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

• Oct 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 $\bar{b} = 2\hat{i} + 3\hat{j} - \hat{k}$
• Oct 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: $\overrightarrow{u_n}=\dfrac{\vec a \times \vec b}{|\vec a \times \vec b|}$
• Dec 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 ??
• Dec 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 $5 \sqrt{3}$.
• Dec 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

$\frac{-5i+5j+5k}{5 \sqrt{3}}$
• Dec 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: $- \frac{{\sqrt 3 }}
{3}i + \frac{{\sqrt 3 }}
{3}j + \frac{{\sqrt 3 }}
{3}k$
• Dec 22nd 2009, 08:48 PM
zorro
Quote:

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

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

Originally Posted by zorro

Basic arithmetic. Cancel the common factor of 5.
• Dec 23rd 2009, 01:15 AM
zorro
how did u get $\sqrt{3}$ at the denominator
• Dec 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 ....?)
• Dec 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??
• Dec 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.