Find the equation of the line through the point `hati+hatj-3hatk` and perpendicular to the lines `vecr=2hati-3hatj+s(2hati+hatj-3hatk)andvecr=3hati-5hatk+t(hati+hatj+hatk)`

Options

A. `vecr=(hati+hatj+3hatk)+lamda(4hati-5hatj-hatk)`
B. `vecr=(hati-hatj+3hatk)+lamda(4hati+5hatj+hatk)`
C. `vecr=(hati-hatj-3hatk)+lamda(4hati-5hatj+hatk)`
D. `vecr=(hati+hatj-3hatk)+lamda(4hati-5hatj+hatk)`

Answer

To find the equation of the line through the point \( \mathbf{a} = \hat{i} + \hat{j} - 3\hat{k} \) and perpendicular to the given lines, we will follow these steps:

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

The first line is given by:
\[ \mathbf{r} = 2\hat{i} - 3\hat{j} + s(2\hat{i} + \hat{j} - 3\hat{k}) \]
The direction ratios of this line can be extracted from the coefficient of \( s \):
\[ \mathbf{d_1} = 2\hat{i} + \hat{j} - 3\hat{k} \quad \Rightarrow \quad (2, 1, -3) \]

The second line is given by:
\[ \mathbf{r} = 3\hat{i} - 5\hat{k} + t(\hat{i} + \hat{j} + \hat{k}) \]
The direction ratios of this line can be extracted from the coefficient of \( t \):
\[ \mathbf{d_2} = \hat{i} + \hat{j} + \hat{k} \quad \Rightarrow \quad (1, 1, 1) \]

### Step 2: Find the direction ratios of the line perpendicular to both lines

To find a line that is perpendicular to both given lines, we need to take the cross product of their direction ratios \( \mathbf{d_1} \) and \( \mathbf{d_2} \).

Let:
\[ \mathbf{d_1} = (2, 1, -3) \quad \text{and} \quad \mathbf{d_2} = (1, 1, 1) \]

The cross product \( \mathbf{d_1} \times \mathbf{d_2} \) is calculated using the determinant:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 1 & -3 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 1 & -3 \\ 1 & 1 \end{vmatrix} = (1)(1) - (-3)(1) = 1 + 3 = 4 \)
2. \( \begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} = (2)(1) - (-3)(1) = 2 + 3 = 5 \)
3. \( \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = (2)(1) - (1)(1) = 2 - 1 = 1 \)

Putting it all together:
\[ \mathbf{d} = 4\hat{i} - 5\hat{j} + 1\hat{k} \quad \Rightarrow \quad (4, -5, 1) \]

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

The equation of a line in vector form can be expressed as:
\[ \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \]
where \( \mathbf{a} \) is a point on the line, \( \mathbf{b} \) is the direction vector, and \( \lambda \) is a scalar parameter.

Substituting \( \mathbf{a} = \hat{i} + \hat{j} - 3\hat{k} \) and \( \mathbf{b} = 4\hat{i} - 5\hat{j} + 1\hat{k} \):
\[ \mathbf{r} = (\hat{i} + \hat{j} - 3\hat{k}) + \lambda(4\hat{i} - 5\hat{j} + 1\hat{k}) \]

### Final Equation

Thus, the equation of the line is:
\[ \mathbf{r} = \hat{i} + \hat{j} - 3\hat{k} + \lambda(4\hat{i} - 5\hat{j} + 1\hat{k}) \]