Question

Find vector equation of line passing through $(3,-1,2)$ and perpendicular to the lines

$\overline{r}=\overline{i}+\overline{j}-\overline{k}+\lambda (2\overline{i}-2\overline{j}+2\overline{k})$ and $\overline{r}=2\overline{i}+\overline{j}-3\overline{k}+\mu (\overline{i}-2\overline{j}+2\overline{k})$

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

Answer 1

Answer: (D)

$\overline{r}=3\overline{i}-\overline{j}+2\overline{k}+\lambda (-2\overline{i}-3\overline{j}-2\overline{k})$

the question has error

the eq of second line will be

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

in place of

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

required line is passing through A(3,-1,2)

position vector become

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

So eq of line become

$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$-------------(1)

required line is perpendicular to lines

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

normal vector of above line

$\overrightarrow{n_1}=2\hat{i}-2\hat{j}+2\hat{k}$

and required line is also perpendicular to lines

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

normal vector of above line

$\overrightarrow{n_2}=\hat{i}-2\hat{j}+4\hat{k}$

line is perpendicular to both lines so

$\overrightarrow{b}=\overrightarrow{n_1}\times \overrightarrow{n_2}$

$\overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -2& 2\\ 1& -2& 4\end{array}\right|$

$\overrightarrow{b}=\hat{i}(-8+4)-\hat{j}(8-2)+\hat{k}(-4+2)$

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

putting a and b in eq (1)

So required line of eq

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

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

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

This is required eq