Question

Find the vector equation of the line which passes through the point \( (5,2,-4) \) and is parallel to the vector \( 3\hat{i} + 2\hat{j} - 8\hat{k} \).

Hint:

The vector equation of a line is \( \mathbf{r} = \mathbf{r_0} + \lambda \mathbf{b} \), where \( \mathbf{r_0} \) is a point on the line and \( \mathbf{b} \) is the direction vector.

Answer 1

Step 1: Use vector equation formula. The vector equation of a line passing through point \( (x_0, y_0, z_0) \) and parallel to vector \( \mathbf{b} \) is: \[ \mathbf{r} = \mathbf{r_0} + \lambda \mathbf{b} \] where: \[ \mathbf{r_0} = 5\hat{i} + 2\hat{j} - 4\hat{k}, \quad \mathbf{b} = 3\hat{i} + 2\hat{j} - 8\hat{k} \] Step 2: Write final equation. \[ \mathbf{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda (3\hat{i} + 2\hat{j} - 8\hat{k}) \] Step 3: Expand components. \[ \mathbf{r} = (5 + 3\lambda) \hat{i} + (2 + 2\lambda) \hat{j} + (-4 - 8\lambda) \hat{k} \]