The cartesian equation of the line passing through the point (− 4, 3, −2) and parallel to the line \( \vec{r}=5\hat{\imath}-\hat{k}+\lambda(\hat{\imath}-\hat{\jmath}+2\hat{k}) \) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{x+4}{1}=\frac{y-3}{-1}=\frac{z+2}{2} \)
A line parallel to the given line has the same direction vector, the vector multiplied by λ: \( \hat{\imath}-\hat{\jmath}+2\hat{k} \), with direction ratios 1, −1, 2. (The vector \( 5\hat{\imath}-\hat{k} \) is only the position vector of a point on the given line.)
The line through \( (x_1,y_1,z_1)=(-4,3,-2) \) with direction ratios a, b, c is
\( \displaystyle \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} \)
\( \displaystyle \Rightarrow\ \frac{x+4}{1}=\frac{y-3}{-1}=\frac{z+2}{2} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{x-4}{1}=\frac{y-3}{-1}=\frac{z-2}{2} \) — Incorrect. The direction ratios are right, but the signs of the point are wrong: x − 4 and z − 2 correspond to the point (4, 3, 2).
Option (3) \( \displaystyle\frac{x+4}{5}=\frac{y-3}{0}=\frac{z+2}{-1} \) — Incorrect. The direction ratios 5, 0, −1 come from the point \( 5\hat{\imath}-\hat{k} \), not from the direction vector.
Option (4) \( \displaystyle\frac{x-4}{5}=\frac{y-3}{0}=\frac{z-2}{-1} \) — Incorrect. This uses the point vector \( 5\hat{\imath}-\hat{k} \) as the direction and also has the wrong signs for the point.