A line passes through the point (3,4,5) and is parallel to the vector `(2hat(i) +2hat(j) -3hat(k))` . Find the equations of the line in the vector as well as Cartesian forms.

Answer

To find the equations of the line that passes through the point (3, 4, 5) and is parallel to the vector \( \mathbf{b} = 2\hat{i} + 2\hat{j} - 3\hat{k} \), we will derive both the vector form and the Cartesian form of the line.

### Step 1: Write the vector form of the line

The vector form of a line can be expressed as:

\[ \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \]

where:
- \( \mathbf{r} \) is the position vector of any point on the line,
- \( \mathbf{a} \) is the position vector of a known point on the line,
- \( \lambda \) is a scalar parameter,
- \( \mathbf{b} \) is the direction vector of the line.

Given:
- The point \( (3, 4, 5) \) corresponds to the position vector \( \mathbf{a} = 3\hat{i} + 4\hat{j} + 5\hat{k} \).
- The direction vector \( \mathbf{b} = 2\hat{i} + 2\hat{j} - 3\hat{k} \).

Thus, the vector equation of the line is:

\[ \mathbf{r} = (3\hat{i} + 4\hat{j} + 5\hat{k}) + \lambda (2\hat{i} + 2\hat{j} - 3\hat{k}) \]

### Step 2: Simplify the vector equation

Distributing \( \lambda \) gives:

\[ \mathbf{r} = 3\hat{i} + 4\hat{j} + 5\hat{k} + \lambda (2\hat{i} + 2\hat{j} - 3\hat{k}) \]

This can be rewritten as:

\[ \mathbf{r} = (3 + 2\lambda)\hat{i} + (4 + 2\lambda)\hat{j} + (5 - 3\lambda)\hat{k} \]

### Step 3: Write the Cartesian form of the line

To convert the vector form into Cartesian form, we equate the components:

Let:
- \( x = 3 + 2\lambda \)
- \( y = 4 + 2\lambda \)
- \( z = 5 - 3\lambda \)

Now, we can express \( \lambda \) in terms of \( x \), \( y \), and \( z \):

From the first equation:

\[ \lambda = \frac{x - 3}{2} \]

From the second equation:

\[ \lambda = \frac{y - 4}{2} \]

From the third equation:

\[ \lambda = \frac{5 - z}{3} \]

### Step 4: Set the equations equal to each other

Now, we can set the expressions for \( \lambda \) equal to each other:

1. From \( \frac{x - 3}{2} = \frac{y - 4}{2} \):

\[ x - 3 = y - 4 \implies x - y + 1 = 0 \quad \text{(Equation 1)} \]

2. From \( \frac{y - 4}{2} = \frac{5 - z}{3} \):

Cross-multiplying gives:

\[ 3(y - 4) = 2(5 - z) \implies 3y - 12 = 10 - 2z \implies 3y + 2z - 22 = 0 \quad \text{(Equation 2)} \]

3. From \( \frac{x - 3}{2} = \frac{5 - z}{3} \):

Cross-multiplying gives:

\[ 3(x - 3) = 2(5 - z) \implies 3x - 9 = 10 - 2z \implies 3x + 2z - 19 = 0 \quad \text{(Equation 3)} \]

### Final Result

The equations of the line in vector form are:

\[ \mathbf{r} = (3 + 2\lambda)\hat{i} + (4 + 2\lambda)\hat{j} + (5 - 3\lambda)\hat{k} \]

And in Cartesian form, the equations are:

1. \( x - y + 1 = 0 \)
2. \( 3y + 2z - 22 = 0 \)
3. \( 3x + 2z - 19 = 0 \)