# Vector problem

• May 15th 2010, 12:02 AM
kandyfloss
Vector problem
I'm sorry if i'm posting this in the wrong section,i couldn't figure out which section vectors would go in to.

Here is the question :-
A,B and C are the point (0,1,2),(3,2,1) and (1,-1,0) respectively.Find the unit vector perpendicular to the plane ABC.
I've just started learning this topic,and i'm not very strong at it.It'd be great if someone can help me out here.
Thanks.
• May 15th 2010, 12:49 AM
Prove It
Quote:

Originally Posted by kandyfloss
I'm sorry if i'm posting this in the wrong section,i couldn't figure out which section vectors would go in to.

Here is the question :-
A,B and C are the point (0,1,2),(3,2,1) and (1,-1,0) respectively.Find the unit vector perpendicular to the plane ABC.
I've just started learning this topic,and i'm not very strong at it.It'd be great if someone can help me out here.
Thanks.

A plane has equation

$\displaystyle ax + by + cz = d$.

$\displaystyle b + 2c = d$

$\displaystyle 3a + 2b + c = d$

$\displaystyle a - b = d$.

This gives the system

$\displaystyle \left[\begin{matrix}1&-1&\phantom{-}0\\0&\phantom{-}1&\phantom{-}2\\3&\phantom{-}2&\phantom{-}1\end{matrix}\right]\left[\begin{matrix}a\\b\\c\end{matrix}\right] = \left[\begin{matrix}d\\d\\d\end{matrix}\right]$ (I switched order of the rows)

$\displaystyle R_3 - 3R_1 \to R_3$

$\displaystyle \left[\begin{matrix}1&-1&\phantom{-}0\\0&\phantom{-}1&\phantom{-}2\\0&\phantom{-}5&\phantom{-}1\end{matrix}\right]\left[\begin{matrix}a\\b\\c\end{matrix}\right] = \left[\begin{matrix}\phantom{-}d\\\phantom{-}d\\-2d\end{matrix}\right]$

$\displaystyle R_3 - 5R_2 \to R_3$

$\displaystyle \left[\begin{matrix}1&-1&\phantom{-}0\\0&\phantom{-}1&\phantom{-}2\\0&\phantom{-}0&-9\end{matrix}\right]\left[\begin{matrix}a\\b\\c\end{matrix}\right] = \left[\begin{matrix}\phantom{-}d\\\phantom{-}d\\-7d\end{matrix}\right]$

You can see $\displaystyle -9c = -7d$ so $\displaystyle c = \frac{7}{9}d$.

You can see $\displaystyle b + 2c = d$

$\displaystyle b + 2\left(\frac{7}{9}d\right) = d$

$\displaystyle b + \frac{14}{9}d = d$

$\displaystyle b = -\frac{5}{9}d$.

You can see $\displaystyle a - b = d$

$\displaystyle a - \left(-\frac{5}{9}d\right) = d$

$\displaystyle a + \frac{5}{9}d = d$

$\displaystyle a = \frac{4}{9}d$.

$\displaystyle d$ is a free variable, you can let it be whatever you like to get a consistent solution set. So why not let $\displaystyle d = 9$.

Therefore a correct equation of the plane is

$\displaystyle 4x - 5y + 7z = 9$.

The vectors normal to the plane has the same coefficients as the coefficients of the plane itself.

Therefore a vector normal to the plane is

$\displaystyle \mathbf{n} = 4\mathbf{i} - 5\mathbf{j} + 7\mathbf{k}$.

To find the unit vector, divide by its length.

$\displaystyle \frac{\mathbf{n}}{|\mathbf{n}|} = \frac{4\mathbf{i} - 5\mathbf{j} + 7\mathbf{k}}{\sqrt{4^2 + (-5)^2 + 7^2}}$

$\displaystyle = \frac{4\mathbf{i} - 5\mathbf{j} + 7\mathbf{k}}{\sqrt{16 + 25 + 49}}$

$\displaystyle = \frac{4\mathbf{i} - 5\mathbf{j} + 7\mathbf{k}}{\sqrt{90}}$

$\displaystyle = \frac{4\mathbf{i} - 5\mathbf{j} + 7\mathbf{k}}{3\sqrt{10}}$

$\displaystyle = \frac{4\sqrt{10}}{30}\mathbf{i} - \frac{5\sqrt{10}}{30}\mathbf{j} + \frac{7\sqrt{10}}{30}\mathbf{k}$

$\displaystyle = \frac{2\sqrt{10}}{15}\mathbf{i} - \frac{\sqrt{10}}{6}\mathbf{j} + \frac{7\sqrt{10}}{30}\mathbf{k}$.
• May 15th 2010, 01:00 AM
Sudharaka
Quote:

Originally Posted by kandyfloss
I'm sorry if i'm posting this in the wrong section,i couldn't figure out which section vectors would go in to.

Here is the question :-
A,B and C are the point (0,1,2),(3,2,1) and (1,-1,0) respectively.Find the unit vector perpendicular to the plane ABC.
I've just started learning this topic,and i'm not very strong at it.It'd be great if someone can help me out here.
Thanks.

Dear kandyfloss,

Another approch,

Find $\displaystyle \overline{AB}$ and $\displaystyle \overline{CB}$

Now, $\displaystyle \overline{AB}\times\overline{CB}$ is a perpendicular vector to the plane. (by the definition of cross product.)

Hence $\displaystyle \frac{\overline{AB}\times\overline{CB}}{\mid\overl ine{AB}\times\overline{CB}\mid}$ is a unit perpenducular vector to the plane.

• May 15th 2010, 01:04 AM
kandyfloss
Wow! That was really helpful (Clapping)..got a bit confused in the row reduction part,but its all clear now.
Thanks a lot :)
• May 15th 2010, 01:06 AM
kandyfloss
Quote:

Originally Posted by Sudharaka
Dear kandyfloss,

Another approch,

Find $\displaystyle \overline{AB}$ and $\displaystyle \overline{CB}$

Now, $\displaystyle \overline{AB}\times\overline{CB}$ is a perpendicular vector to the plane. (by the definition of cross product.)

Hence $\displaystyle \frac{\overline{AB}\times\overline{CB}}{\mid\overl ine{AB}\times\overline{CB}\mid}$ is a unit perpenducular vector to the plane.

Hey you method looks quite easy..lemme try it out,if i get it right.Thanks.
• May 15th 2010, 01:29 AM
Sudharaka
Quote:

Originally Posted by kandyfloss
Hey you method looks quite easy..lemme try it out,if i get it right.Thanks.

Dear kandyfloss,