Find all vectors in $\Bbb{R}^3$ which are orthogonal to the plane

Question

Consider the vector space $\Bbb{R}^3$ with coordinates $(x_1, x_2, x_3)$ equipped with the inner product $$\langle(a_1, a_2, a_3),(b_1, b_2, b_3)\rangle= 2(a_1b_1 + a_2b_2 + a_3b_3) − (a_1b_2 + a_2b_1 + a_2b_3 + a_3b_2).$$

Write down all vectors in $\Bbb{R}^3$ which are orthogonal to the plane $x_1 − 2x_2 + 2x_3 = 0$ and have norm $1$... This is the original question

My attempt: I considered $A = \begin{bmatrix}2&2&2\\-1&-1&-1\end{bmatrix}\begin{bmatrix}1\\-2\\2\end{bmatrix} =0$. Now I cannot proceed further......

Please help me...

Answer 1

Note that your plane contains $(0,0,0)$, which is handy.

Find 2 linearly independent vectors in the plane, for example $$v_1=(0,1,1)$$ and $$v_2=(2,0,-1)$$ All we have to do is find a vector $v=(x_1,x_2,x_3)$ perpendicular to $v_1$ and $v_2$ $$\langle(x_1,x_2,x_3),v_1\rangle=0$$ $$\langle(x_1,x_2,x_3),v_2\rangle=0$$ Solve for $(x_1,x_2,x_3)$. Two equations in three unknowns gives you a one-dimensional space of vectors perpendicular to the plane.

Answer 2

Note that $$\begin{bmatrix}1\\-2\\2\end{bmatrix} $$ is not your normal vector to the plane any more .

You need to find the normal by first finding two linearly independent vectors in the plane, such as $(0,1,1)$ and $(-2,0,1)$ and find the normal vector perpendicular to these vectors with the new inner product.