Question

The cartesian equation of a line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$. Its vector form is:

• A. $\overrightarrow{r}=(5\hat{i}-4\hat{j}+6\hat{k})+\lambda (3\hat{i}+7\hat{j}+2\hat{k})$
• B. $\overrightarrow{r}=(5\hat{i}+4\hat{j}-6\hat{k})+\lambda (3\hat{i}+7\hat{j}+2\hat{k})$
• C. $\overrightarrow{r}=(5\hat{i}-4\hat{j}+6\hat{k})+\lambda (3\hat{i}-7\hat{j}-2\hat{k})$
• D. $\overrightarrow{r}=(5\hat{i}-4\hat{j}+6\hat{k})+\lambda (3\hat{i}+7\hat{j}-2\hat{k})$

Answer 1

The correct option is A $\overrightarrow{r}=(5\hat{i}-4\hat{j}+6\hat{k})+\lambda (3\hat{i}+7\hat{j}+2\hat{k})$

The Cartesian equation of the line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$
The given line passes through the point $(5,-4,6)$.
The direction ratios of the given line are $3,7$ and $2$.
This means that the line is in the direction of vector, $3\hat{i}+7\hat{j}+2\widehat{k.}$
Hence the vector form is:
$\overrightarrow{r}=(5\hat{i}-4\hat{j}+6\hat{k})+\lambda (3\hat{i}+7\hat{j}+2\hat{k})$