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Question: Find a vector that is perpendicular to both $2\underset{\sim}{i}+\underset{\sim}{j}-\underset{\sim}{k}$ and $\underset{\sim}{i}-2\underset{\sim}{j}+\underset{\sim}{k}$.

This is for the 3D topic of the Extension 2 HSC maths course. This belongs to a section introducing the scaler product.

I have tried using creating simultaneous equations with it:

let $\underset{\sim}{u}=2\underset{\sim}{i}+\underset{\sim}{j}-\underset{\sim}{k}$ and $\underset{\sim}{v}=\underset{\sim}{i}-2\underset{\sim}{j}+\underset{\sim}{k}$

let $\underset{\sim}{w}$ be the vector that I'm trying to find, therefore $\underset{\sim}{w}=a\underset{\sim}{i}+b\underset{\sim}{j}+c\underset{\sim}{k}$.

  • $\underset{\sim}{u} \times \underset{\sim}{w}=2a+b-c=0$
  • $\underset{\sim}{v} \times \underset{\sim}{w}=a-2b+c=0$

but that doesn't seem to get me anywhere.

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Summing the two equations, we get $3a-b=0$ or $b=3a$. Substituting to the first equation to obtain $2a+3a-c=0$ or $c=5a$.

Therefore Your vector is $a(1 \hat{i} + 3 \hat{j} + 5 \hat{k})$

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  • $\begingroup$ Thank you so much! $\endgroup$ – Jieshun Wang Feb 28 at 8:37
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I'd advise you against using x as the scalar product of two vectors, for it usually denotes the cross product. Mainly a dot is written or is used. You got two linear equation and three parameters, hence you will get a 1 parameter family of solutions, which is fine, since if two vectors x,y are perpendicular then ax and ay are also perpendicular, where a,b are arbitrary scalars. So if you add the two bottom eqns you get $3a-b =0$, hence $3a=b$. Plugging back in the first one reults in $c=5a$. Choose e.g. $a=1$, thus the vector $(1,3,5)$ would be perp. to u and w.

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  • $\begingroup$ Thank you so much! $\endgroup$ – Jieshun Wang Feb 28 at 8:39

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