The position vector of the point of intersection of the line $\bar{r}=(2 \hat{\imath}+\hat{\jmath}-4…
The position vector of the point of intersection of the line
$\bar{r}=(2 \hat{\imath}+\hat{\jmath}-4 \hat{k})+\lambda(\hat{\imath}-2 \hat{\jmath}+2 \hat{k})$ and XOY-Plane is
$4 \hat{\imath}+3 \hat{k}$
$4 \hat{\imath}+3 \hat{\jmath}$
$4 \hat{\imath}-3 \hat{k}$
$4 \hat{\imath}-3 \hat{\jmath}$
Solution
We have line $\bar{r}=(2 \hat{i}+\hat{j}-4 \hat{k})+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$
Hence coordinates of any point on this line are $(\lambda+2,-2 \lambda+1,2 \lambda-4)$
This point lies on XOY plane whose equation is $z=0$ $\therefore \quad 2 \lambda-4=0 \Rightarrow \lambda=2$
Hence point of intersection is $(4,-3,0)$. Thus $\mathrm{pv}$ is $4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}$