Question

Find the vector equation of the line $\dfrac{x+3}{2} = \dfrac{y-5}{4} = \dfrac{z+6}{2}$.

Hint:

Vector equation of a line: $\vec{r} = \vec{a} + \lambda \vec{d}$, where $\vec{a}$ is a point and $\vec{d}$ is direction vector.

Answer 1

Step 1: Compare with symmetric form. A line in symmetric form is: \[ \frac{x-x_{1}}{a} = \frac{y-y_{1}}{b} = \frac{z-z_{1}}{c} \] Here, \[ \frac{x+3}{2} = \frac{y-5}{4} = \frac{z+6}{2} \] So, \[ (x_{1}, y_{1}, z_{1}) = (-3, 5, -6), \vec{d} = (2, 4, 2) \]

Step 2: Write vector equation of line. General form: \[ \vec{r} = \vec{a} + \lambda \vec{d} \] where $\vec{a}$ is position vector of a point, $\vec{d}$ is direction vector.

Step 3: Substitute values. \[ \vec{a} = -3\hat{i} + 5\hat{j} - 6\hat{k}, \vec{d} = 2\hat{i} + 4\hat{j} + 2\hat{k} \] So, \[ \vec{r} = (-3\hat{i} + 5\hat{j} - 6\hat{k}) + \lambda(2\hat{i} + 4\hat{j} + 2\hat{k}) \]

Final Answer: \[ \boxed{\vec{r} = (-3\hat{i} + 5\hat{j} - 6\hat{k}) + \lambda(2\hat{i} + 4\hat{j} + 2\hat{k})} \]