Find the unit vector perpendicular to the plane of 2 vector

**zorro** · October 24th 2009, 11:28 PM

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

**earboth** · October 24th 2009, 11:34 PM

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: $\overrightarrow{u_n}=\dfrac{\vec a \times \vec b}{|\vec a \times \vec b|}$

**zorro** · December 22nd 2009, 01:06 AM

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 ??

**mr fantastic** · December 22nd 2009, 02:24 AM

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 $5 \sqrt{3}$ .

**zorro** · December 22nd 2009, 12:58 PM

Originally Posted by mr fantastic

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

The result should be then

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

**Plato** · December 22nd 2009, 01:14 PM

Originally Posted by zorro

The result should be then

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

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

**zorro** · December 22nd 2009, 08:48 PM

Originally Posted by Plato

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

Could u please explain how u got this answer.......

**mr fantastic** · December 23rd 2009, 01:13 AM

Originally Posted by zorro

Could u please explain how u got this answer.......

Basic arithmetic. Cancel the common factor of 5.

**zorro** · December 23rd 2009, 01:15 AM

how did u get $\sqrt{3}$ at the denominator

**mr fantastic** · December 23rd 2009, 01:18 AM

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 ....?)

**zorro** · December 23rd 2009, 01:26 AM

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??

**mr fantastic** · December 23rd 2009, 01:30 AM

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.

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Find the unit vector perpendicular to the plane of 2 vector

Is this correct?

okay

Could u please explain

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