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

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

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

3. ## Is this correct?

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 ??

4. 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 $5 \sqrt{3}$.

5. ## okay

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

The result should be then

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

6. Originally Posted by zorro
The result should be then

$\frac{-5i+5j+5k}{5 \sqrt{3}}$
Well it should be written as: $- \frac{{\sqrt 3 }}
{3}i + \frac{{\sqrt 3 }}
{3}j + \frac{{\sqrt 3 }}
{3}k$

7. ## Could u please explain

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

8. Originally Posted by zorro
Basic arithmetic. Cancel the common factor of 5.

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

10. 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 ....?)

11. 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??

12. 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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# finding unit vector perpendicular to plane

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