Animated Solution for Mathematics - Vector Algebra: A unit vector perpendicular to the plane determined by the points $P(1,-1,2)$, $Q(2,0,-1)$ and $R(0,2,1)$ is .........

To define the orientation of our plane, we need two vectors that lie flat on its surface. We can create these by connecting our given points: $P(1,-1,2)$, $Q(2,0,-1)$, and $R(0,2,1)$.

Let us pick point $Q$ as our anchor. We create vector $\overrightarrow{QP}$ by subtracting the position vector of $Q$ from $P$, and vector $\overrightarrow{QR}$ by subtracting $Q$ from $R$.

For $\overrightarrow{QP}$, we calculate:

For $\overrightarrow{QR}$, we calculate:

These two vectors, $\overrightarrow{QP}$ and $\overrightarrow{QR}$, now act as our foundation, spanning the plane.

To find a vector perpendicular to both $\overrightarrow{QP}$ and $\overrightarrow{QR}$, we utilize the cross product. We define our normal vector $\vec{n}$ as $\vec{n}=\overrightarrow{QP}\times \overrightarrow{QR}$.

We set this up as a determinant:

Expanding this determinant, we calculate the components:

For the $\hat{i}$ component:

For the $\hat{j}$ component (remembering the negative sign):

For the $\hat{k}$ component:

Thus, our normal vector is $\vec{n}=-8\hat{i}-4\hat{j}-4\hat{k}$.

To make this a unit vector, we must scale it down to a length of one. First, we find the magnitude of $\vec{n}$:

Since $\sqrt{96}=\sqrt{16\times 6}=4\sqrt{6}$, our magnitude is $4\sqrt{6}$. Now, we divide our vector $\vec{n}$ by this magnitude to find the unit vector $\hat{n}$:

Simplifying this by dividing each component by 4, we obtain:

We can absorb the negative sign into the $\pm$ to write the final answer as:

This vector is the mathematical embodiment of the direction perpendicular to your plane. You have successfully navigated the 3D landscape and pinned down the orientation of the surface.