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Considering $\mathbb{R}$ with the inner product $$\langle(a_1,a_2,a_3),(b_1,b_2,b_3)\rangle=2(a_1b_1+a_2b_2+a_3b_3)-(a_1b_2+a_2b_1+a_2b_3+a_3b_2)$$

Then, how could we find the set of vectors orthogonal(with respect to the above inner product) to the plane given by the equation $x_1-2x_2+2x_3=0$?

I think we have to rewrite the equation of plane with respect to the above inner product. Any hints Thanks beforehand.

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Let $(a,b,c)$ be an orthogonal vector.

Thus, $$k(x_1-2x_2+2x_3)=2(ax_1+bx_2+cx_3)-(ax_2+bx_1+bx_3+cx_2)=$$ $$=(2a-b)x_1+(2b-a-c)x_2+(2c-b)x_3,$$ which gives $$2a-b=k,$$ $$2b-a-c=-2k$$ and $$2c-b=2k,$$ which gives $$(a,b,c)||(1,-2,3).$$

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  • $\begingroup$ so the vectors parallel in one basis are also parallel in any other basis, is this what your answer says? $\endgroup$
    – vidyarthi
    Apr 9, 2019 at 11:46
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    $\begingroup$ @vidyarth $(a,b,c)$ this is a normal to the plane respect to our inner product. $\endgroup$
    – Michael Rozenberg
    Apr 9, 2019 at 12:05

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