Question

The vector equation of a line which passes through the point (2, -4, 5) and is parallel to the line $\displaystyle \frac{x+3}{3} = \frac{4-y}{2} = \frac{z+8}{6}$ is:

• A. $\vec{r} = (-2\hat{i} + 4\hat{j} - 5\hat{k}) + \lambda (3\hat{i} + 2\hat{j} + 6\hat{k})$
• B. $\vec{r} = (2\hat{i} - 4\hat{j} + 5\hat{k}) + \lambda (3\hat{i} - 2\hat{j} + 6\hat{k})$
• C. $\vec{r} = (2\hat{i} - 4\hat{j} + 5\hat{k}) + \lambda (3\hat{i} + 2\hat{j} + 6\hat{k})$
• D. $\vec{r} = (-2\hat{i} + 4\hat{j} - 5\hat{k}) + \lambda (3\hat{i} - 2\hat{j} - 6\hat{k})$

Hint:

For a line, use \textit{Point + Direction Vector}. Pick sign of direction carefully.

Answer 1

The Correct Option is B

The given line is in symmetric form: \[ \frac{x+3}{3} = \frac{4-y}{2} = \frac{z+8}{6}. \] Direction ratios are: \[ \text{DRs} = (3, -2, 6) (\text{note: } 4-y = t \implies y = 4 - t \implies \text{coefficient of } y \text{ is } -2). \] Required line must pass through (2, -4, 5) and have same direction vector: \[ \vec{r} = (2\hat{i} -4\hat{j} +5\hat{k}) + \lambda (3\hat{i} -2\hat{j} + 6\hat{k}). \]