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

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

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- Oct 24th 2009, 11:28 PMzorroFind the unit vector perpendicular to the plane of 2 vector
Question : Find the unit vector perpendicular to the plane of two vectors $\displaystyle \bar{a}$ and $\displaystyle \bar{b}$ ,

$\displaystyle \bar{a} = \hat{i} - \hat{j} + 2\hat{k}$ and $\displaystyle \bar{b} = 2\hat{i} + 3\hat{j} - \hat{k}$ - Oct 24th 2009, 11:34 PMearboth
- Dec 22nd 2009, 01:06 AMzorroIs this correct?

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

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

$\displaystyle u_n$ = $\displaystyle \frac{\bar a \times \bar b}{|\bar a \times \bar b|}$ = $\displaystyle -i + j + k$..............Is this correct ?? - Dec 22nd 2009, 02:24 AMmr fantastic
- Dec 22nd 2009, 12:58 PMzorrookay
- Dec 22nd 2009, 01:14 PMPlato
- Dec 22nd 2009, 08:48 PMzorroCould u please explain
- Dec 23rd 2009, 01:13 AMmr fantastic
- Dec 23rd 2009, 01:15 AMzorro
how did u get $\displaystyle \sqrt{3}$ at the denominator

- Dec 23rd 2009, 01:18 AMmr fantastic
- Dec 23rd 2009, 01:26 AMzorro
- Dec 23rd 2009, 01:30 AMmr fantastic
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.