Question

Vector equation of the line $6x-3 = 3y + 4 = 2z-2$ is

• A. $\vec{r}=\hat{i}-\hat{j}+\hat{k}+\lambda\left(6\hat{i}+\hat{j}+\hat{k}\right)$
• B. $\vec{r}=6\hat{i}+3\hat{j}+2\hat{k}+\lambda\left(3\hat{i}+4\hat{j}-2\hat{k}\right)$
• C. $\vec{r}=\frac{1}{2}\hat{i}-\frac{4}{3}\hat{j}+\hat{k}+\lambda\left(\frac{1}{6}\hat{i}+\frac{1}{3}\hat{j}+\frac{1}{2}\hat{k}\right)$
• D. None of these

Answer 1

Answer: (C)

Given equation can be written in symmetric form as
$\frac{x-\frac{1}{2}}{\frac{1}{6}}=\frac{y-\left(-\frac{4}{3}\right)}{\frac{1}{3}}=\frac{z-1}{\frac{1}{2}}$
$\therefore $ Its vector equation is
$\vec{r}=\frac{1}{2}\hat{i}-\frac{4}{3}\hat{j}++\hat{k}+\lambda\left(\frac{1}{6}\hat{i}+\frac{1}{3}\hat{j}+\frac{1}{2}\hat{k}\right)$.