Question

Equation of the line which passes through the point with position vector $(2,1,0)$ and perpendicular to the plane containing the vectors $i+j$ and $j+k$ is

• A. $\overrightarrow{r}=(2,1,0)+t(1,-1,1)$
• B. $\overrightarrow{r}=(2,1,0)+t(-1,1,1)$
• C. $\overrightarrow{r}=(2,1,0)+t(1,1,-1)$
• D. $\overrightarrow{r}=(2,1,0)+t(1,1,1)$

Answer 1

The correct option is A $\overrightarrow{r}=(2,1,0)+t(1,-1,1)$

$\overrightarrow{r}=positionvector\overrightarrow{a}+t\left(normalvector\overrightarrow{n}\right)$

normal vector of plane

Given $\overrightarrow{a}=\hat{i}+\hat{j}$

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

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

$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 0\\ 0& 1& 1\end{array}\right|$

$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=\hat{i}(1-0)-\hat{j}(1-0)+\hat{k}(1-0)$

$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$

eq become

$\Rightarrow \overrightarrow{r}=(2,1,0)+t(1,-1,1)$