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I need orthogonal basis for R3. I am given v1 = (1,1,1), so I need so I need other two vectors in this basis but how do i find the other two?

at first i thought i would use gram schmidt but that doesn't seem plausible with just one vector.

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  • $\begingroup$ Select two other vectors, so that the three are independent. Then G-S them. $\endgroup$ – David Mitra Mar 14 '14 at 11:27
  • $\begingroup$ You can start by finding two linearly indepedent vectors orthogonal to $v_1$ by solving a system. Then you'll have a basis of $\mathbb R^3$ and you can apply G-S. Or you can just 'guess' that with $v_2=(0,1,0)$ and $v_3=(0,0,1)$, the set $\{v_1, v_2, v_3\}$ is linearly independent and you can apply G-S. $\endgroup$ – Git Gud Mar 14 '14 at 11:27
  • $\begingroup$ Gram-Schmidt really is the way you'd want to go about this (because it works in any dimension), but since we are in $\mathbb{R}^3$ there is also a funny and simple alternative: take any non-zero vector orthogonal to $(1,1,1)$ (this can be found very easily) and then simply take the cross product of the two vectors. Pairwise orthogonality gives you, that the resulting set is linearly independent. $\endgroup$ – walcher Mar 14 '14 at 11:41
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Pick any vector $\mathbf u$ that is not a multiple of $\mathbf v_1$.
Set $\mathbf v_2 = \mathbf v_1 \times \mathbf u$.
Set $\mathbf v_3 = \mathbf v_1 \times \mathbf v_2$.

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Take for example:

$(1,-1,0)$

and

$(1,1,-2)$

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  • $\begingroup$ Maybe you should some more details, so the OP can better understand your solution $\endgroup$ – M Turgeon Mar 14 '14 at 11:59
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    $\begingroup$ It might be useful to explain how you got those vectors :) For the OPs benefit: for the first vector, we can find a vector in the plane orthogonal to (a,b,c) by selecting (b,-a,0) (take their dot product to see this), so we get (1,-1,0). For the third vector, take the cross-product of the two you now have; that gives you a vector orthogonal to the first two (i.e. (1,1,-2) here). $\endgroup$ – postmortes Mar 14 '14 at 11:59
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The "standard" way is to find two more vectors then use Gram-Schmidt, as suggested by others. For a short cut using trial and (not much) error, take $(1,-1,0)$ which is orthogonal to your first vector. Then $(1,1,a)$. This is orthogonal to the second vector, and also to the first if you choose $a$ suitably. I expect this is how Yiorgos obtained his answer.

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Just pick two more vectors arbitrarily and apply Gram-Schmidt. Gram-Schmidt preserves the direction of the first input vector, so you'll get a normalized copy of $(1, 1, 1)$.

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  • $\begingroup$ Not arbitrarily. G-S requires the input to be a linearly independent set. $\endgroup$ – Git Gud Mar 14 '14 at 11:37

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