# How do you find a unit vector perpendicular to a 3-D plane formed by points (1,01),(0,2,2) and (3,3,0)?

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Oct 17, 2016

$\left(- \frac{1}{\sqrt{3}} , \frac{1}{5 \sqrt{3}} , - \frac{7}{5 \sqrt{3}}\right)$

#### Explanation:

Our strategy will be to find two vectors in the plane, take their cross product to find a vector perpendicular to both of them (and thus to the plane), and then divide that vector by its measure to make it a unit vector.

Step 1) Find two vectors in the plane.

We will do this by finding the vector from $\left(1 , 0 , 1\right)$ to $\left(0 , 2 , 2\right)$ and from $\left(1 , 0 , 1\right)$ to $\left(3 , 3 , 0\right)$. As all three points are in the plane, so will each of those vectors.

$\vec{{v}_{1}} = \left(0 , 2 , 2\right) - \left(1 , 0 , 1\right) = \left(- 1 , 2 , 1\right)$

$\vec{{v}_{2}} = \left(3 , 3 , 0\right) - \left(1 , 0 , 1\right) = \left(2 , 3 , - 1\right)$

Step 2) Find a vector perpendicular to the plane.

If a vector is perpendicular to two vectors in a plane, it must be perpendicular to the plane itself. As the cross product of two vectors produces a vector perpendicular to both, we will use the cross product of $\vec{{v}_{1}}$ and $\vec{{v}_{2}}$ to find a vector $\vec{u}$ perpendicular to the plane containing them.

$\vec{u} = \vec{{v}_{1}} \times \vec{{v}_{2}}$

$= | \left(\hat{i} , \hat{j} , \hat{k}\right) , \left(- 1 , 2 , 1\right) , \left(2 , 3 , - 1\right) |$

$= \left(2 \left(- 1\right) - 1 \left(3\right)\right) \hat{i} - \left(\left(- 1\right) \left(- 1\right) - \left(1\right) \left(2\right)\right) \hat{j} + \left(\left(- 1\right) \left(3\right) - \left(2\right) \left(2\right)\right) \hat{k}$

$= - 5 \hat{i} + \hat{j} - 7 \hat{k}$

$= \left(- 5 , 1 , - 7\right)$

Step 3) Turn $\vec{u}$ into a unit vector.

A unit vector is a vector whose measure is $1$. Using the fact that for any vector $\vec{v}$ and scalar $c$, we have $| | c \vec{v} | | = c | | \vec{v} | |$, we will find $| | \vec{u} | | = u$, then divide by $u$.

$| | \frac{\vec{u}}{u} | | = | | \vec{u} | \frac{|}{u} = \frac{u}{u} = 1$

As multiplying by a scalar does not change the direction of a vector, this will be a unit vector perpendicular to the plane. Proceeding,

$| | \vec{u} | | = \sqrt{{\left(- 5\right)}^{2} + {1}^{2} + {\left(- 7\right)}^{2}} = \sqrt{75} = 5 \sqrt{3}$

Thus, our final result is

$\frac{\vec{u}}{u} = \frac{\left(- 5 \text{,"1",} - 7\right)}{5 \sqrt{3}} = \left(- \frac{1}{\sqrt{3}} , \frac{1}{5 \sqrt{3}} , - \frac{7}{5 \sqrt{3}}\right)$

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