Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The first line is given by $x=y,z=0$. It passes through the origin and its direction vector is ${\vec{b}}_1=\hat{i}+\hat{j}$.

The second line is given by $x+y=0,z=0$. It also passes through the origin and its direction vector is ${\vec{b}}_2=\hat{i}-\hat{j}$.

The normal vector $\vec{n}$ to the plane containing these two intersecting lines is the cross product of their direction vectors:

$\vec{n}={\vec{b}}_1\times {\vec{b}}_2=(\hat{i}+\hat{j})\times (\hat{i}-\hat{j})=-\hat{k}-\hat{k}=-2\hat{k}$

Since the plane passes through the origin, its vector equation is $\vec{r}\cdot \vec{n}=0$.

$\vec{r}\cdot (-2\hat{k})=0$

$\vec{r}\cdot \hat{k}=0$

Alternatively, both lines lie in the $xy$-plane, whose Cartesian equation is $z=0$. The vector form of $z=0$ is $\vec{r}\cdot \hat{k}=0$.

Answer: $\vec{r}\cdot \hat{k}=0$