The Cartesian equation of the line passing through the point (3, 0, −2) and parallel to the line \(\vec{r} = (2-\lambda)\hat{\imath}+(3+2\lambda)\hat{\jmath}+(\lambda-1)\hat{k}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{x-3}{-1}=\frac{y}{2}=\frac{z+2}{1}\)
Separate the given line into a fixed part and a part multiplied by \(\lambda\):
\(\vec{r} = (2\hat{\imath}+3\hat{\jmath}-\hat{k})+\lambda(-\hat{\imath}+2\hat{\jmath}+\hat{k})\)
The direction is the coefficient of \(\lambda\): direction ratios −1, 2, 1. A parallel line has the same direction ratios.
Through (3, 0, −2):
\( \displaystyle \frac{x-3}{-1}=\frac{y-0}{2}=\frac{z+2}{1} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{x-3}{2}=\frac{y}{3}=\frac{z+2}{-1}\) — Incorrect. It uses the constant terms 2, 3, −1 as the direction instead of the coefficients of \(\lambda\).
Option (2) \(\displaystyle\frac{x+3}{-1}=\frac{y}{2}=\frac{z-2}{1}\) — Incorrect. This passes through (−3, 0, 2), not (3, 0, −2).
Option (3) \(\displaystyle\frac{x-2}{-1}=\frac{y-3}{2}=\frac{z+1}{1}\) — Incorrect. This is the given line itself, through (2, 3, −1); (3, 0, −2) does not lie on it.