# Find the unit vector perpendicular to the plane of 2 vector

• October 24th 2009, 11:28 PM
zorro
Find the unit vector perpendicular to the plane of 2 vector
Question : Find the unit vector perpendicular to the plane of two vectors $\bar{a}$ and $\bar{b}$ ,

$\bar{a} = \hat{i} - \hat{j} + 2\hat{k}$ and $\bar{b} = 2\hat{i} + 3\hat{j} - \hat{k}$
• October 24th 2009, 11:34 PM
earboth
Quote:

Originally Posted by zorro
Question : Find the unit vector perpendicular to the plane of two vectors $\bar{a}$ and $\bar{b}$ ,

$\bar{a} = \hat{i} - \hat{j} + 2\hat{k}$ and $\bar{b} = 2\hat{i} + 3\hat{j} - \hat{k}$

Use the formula: $\overrightarrow{u_n}=\dfrac{\vec a \times \vec b}{|\vec a \times \vec b|}$
• December 22nd 2009, 01:06 AM
zorro
Is this correct?
Quote:

Originally Posted by earboth
Use the formula: $\overrightarrow{u_n}=\dfrac{\vec a \times \vec b}{|\vec a \times \vec b|}$

$| \bar a \times \bar b|$ = $\begin{vmatrix}i & j & k \\ 1 & -1 & 2 \\ 2 & 3 & -1 \end{vmatrix}$ = $-5i + 5j +5k$

$| \bar a \times \bar b|$ = $\sqrt{ (- 5)^2 + 5^2 + 5^2 }$ = $\sqrt{75}$ = $5 \sqrt{3}$

$u_n$ = $\frac{\bar a \times \bar b}{|\bar a \times \bar b|}$ = $-i + j + k$..............Is this correct ??
• December 22nd 2009, 02:24 AM
mr fantastic
Quote:

Originally Posted by zorro
$| \bar a \times \bar b|$ = $\begin{vmatrix}i & j & k \\ 1 & -1 & 2 \\ 2 & 3 & -1 \end{vmatrix}$ = $-5i + 5j +5k$ Mr F says: Correct.

$| \bar a \times \bar b|$ = $\sqrt{ (- 5)^2 + 5^2 + 5^2 }$ = $\sqrt{75}$ = $5 \sqrt{3}$ Mr F says: Correct.

$u_n$ = $\frac{\bar a \times \bar b}{|\bar a \times \bar b|}$ = $-i + j + k$..............Is this correct ?? Mr F says: No.

You divided by 5. You're meant to divide by $5 \sqrt{3}$.
• December 22nd 2009, 12:58 PM
zorro
okay
Quote:

Originally Posted by mr fantastic
You divided by 5. You're meant to divide by $5 \sqrt{3}$.

The result should be then

$\frac{-5i+5j+5k}{5 \sqrt{3}}$
• December 22nd 2009, 01:14 PM
Plato
Quote:

Originally Posted by zorro
The result should be then

$\frac{-5i+5j+5k}{5 \sqrt{3}}$

Well it should be written as: $- \frac{{\sqrt 3 }}
{3}i + \frac{{\sqrt 3 }}
{3}j + \frac{{\sqrt 3 }}
{3}k$
• December 22nd 2009, 08:48 PM
zorro
Quote:

Originally Posted by Plato
Well it should be written as: $- \frac{{\sqrt 3 }}
{3}i + \frac{{\sqrt 3 }}
{3}j + \frac{{\sqrt 3 }}
{3}k$

• December 23rd 2009, 01:13 AM
mr fantastic
Quote:

Originally Posted by zorro

Basic arithmetic. Cancel the common factor of 5.
• December 23rd 2009, 01:15 AM
zorro
how did u get $\sqrt{3}$ at the denominator
• December 23rd 2009, 01:18 AM
mr fantastic
Quote:

Originally Posted by zorro
how did u get $\sqrt{3}$ at the denominator

$\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ (you must have been taught how to rationalise the denominator of surds ....?)
• December 23rd 2009, 01:26 AM
zorro
Quote:

Originally Posted by mr fantastic
$\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ (you must have been taught how to rationalise the denominator of surds ....?)

okay but is that neccessary ..........my answer is correct isnt it??
• December 23rd 2009, 01:30 AM
mr fantastic
Quote:

Originally Posted by zorro
okay but is that neccessary ..........my answer is correct isnt it??

Yes, it is. But whether or not it got full marks would depend on whether the person marking it wanted a simplified answer. Certainly at this level the common factor of 5 should be cancelled. It is personal taste whether or not to rationalise the denominator. You should discuss this further with your instructor.