Prove that lines `vec( r)= (hat(i)+ hat(j) + hat(k))+ t(hat(i) - hat(j) + hat(k)) and vec(r ) = (3hat(i) - hat(k)) + s(4hat(j) - 16hat(k))` intersect. Also find the position vector of their point of intersection

Answer

To prove that the lines given by the equations

\[ \vec{r_1} = (\hat{i} + \hat{j} + \hat{k}) + t(\hat{i} - \hat{j} + \hat{k}) \]

and

\[ \vec{r_2} = (3\hat{i} - \hat{k}) + s(4\hat{j} - 16\hat{k}) \]

intersect, we will express both lines in terms of their coordinates and then set them equal to find the values of parameters \(t\) and \(s\).

### Step 1: Express the first line in terms of coordinates

From the first line equation:

\[ \vec{r_1} = (\hat{i} + \hat{j} + \hat{k}) + t(\hat{i} - \hat{j} + \hat{k}) \]

We can expand this:

\[ \vec{r_1} = (1 + t)\hat{i} + (1 - t)\hat{j} + (1 + t)\hat{k} \]

Thus, the coordinates of the first line are:

\[ x_1 = 1 + t, \quad y_1 = 1 - t, \quad z_1 = 1 + t \]

### Step 2: Express the second line in terms of coordinates

From the second line equation:

\[ \vec{r_2} = (3\hat{i} - \hat{k}) + s(4\hat{j} - 16\hat{k}) \]

We can expand this:

\[ \vec{r_2} = 3\hat{i} + (4s)\hat{j} + (-1 - 16s)\hat{k} \]

Thus, the coordinates of the second line are:

\[ x_2 = 3, \quad y_2 = 4s, \quad z_2 = -1 - 16s \]

### Step 3: Set the coordinates equal to find \(t\) and \(s\)

For the lines to intersect, we must have:

1. \(1 + t = 3\)
2. \(1 - t = 4s\)
3. \(1 + t = -1 - 16s\)

#### Solving the first equation:

From \(1 + t = 3\):

\[ t = 3 - 1 = 2 \]

#### Solving the second equation:

Substituting \(t = 2\) into \(1 - t = 4s\):

\[ 1 - 2 = 4s \implies -1 = 4s \implies s = -\frac{1}{4} \]

#### Solving the third equation:

Now, substituting \(t = 2\) and \(s = -\frac{1}{4}\) into \(1 + t = -1 - 16s\):

\[ 1 + 2 = -1 - 16\left(-\frac{1}{4}\right) \]
\[ 3 = -1 + 4 \]
\[ 3 = 3 \]

This confirms that both equations are satisfied.

### Step 4: Find the point of intersection

Now, we can find the coordinates of the point of intersection by substituting \(t = 2\) into the first line's coordinates:

\[ x = 1 + t = 1 + 2 = 3 \]
\[ y = 1 - t = 1 - 2 = -1 \]
\[ z = 1 + t = 1 + 2 = 3 \]

Thus, the point of intersection is:

\[ (3, -1, 3) \]

### Final Result

The lines intersect at the point:

\[ \vec{P} = 3\hat{i} - \hat{j} + 3\hat{k} \]