The Cartesian equation of a line is \( 6x-2=3y+1=2z-2 \) . Then its vector form is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \vec{r}=\left(\displaystyle\frac{1}{3}\hat{\imath}-\displaystyle\frac{1}{3}\hat{\jmath}+\hat{k}\right)+\lambda\left(\hat{\imath}+2\hat{\jmath}+3\hat{k}\right) \)
The point and the direction can be read only from the symmetric form \( \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} \), where x, y and z have coefficient 1:
\( \displaystyle 6\left(x-\frac{1}{3}\right)=3\left(y+\frac{1}{3}\right)=2(z-1) \)
Dividing each part by 6:
\( \displaystyle \frac{x-\frac{1}{3}}{1}=\frac{y+\frac{1}{3}}{2}=\frac{z-1}{3} \)
So the line passes through \( \left(\frac{1}{3},-\frac{1}{3},1\right) \) and has direction ratios 1, 2, 3:
\( \displaystyle \vec{r}=\left(\frac{1}{3}\hat{\imath}-\frac{1}{3}\hat{\jmath}+\hat{k}\right)+\lambda\left(\hat{\imath}+2\hat{\jmath}+3\hat{k}\right) \)
Why the other options are incorrect
Option (1) \( \vec{r}=\left(2\hat{\imath}-\hat{\jmath}+2\hat{k}\right)+\lambda\left(6\hat{\imath}+3\hat{\jmath}+2\hat{k}\right) \) — Incorrect. 6, 3, 2 are the coefficients of x, y, z, not the direction ratios; the direction ratios are \( \frac{1}{6}:\frac{1}{3}:\frac{1}{2} \), that is 1 : 2 : 3.
Option (2) \( \vec{r}=\left(\hat{\imath}+2\hat{\jmath}+3\hat{k}\right)+\lambda\left(\displaystyle\frac{1}{3}\hat{\imath}-\displaystyle\frac{1}{3}\hat{\jmath}+\hat{k}\right) \) — Incorrect. The point and the direction have been interchanged.
Option (3) \( \vec{r}=\left(-\displaystyle\frac{1}{3}\hat{\imath}+\displaystyle\frac{1}{3}\hat{\jmath}-\hat{k}\right)+\lambda\left(\hat{\imath}+2\hat{\jmath}+3\hat{k}\right) \) — Incorrect. The point has the signs reversed; \( \left(-\frac{1}{3},\frac{1}{3},-1\right) \) gives 6x − 2 = −4 but 3y + 1 = 2, so it is not on the line.