Question

Find vector equation of line passing through $(1,4,1),(1,2,2)$.

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

Answer 1

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

Given points

$A(1,4,1)$ and $B(1,2,2)$

position vector of point A

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

position vector of point B

$\overrightarrow{b}=\hat{i}+2\hat{j}+2\hat{k}$

normal vector of line be

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

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

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

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

vector eq of line

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

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

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

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