From this, we can identify the coefficients \( A = 3, B = -1, C = -1 \). The normal vector \( \vec{n} \) to the plane is given by:
\[ \vec{n} = (3, -1, -1) \]

### Step 2: Write the parametric equations of the line
The line through the point \( P(2, 3, 7) \) in the direction of the normal vector can be expressed parametrically as:
\[ \begin{align*} x &= 2 + 3t \\ y &= 3 - t \\ z &= 7 - t \end{align*} \]
where \( t \) is a parameter.

### Step 3: Substitute the parametric equations into the plane equation
To find the foot of the perpendicular, we substitute the parametric equations into the plane equation:
\[ 3(2 + 3t) - (3 - t) - (7 - t) = 7 \]
Expanding this gives:
\[ 6 + 9t - 3 + t - 7 + t = 7 \]
Simplifying, we have:
\[ 9t + 3t - 4 = 7 \implies 10t - 4 = 7 \implies 10t = 11 \implies t = \frac{11}{10} \]

### Step 4: Find the coordinates of the foot of the perpendicular
Now we substitute \( t = \frac{11}{10} \) back into the parametric equations:
\[ \begin{align*} x &= 2 + 3\left(\frac{11}{10}\right) = 2 + \frac{33}{10} = \frac{20}{10} + \frac{33}{10} = \frac{53}{10} \\ y &= 3 - \left(\frac{11}{10}\right) = \frac{30}{10} - \frac{11}{10} = \frac{19}{10} \\ z &= 7 - \left(\frac{11}{10}\right) = \frac{70}{10} - \frac{11}{10} = \frac{59}{10} \end{align*} \]
Thus, the coordinates of the foot of the perpendicular \( Q \) are:
\[ Q\left(\frac{53}{10}, \frac{19}{10}, \frac{59}{10}\right) \]

### Step 5: Calculate the length of the perpendicular
To find the length of the perpendicular from point \( P(2, 3, 7) \) to point \( Q\left(\frac{53}{10}, \frac{19}{10}, \frac{59}{10}\right) \), we use the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Substituting the coordinates:
\[ \begin{align*} d &= \sqrt{\left(\frac{53}{10} - 2\right)^2 + \left(\frac{19}{10} - 3\right)^2 + \left(\frac{59}{10} - 7\right)^2} \\ &= \sqrt{\left(\frac{53}{10} - \frac{20}{10}\right)^2 + \left(\frac{19}{10} - \frac{30}{10}\right)^2 + \left(\frac{59}{10} - \frac{70}{10}\right)^2} \\ &= \sqrt{\left(\frac{33}{10}\right)^2 + \left(-\frac{11}{10}\right)^2 + \left(-\frac{11}{10}\right)^2} \\ &= \sqrt{\frac{1089}{100} + \frac{121}{100} + \frac{121}{100}} \\ &= \sqrt{\frac{1331}{100}} \\ &= \frac{\sqrt{1331}}{10} \end{align*} \]

### Final Result
The coordinates of the foot of the perpendicular are:
\[ Q\left(\frac{53}{10}, \frac{19}{10}, \frac{59}{10}\right) \]
And the length of the perpendicular is:
\[ \frac{\sqrt{1331}}{10} \]