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Find a unit vector that is parallel to both the plane $8x+y+z = 1$ and the plane $x-y-z=0$.

I found the normal vectors to be: $(8,1,1)$ and $(1,-1,-1)$

I took the cross product.

$(8,1,1)\times(1,-1,-1) = \begin{vmatrix} i & j & k \\ 8 & 1 & 1 \\ 1 & -1 & -1 \end{vmatrix}$ = $(2,-7,-9)$ right?

I was checking if $(2,-7,-9)$ was orthogonal to my two normal vectors and found it was not orthogonal to $(1,-1,-1)$.

dfeuer
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Ozera
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1 Answers1

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It's a simple miscalculation. $(8,1,1)\times(1,-1,-1) = (0,9,-9)$, and normalized $$\frac{1}{\sqrt{2}}(0,1,-1).$$

Jonathan Y.
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  • $(0,9,-9)$? My process is: $i(1-1 - 1-1) - j(8-1 -9) +k(8-1 - 1*1)$ which gives... $0i +9j - 9k$. Ah. However , what did you mean by "and normalized..." though? – Ozera Sep 02 '13 at 06:05
  • Since we were asked for a unit vector, we ought to normalize the vector we get. That leaves the result unique up to direction (a factor of $(-1)$). – Jonathan Y. Sep 02 '13 at 06:08
  • I thought the unit vector of a vector $a$ was defined as: $\vec{a} / ||a||$ which would explain how we get $2^{1/2}$, but $1$ in the numerator? – Ozera Sep 02 '13 at 06:13
  • Just another way of writing $(0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$ (a scalar multiplied with a vector). – Jonathan Y. Sep 02 '13 at 06:23
  • But.. $( 0, \frac{9}{\sqrt(2)}, \frac{-9}{\sqrt(2)} )$ is not equivalent to what you said. – Ozera Sep 02 '13 at 06:25
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    But $(0,\frac{9}{9\sqrt{2}},-\frac{9}{9\sqrt{2}})$ does equal it. Alternatively, you can verify that the given vector has unit norm and is parallel to both planes (hence, unique up to direction). – Jonathan Y. Sep 02 '13 at 06:30