Vector problem

**kandyfloss** · May 15th 2010, 12:02 AM

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.

**Prove It** · May 15th 2010, 12:49 AM

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

$ax + by + cz = d$ .

Substituting your points gives

$b + 2c = d$

$3a + 2b + c = d$

$a - b = d$ .

This gives the system

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

$R_3 - 3R_1 \to R_3$

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

$R_3 - 5R_2 \to R_3$

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

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

You can see $b + 2c = d$

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

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

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

You can see $a - b = d$

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

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

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

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

Therefore a correct equation of the plane is

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

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

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

To find the unit vector, divide by its length.

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

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

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

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

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

$= \frac{2\sqrt{10}}{15}\mathbf{i} - \frac{\sqrt{10}}{6}\mathbf{j} + \frac{7\sqrt{10}}{30}\mathbf{k}$ .

**Sudharaka** · May 15th 2010, 01:00 AM

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

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

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

Hope these ideas will help you to solve the problem.

**kandyfloss** · May 15th 2010, 01:04 AM

Wow! That was really helpful

..got a bit confused in the row reduction part,but its all clear now.
Thanks a lot

**kandyfloss** · May 15th 2010, 01:06 AM

Originally Posted by Sudharaka

Dear kandyfloss,

Another approch,

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

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

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

Hope these ideas will help you to solve the problem.

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

**Sudharaka** · May 15th 2010, 01:29 AM

Originally Posted by kandyfloss

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

Dear kandyfloss,

Sure it does. If you have any questions about this mehtod, please don't hesitate to ask me.

**kandyfloss** · May 15th 2010, 01:33 AM

yes i did get the right answer..i've just posted another vector problem,can you help me out?

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