Find the co-ordinates of the foot of perpendicular and the length of the perpendicular drawn from the point P (5,4,2) to the line :
`vec(r) = hati + 3 hat(j) + hat(k) + lambda ( 2 hat(i) + 3 hatj - hatk )`.
Also, find the image of P in this line.

Answer

To find the coordinates of the foot of the perpendicular from the point \( P(5, 4, 2) \) to the line given by the vector equation:

\[ \vec{r} = \hat{i} + 3\hat{j} + \hat{k} + \lambda(2\hat{i} + 3\hat{j} - \hat{k}), \]

we will follow these steps:

### Step 1: Identify the line's point and direction vector
From the line equation, we can identify a point on the line \( A(1, 3, 1) \) and the direction vector \( \vec{d} = (2, 3, -1) \).

### Step 2: Write the coordinates of point B on the line
Let the coordinates of the foot of the perpendicular be \( B(a, b, c) \). Since \( B \) lies on the line, we can express its coordinates in terms of \( \lambda \):

\[ B = (1 + 2\lambda, 3 + 3\lambda, 1 - \lambda). \]

### Step 3: Set up the perpendicular condition
The vector \( \vec{BP} \) from \( B \) to \( P \) is given by:

\[ \vec{BP} = (5 - (1 + 2\lambda), 4 - (3 + 3\lambda), 2 - (1 - \lambda)) = (4 - 2\lambda, 1 - 3\lambda, 1 + \lambda). \]

Since \( BP \) is perpendicular to the line, the dot product of \( \vec{BP} \) and the direction vector \( \vec{d} \) should be zero:

\[ (4 - 2\lambda, 1 - 3\lambda, 1 + \lambda) \cdot (2, 3, -1) = 0. \]

### Step 4: Calculate the dot product
Calculating the dot product:

\[ (4 - 2\lambda) \cdot 2 + (1 - 3\lambda) \cdot 3 + (1 + \lambda) \cdot (-1) = 0. \]

Expanding this gives:

\[ 8 - 4\lambda + 3 - 9\lambda - 1 - \lambda = 0. \]

Combining like terms:

\[ 10 - 14\lambda = 0. \]

### Step 5: Solve for \( \lambda \)
Solving for \( \lambda \):

\[ 14\lambda = 10 \implies \lambda = \frac{10}{14} = \frac{5}{7}. \]

### Step 6: Find coordinates of point B
Substituting \( \lambda = \frac{5}{7} \) back into the equations for \( B \):

\[ a = 1 + 2\left(\frac{5}{7}\right) = 1 + \frac{10}{7} = \frac{17}{7}, \]
\[ b = 3 + 3\left(\frac{5}{7}\right) = 3 + \frac{15}{7} = \frac{36}{7}, \]
\[ c = 1 - \left(\frac{5}{7}\right) = \frac{2}{7}. \]

Thus, the coordinates of point \( B \) are:

\[ B\left(\frac{17}{7}, \frac{36}{7}, \frac{2}{7}\right). \]

### Step 7: Calculate the length of the perpendicular
To find the length of the perpendicular \( d \), we can use the distance formula between points \( P(5, 4, 2) \) and \( B\left(\frac{17}{7}, \frac{36}{7}, \frac{2}{7}\right) \):

\[ d = \sqrt{\left(5 - \frac{17}{7}\right)^2 + \left(4 - \frac{36}{7}\right)^2 + \left(2 - \frac{2}{7}\right)^2}. \]

Calculating each component:

1. \( 5 - \frac{17}{7} = \frac{35}{7} - \frac{17}{7} = \frac{18}{7} \)
2. \( 4 - \frac{36}{7} = \frac{28}{7} - \frac{36}{7} = -\frac{8}{7} \)
3. \( 2 - \frac{2}{7} = \frac{14}{7} - \frac{2}{7} = \frac{12}{7} \)

Now substituting back into the distance formula:

\[ d = \sqrt{\left(\frac{18}{7}\right)^2 + \left(-\frac{8}{7}\right)^2 + \left(\frac{12}{7}\right)^2} = \sqrt{\frac{324}{49} + \frac{64}{49} + \frac{144}{49}} = \sqrt{\frac{532}{49}} = \frac{\sqrt{532}}{7} = \frac{2\sqrt{133}}{7}. \]

### Step 8: Find the image of point P
The image of point \( P \) across the line can be found using the midpoint formula. The midpoint \( M \) of \( P \) and its image \( P' \) is point \( B \):

\[ M = \left(\frac{5 + x'}{2}, \frac{4 + y'}{2}, \frac{2 + z'}{2}\right) = B\left(\frac{17}{7}, \frac{36}{7}, \frac{2}{7}\right). \]

Setting up the equations:

1. \( \frac{5 + x'}{2} = \frac{17}{7} \)
2. \( \frac{4 + y'}{2} = \frac{36}{7} \)
3. \( \frac{2 + z'}{2} = \frac{2}{7} \)

Solving for \( x', y', z' \):

1. \( 5 + x' = \frac{34}{7} \implies x' = \frac{34}{7} - 5 = \frac{34}{7} - \frac{35}{7} = -\frac{1}{7} \)
2. \( 4 + y' = \frac{72}{7} \implies y' = \frac{72}{7} - 4 = \frac{72}{7} - \frac{28}{7} = \frac{44}{7} \)
3. \( 2 + z' = \frac{4}{7} \implies z' = \frac{4}{7} - 2 = \frac{4}{7} - \frac{14}{7} = -\frac{10}{7} \)

Thus, the image of point \( P \) is:

\[ P' \left(-\frac{1}{7}, \frac{44}{7}, -\frac{10}{7}\right). \]

### Summary of Results
- Coordinates of foot of the perpendicular \( B \): \( \left(\frac{17}{7}, \frac{36}{7}, \frac{2}{7}\right) \)
- Length of the perpendicular \( d \): \( \frac{2\sqrt{133}}{7} \)
- Image of point \( P \): \( P' \left(-\frac{1}{7}, \frac{44}{7}, -\frac{10}{7}\right) \)