Find the equation of the plane that passes through the point and has as a normal.

Question

Find the equation of the plane that passes through the point  and has  as a normal.

Quizplus · Accepted Answer

The equation of a plane in point-normal form is given by:
$(\vec{r}-\vec{a})\cdot\vec{n}=0$
where $\vec{r}$ is the position vector of a point on the plane, $\vec{a}$ is a known point on the plane, and $\vec{n}$ is a normal vector to the plane.
In this case, our known point is $(11ead0bc\_8e1f\_82ac\_99a0\_5b738928e43a\_TB6988\_11)$ and our normal vector is $(11ead0bc\_8e1f\_a9bd\_99a0\_7f319fc61dc7\_TB6988\_11)$.
Plugging in the values, we get:
$(\vec{r}-\begin{pmatrix}11ead0bc\_8e1f\_82ac\_99a0\_5b738928e43a\_TB6988\_11\end{pmatrix})\cdot\begin{pmatrix}11ead0bc\_8e1f\_a9bd\_99a0\_7f319fc61dc7\_TB6988\_11\end{pmatrix}=0$
Simplifying, we get:
$11ead0bc\_8e1f\_d0d0\_99a0\_eb28078cf679\_TB6988\_11 \cdot \vec{r} -11ead0bc\_8e1f\_d0d0\_99a0\_eb28078cf679\_TB6988\_11 \cdot \begin{pmatrix}11ead0bc\_8e1f\_82ac\_99a0\_5b738928e43a\_TB6988\_11\end{pmatrix}=0$
Therefore, the equation of the plane is:
$11ead0bc\_8e1f\_d0d0\_99a0\_eb28078cf679\_TB6988\_11 \cdot \vec{r}=11ead0bc\_8e1f\_d0d0\_99a0\_eb28078cf679\_TB6988\_11 \cdot \begin{pmatrix}11ead0bc\_8e1f\_82ac\_99a0\_5b738928e43a\_TB6988\_11\end{pmatrix}$.
The correct choice is (C)

Find the Equation of the Plane That Passes Through the Point