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Determine the components of a unit vector perpendicular to (0, 3, -5) and (2, 3, 1). I think I should be using either cross or dot product, but am unsure on what to do from there.

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First calculate the determinant $$ \left| \begin{array}{ccc} i & j & k \\ 0 & 3& -5 \\ 2 & 3 & 1 \end{array} \right| $$ You get a vector perpendicular to both given vectors then normalize it by dividing the components by the length of it.

Edit: See this link from wikipedia.

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  • $\begingroup$ The determinant is a number, not a vector. $\endgroup$ – Clement C. May 21 '14 at 22:02
  • $\begingroup$ No. A determinant is not always a number and depends on the entries. Here the determinant involves $i, j$ and $k$. Therefore it will be of the form $ai+bj+ck$ $\endgroup$ – Fermat May 22 '14 at 5:58
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    $\begingroup$ Oh, saw the link you added. I missed the last step, with the coefficients of the three variables of the polynomial giving the coordinates. $\endgroup$ – Clement C. May 22 '14 at 10:26
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"In mathematics, the cross product or vector product is a binary operation on two vectors in three-dimensional space. It results in a vector which is perpendicular to both and therefore normal to the plane containing them."

http://en.wikipedia.org/wiki/Cross_product

(furthermore, once you have the vector, you need to unitize it - http://en.wikipedia.org/wiki/Unit_vector)

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  • $\begingroup$ (and so $(0,3,-5) \times (2,3,1)=(18,-10,-6)$) $\endgroup$ – rbm May 21 '14 at 21:45
  • $\begingroup$ One more link for you $\endgroup$ – rbm May 21 '14 at 21:59
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Using the dot product: let $u=(a,b,c)\in\mathbb{R}^3$ be such a vector, and write $x=(0,3,-5)$, $y=(2,3,1)$. Then, the orthogonality conditions mean that $$ \begin{align*} 0 &= \langle u,x\rangle = 0\cdot a + 3\cdot b+(-5)\cdot c = 3b-5c \\ 0 &= \langle u,y\rangle = 2\cdot a + 3\cdot b+1\cdot c = 2a+3b+c \end{align*} $$ and the unit vector condition further imposes $$ \begin{align*} 1 &= a^2+b^2+c^2 \end{align*} $$ You thus have 3 constraints (two of them linear), and 3 unknowns. This will be enough to find solutions for $u$ (there is no unicity, as if $u$ is solution, so is $-u$, for instance).

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