Question

Find the vector equation of line joining the points $(2,1,3)$ and $(-4,3,-1)$

• A. $\overline{r}=2(1-3\lambda )\overline{i}-(1+2\lambda )\overline{j}-(3-4\lambda )\overline{k}$
• B. $\overline{r}=2(1-3\lambda )\overline{i}-(1+2\lambda )\overline{j}+(3-4\lambda )\overline{k}$
• C. $\overline{r}=2(1-3\lambda )\overline{i}+(1+2\lambda )\overline{j}+(3-4\lambda )\overline{k}$
• D. $\overline{r}=2(1+3\lambda )\overline{i}+(1+2\lambda )\overline{j}+(3+4\lambda )\overline{k}$

Answer 1

The correct option is C $\overline{r}=2(1-3\lambda )\overline{i}+(1+2\lambda )\overline{j}+(3-4\lambda )\overline{k}$

given Points

A(2,1,3) and B(-4,3,-1)

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

Position vector of line

$A=\overrightarrow{a}=2\hat{i}+\hat{j}+3\hat{k}$

normal vector can be calculated by

$\overrightarrow{b}-\overrightarrow{a}=\overrightarrow{n}=-4\hat{i}+3\hat{j}-\hat{k}-(2\hat{i}+\hat{j}+3\hat{k})$

$\overrightarrow{n}=-4\hat{i}+3\hat{j}-\hat{k}-2\hat{i}-\hat{j}-3\hat{k}$

$\overrightarrow{n}=-6\hat{i}+2\hat{j}-4\hat{k}$

eq of line

$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{n}$

$\overrightarrow{r}=2\hat{i}+\hat{j}+3\hat{k}+\lambda (-6\hat{i}+2\hat{j}-4\hat{k})$

$\overrightarrow{r}=2\hat{i}+\hat{j}+3\hat{k}-6\lambda \hat{i}+2\lambda \hat{j}-4\lambda \hat{k}$

$\overrightarrow{r}=(2-6\lambda )\hat{i}+(1+2\lambda )\hat{j}+(3-4\lambda )\hat{k}$

$\overrightarrow{r}=2(1-3\lambda )\hat{i}+(1+2\lambda )\hat{j}+(3-4\lambda )\hat{k}$