Question

Find vector equation for the line passing through the points $3\overline{i}+4\overline{j}-7\overline{k},\overline{i}-\overline{j}+6\overline{k}$.

• A. $\overline{r}=(3-2\lambda )\overline{i}+(4-5\lambda )\overline{j}+(-7+13\lambda )\overline{k}$
• B. $\overline{r}=\left(2\lambda \right)\overline{i}+(4+5\lambda )\overline{j}+(-7-13\lambda )\overline{k}$
• C. $\overline{r}=(3-2\lambda )\overline{i}-(4-5\lambda )\overline{j}+(-7+13\lambda )\overline{k}$
• D. $\overline{r}=(3-2\lambda )\overline{i}+(4-5\lambda )\overline{j}-(-7+13\lambda )\overline{k}$

Answer 1

The correct option is A $\overline{r}=(3-2\lambda )\overline{i}+(4-5\lambda )\overline{j}+(-7+13\lambda )\overline{k}$

Given points

$\overrightarrow{a}=3\hat{i}+4\hat{j}-7\hat{k}$

$\overrightarrow{b}=\hat{i}-\hat{j}+6\hat{k}$

normal vector of line be

$\overrightarrow{n}=\overrightarrow{b}-\overrightarrow{a}$

$\overrightarrow{n}=\hat{i}-\hat{j}+6\hat{k}-(3\hat{i}+4\hat{j}-7\hat{k})$

$\overrightarrow{n}=\hat{i}-\hat{j}+6\hat{k}-3\hat{i}-4\hat{j}+7\hat{k}$

$\overrightarrow{n}=-2\hat{i}-5\hat{j}+13\hat{k}$

vector eq of line

$\overrightarrow{r}=\overrightarrow{a}+\lambda \left(\overrightarrow{n}\right)$

$\overrightarrow{r}=3\hat{i}+4\hat{j}-7\hat{k}+\lambda (-2\hat{i}-5\hat{j}+13\hat{k})$

$\overrightarrow{r}=3\hat{i}+4\hat{j}-7\hat{k}-2\lambda \hat{i}-5\lambda \hat{j}+13\lambda \hat{k})$

$\overrightarrow{r}=(3-2\lambda )\hat{i}+(4-5\lambda )\hat{j}+(-7+13\lambda )\hat{k}$