# Find a unit vector that is orthogonal to the plane

• Mar 17th 2010, 12:00 AM
nemixus
Find a unit vector that is orthogonal to the plane
I was given a vector A = [2 2 2] and a vector B = [4 6 5] from a plane and supposed to find a vector orthogonal to this plane. The vector has to intersect with the plane at [0 0 0]
and have length 1. I was given a hint that the dot product of these two vectors with the orthogonal vector should be zero. How do I do that?

Also, can the desired vector be generated by linear transformation?

• Mar 17th 2010, 12:54 AM
Prove It
Quote:

Originally Posted by nemixus
I was given a vector A = [2 2 2] and a vector B = [4 6 5] from a plane and supposed to find a vector orthogonal to this plane. The vector has to intersect with the plane at [0 0 0]
and have length 1. I was given a hint that the dot product of these two vectors with the orthogonal vector should be zero. How do I do that?

Also, can the desired vector be generated by linear transformation?

If you take the cross product of the two vectors, you will get a vector that is orthogonal to both of the vectors, and as such, is also orthogonal to the plane they lie in.

So $A \times B = \left|\begin{matrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 2 & 2 & 2 \\ 4 & 6 & 5\end{matrix}\right|$

$= \mathbf{i}\left|\begin{matrix}2&2\\6&5\end{matrix} \right| - \mathbf{j}\left|\begin{matrix}2&2\\4&5\end{matrix} \right| + \mathbf{k}\left|\begin{matrix}2&2\\4&6\end{matrix} \right|$

$= \mathbf{i}(2\cdot 5 - 2\cdot 6) - \mathbf{j}(2\cdot 5 - 2\cdot 4) + \mathbf{k}(2\cdot 6 - 2\cdot 4)$

$= -2\mathbf{i} - 2\mathbf{j} + 4\mathbf{j}$.

Now, if you were to take the dot product of this new vector with either $A$ or $B$, you should get 0.

You need a UNIT vector in this direction. So divide it by its length.

$|A\times B| = \sqrt{(-2)^2 + (-2)^2 + 4^2}$

$= \sqrt{4 + 4 + 16}$

$= \sqrt{24}$

$= 2\sqrt{6}$.

So the unit vector is

$\frac{A\times B}{|A \times B|} = \frac{-2\mathbf{i} - 2\mathbf{j} + 4\mathbf{j}}{2\sqrt{6}}$

$= -\frac{1}{\sqrt{6}}\mathbf{i} - \frac{1}{\sqrt{6}}\mathbf{j} + \frac{2}{\sqrt{6}}\mathbf{k}$

$= -\frac{\sqrt{6}}{6}\mathbf{i} - \frac{\sqrt{6}}{6}\mathbf{j} + \frac{\sqrt{6}}{3}\mathbf{k}$.
• Mar 17th 2010, 01:12 AM
earboth
Quote:

Originally Posted by nemixus
I was given a vector A = [2 2 2] and a vector B = [4 6 5] from a plane and supposed to find a vector orthogonal to this plane. The vector has to intersect with the plane at [0 0 0]
and have length 1. I was given a hint that the dot product of these two vectors with the orthogonal vector should be zero. How do I do that?

Also, can the desired vector be generated by linear transformation?

Let $\vec n = (n_1, n_2, n_3)$ denote the orthogonal vector to the plane.

Then according to the hint:

$2n_1+2n_2+2n_3=0$
$4n_1+6n_2+5n_3=0$

Multiply the 1st equation by 2 and subtract it from the 2nd one:

$2n_2+n_3=0~\implies~n_3=-2n_2$

Plug in this term into one of the original equations and you'll get:

$n_1=n_2$

Set $n_2 = t$ and you'll have:

$\vec n=\left\{\begin{array}{l}n_1=t\\n_2=t\\n_3=-2t\end{array}\right.$

Therefore $\vec n = (1,1,-2)$

Use $\overrightarrow{n^0} = \dfrac{\vec n}{|\vec n|}$ to calculate the components of the unit vector.
• Mar 18th 2010, 03:16 PM
nemixus
Find a unit vector that is orthogonal to the plane by Linear Transformation
Quote:

Originally Posted by nemixus
I was given a vector A = [2 2 2] and a vector B = [4 6 5] from a plane and supposed to find a vector orthogonal to this plane. The vector has to intersect with the plane at [0 0 0]
and have length 1. I was given a hint that the dot product of these two vectors with the orthogonal vector should be zero. How do I do that?

Also, can the desired vector be generated by linear transformation?

Can the desired vector be generated by linear transformation?