Answer
$y-z=0$
Work Step by Step
The equation of the plane passing through the points $O(0,0,0), P(1,0,0)$ and $Q(0,1,1)$ is:
\begin{aligned}
&ax + by + cz = 0 \\
&\Rightarrow x\cdot a + y\cdot b + z\cdot c = 0 \\
&\Rightarrow (1\cdot a + 0\cdot b + 0\cdot c)\cdot x + (0\cdot a + 1\cdot b + 0\cdot c)\cdot y + (0\cdot a + 0\cdot b + 1\cdot c)\cdot z = 0 \\
&\Rightarrow (a,b,c)\cdot (x,y,z) = 0 \\
&\Rightarrow \langle a,b,c \rangle \cdot \langle x,y,z \rangle = 0 \
\end{aligned}
Therefore, the normal vector to the plane is $\langle a,b,c \rangle$.
To find the equation of the plane, we need to find the values of $a$, $b$, and $c$. We can use the fact that the plane passes through the points $O(0,0,0), P(1,0,0)$ and $Q(0,1,1)$ to find these values.
Let $\vec{u} = \overrightarrow{OP} = \langle 1,0,0 \rangle$ and $\vec{v} = \overrightarrow{OQ} = \langle 0,1,1 \rangle$. Then, the normal vector to the plane is:
\begin{aligned}
\vec{n} &= \vec{u} \times \vec{v} \
&= \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{vmatrix} \
&= \langle 0,-1,1 \rangle
\end{aligned}
Since the plane passes through the origin, we can write the equation of the plane as:
\begin{aligned}
\langle a,b,c \rangle \cdot \langle x,y,z \rangle = 0 \
\Rightarrow \langle a,b,c \rangle \cdot \langle 0,0,0 \rangle = 0 \
\Rightarrow 0= 0 \
\end{aligned}
Therefore, the constant term in the equation of the plane is $0$.
Substituting the normal vector and the constant term in the equation of the plane, we get:
\begin{aligned}
\langle a,b,c \rangle \cdot \langle x,y,z \rangle = 0 \
\Rightarrow \langle 0,-1,1 \rangle \cdot \langle x,y,z \rangle = 0 \
\Rightarrow -y+z = 0 \
\Rightarrow y &= z \
\end{aligned}
Thus, the equation of the plane passing through the points $O(0,0,0), P(1,0,0)$ and $Q(0,1,1)$ is $\boxed{y-z=0}$.
