Question

The point of intersection of the lines $\overrightarrow{r}=(\hat{i}+\hat{j}+\hat{k})+\lambda (3\hat{i}-\hat{j})$ and $\overrightarrow{r}=(4\hat{i}-\hat{k})+\mu (2\hat{i}+3\hat{j})$ is

• A. $(4,0,1)$
• B. $(4,0,-1)$
• C. $(-4,0,1)$
• D. $(-4,0,-1)$

Answer 1

Answer: (A)

$(4,0,1)$

$\begin{array}{c}\overrightarrow{r}=\left(\hat{i}+\hat{j}+\hat{k}\right)+\lambda \left(3i-j\right)\overrightarrow{r}=\left(4i-k\right)+\mu \left(2\hat{i}+3\hat{j}\right)\\ Let\overrightarrow{r}=\left(x\hat{i}+y\hat{j}+z\hat{k}\right)\\ \Rightarrow \overrightarrow{r}=\left(3\lambda +1\right)\hat{i}+\left(1-\lambda \right)\hat{j}-\hat{k},\overrightarrow{r}=\left(2\mu +4\right)\hat{i}+3\mu \hat{j}-\hat{k}\\ \Rightarrow \left(1-\lambda \right)\hat{j}=3\mu \hat{j}\Rightarrow 1-\lambda =3\mu \\ 3\mu +\lambda =1\to \left(i\right)\\ \Rightarrow 3\lambda +1=2\mu +4\\ 3\lambda -2\mu =3\to \left(ii\right)\\ Bycrossmultiplication,\\ \Rightarrow \lambda +3\mu -1=0\\ \Rightarrow 3\lambda -2\mu -3=0\\ \Rightarrow \frac{\lambda }{-11}=\frac{\mu }{0}=\frac{1}{-11}\\ \Rightarrow \lambda =1and=0\\ \therefore x=2\mu +4=4,y=3\mu =0\\ \Rightarrow andz=1,pointis\left(4,0,-1\right)\end{array}$