Find the equation of the line perpendicular to the lines :
`vec(r) = ( 3 hat(i) + 2 hat(j) - 4 hat(k)) + lambda (hat(i) + 2 hat(j) - 2 hat(k))`
and `vec(r) = (5 hat(j) - 2 hat(k) + mu (3 hat(i) + 2 hat(j) + 6 hat(k) )` and passing through the point (1,1,1) .

Answer

To find the equation of the line that is perpendicular to the given lines and passes through the point (1, 1, 1), we will follow these steps:

### Step 1: Identify the direction vectors of the given lines

The first line is given by:
\[ \vec{r_1} = (3 \hat{i} + 2 \hat{j} - 4 \hat{k}) + \lambda (\hat{i} + 2 \hat{j} - 2 \hat{k}) \]
The direction vector \( \vec{b_1} \) of the first line is:
\[ \vec{b_1} = \hat{i} + 2\hat{j} - 2\hat{k} \]

The second line is given by:
\[ \vec{r_2} = (5 \hat{j} - 2 \hat{k}) + \mu (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) \]
The direction vector \( \vec{b_2} \) of the second line is:
\[ \vec{b_2} = 3\hat{i} + 2\hat{j} + 6\hat{k} \]

### Step 2: Find the cross product of the direction vectors

To find the direction vector \( \vec{b} \) of the line we are looking for, we take the cross product of \( \vec{b_1} \) and \( \vec{b_2} \):
\[ \vec{b} = \vec{b_1} \times \vec{b_2} \]

Calculating the cross product:
\[ \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -2 \\ 3 & 2 & 6 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{b} = \hat{i}(2 \cdot 6 - (-2) \cdot 2) - \hat{j}(1 \cdot 6 - (-2) \cdot 3) + \hat{k}(1 \cdot 2 - 2 \cdot 3) \]
\[ = \hat{i}(12 + 4) - \hat{j}(6 + 6) + \hat{k}(2 - 6) \]
\[ = 16\hat{i} - 12\hat{j} - 4\hat{k} \]

### Step 3: Write the equation of the line

The equation of the line can be expressed in vector form as:
\[ \vec{r} = \vec{A} + \lambda \vec{b} \]
where \( \vec{A} = (1, 1, 1) \) and \( \vec{b} = (16, -12, -4) \).

Thus, the equation of the line is:
\[ \vec{r} = (1 \hat{i} + 1 \hat{j} + 1 \hat{k}) + \lambda (16 \hat{i} - 12 \hat{j} - 4 \hat{k}) \]

### Step 4: Write the parametric equations

From the vector equation, we can derive the parametric equations:
\[ x = 1 + 16\lambda \]
\[ y = 1 - 12\lambda \]
\[ z = 1 - 4\lambda \]

### Final Answer

The equation of the line perpendicular to the given lines and passing through the point (1, 1, 1) is:
\[ \vec{r} = (1 + 16\lambda) \hat{i} + (1 - 12\lambda) \hat{j} + (1 - 4\lambda) \hat{k} \]

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