Find the vector equation of the plane that contains the line `vec(r) = (hat(i) + hat(j) ) + lambda (hat(i) + 2 hat(j) - hat(k))` and the point (-1, 3,-4). Also , find the length of the perpendicular from the point (2,1,4) to the plane, thus botained.

Answer

To solve the problem, we need to find the vector equation of the plane that contains the given line and the point (-1, 3, -4). We will also calculate the length of the perpendicular from the point (2, 1, 4) to the plane.

### Step 1: Identify the line and point
The line is given by the vector equation:
\[ \vec{r} = (\hat{i} + \hat{j}) + \lambda (\hat{i} + 2\hat{j} - \hat{k}) \]
From this, we can identify:
- A point on the line, \( A(1, 1, 0) \) (when \(\lambda = 0\))
- The direction vector of the line, \( \vec{b} = \hat{i} + 2\hat{j} - \hat{k} \)

The point given is \( P(-1, 3, -4) \).

### Step 2: Find the vector \( \vec{AP} \)
The vector \( \vec{AP} \) from point \( A \) to point \( P \) is calculated as follows:
\[ \vec{AP} = P - A = (-1 - 1, 3 - 1, -4 - 0) = (-2, 2, -4) \]

### Step 3: Find the normal vector \( \vec{n} \)
To find the normal vector \( \vec{n} \) of the plane, we take the cross product of the direction vector \( \vec{b} \) and the vector \( \vec{AP} \):
\[ \vec{b} = (1, 2, -1), \quad \vec{AP} = (-2, 2, -4) \]

Calculating the cross product \( \vec{AP} \times \vec{b} \):
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 2 & -4 \\ 1 & 2 & -1 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{n} = \hat{i} \begin{vmatrix} 2 & -4 \\ 2 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} -2 & -4 \\ 1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} -2 & 2 \\ 1 & 2 \end{vmatrix} \]

Calculating the minors:
1. \( \begin{vmatrix} 2 & -4 \\ 2 & -1 \end{vmatrix} = (2)(-1) - (2)(-4) = -2 + 8 = 6 \)
2. \( \begin{vmatrix} -2 & -4 \\ 1 & -1 \end{vmatrix} = (-2)(-1) - (1)(-4) = 2 + 4 = 6 \)
3. \( \begin{vmatrix} -2 & 2 \\ 1 & 2 \end{vmatrix} = (-2)(2) - (1)(2) = -4 - 2 = -6 \)

Thus, we have:
\[ \vec{n} = 6\hat{i} - 6\hat{j} - 6\hat{k} = 6(\hat{i} - \hat{j} - \hat{k}) \]

### Step 4: Write the equation of the plane
The equation of the plane can be expressed as:
\[ \vec{r} \cdot \vec{n} = \vec{p} \cdot \vec{n} \]
Where \( \vec{p} = (-1, 3, -4) \) and \( \vec{n} = (6, -6, -6) \).

Calculating \( \vec{p} \cdot \vec{n} \):
\[ \vec{p} \cdot \vec{n} = (-1)(6) + (3)(-6) + (-4)(-6) = -6 - 18 + 24 = 0 \]

Thus, the equation of the plane is:
\[ \vec{r} \cdot (6\hat{i} - 6\hat{j} - 6\hat{k}) = 0 \]
Or simplified:
\[ \vec{r} \cdot (\hat{i} - \hat{j} - \hat{k}) = 0 \]

### Step 5: Find the length of the perpendicular from point (2, 1, 4) to the plane
The formula for the distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

Here, the plane equation can be rewritten as:
\[ x - y - z = 0 \quad \text{(where } A = 1, B = -1, C = -1, D = 0\text{)} \]
Now substituting \( (x_0, y_0, z_0) = (2, 1, 4) \):
\[ d = \frac{|1(2) - 1(1) - 1(4) + 0|}{\sqrt{1^2 + (-1)^2 + (-1)^2}} = \frac{|2 - 1 - 4|}{\sqrt{1 + 1 + 1}} = \frac{|-3|}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \]

### Final Answer
The vector equation of the plane is:
\[ \vec{r} \cdot (\hat{i} - \hat{j} - \hat{k}) = 0 \]
The length of the perpendicular from the point (2, 1, 4) to the plane is:
\[ \sqrt{3} \]