1
$\begingroup$

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

enter image description here

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...

$\endgroup$

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ thanks @wouter Sorry for late reply$$\langle(x_1,x_2,x_3),v_1\rangle=0 , 0+x_2 +x_3 =0$$ $$\langle(x_1,x_2,x_3),v_2\rangle=0, 2x_1 + 0 - x_3 =0$$ now i add and got $2x_1 =- x_2 $ as i can not able to proceed further...can u help me??? again $\endgroup$
    – user396850
    Apr 19, 2018 at 14:48
  • 1
    $\begingroup$ You will not find a unique solution, but a one-dimensional space of solutions. The space of solutions to your 2 equations in 3 unknowns is $(x_3/2,-x_3,x_3)$ $\endgroup$
    – Wouter
    Apr 19, 2018 at 16:36
  • $\begingroup$ that mean what is the answer of this (post) question ??@Wouter $\endgroup$
    – user396850
    Apr 19, 2018 at 16:48
  • $\begingroup$ Understand that you are asked to find all vectors orthogonal to the plane. You can answer "the 1D vector space uniquely specified by $x_2+x_3=0,2x_1-x_3=0$". You can answer "$(x/2,-x,x),x\in\mathbb{R}$". You can answer "all vectors parallel to $(1/2,-1,1)$".... all these answers are correct. $\endgroup$
    – Wouter
    Apr 19, 2018 at 18:12
  • 1
    $\begingroup$ You are not looking for a solution, you are looking for a space of solutions, and you have already found it. $\endgroup$
    – Wouter
    Apr 19, 2018 at 18:16
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ thanks @ Mohammad Riazikermani $$\langle(x_1,x_2,x_3),v_1\rangle=0 , 0+x_2 +x_3 =0$$ $$\langle(x_1,x_2,x_3),v_2\rangle=0, -2x_1 + 0 + x_3 =0$$ now i add and got $-2x_1 =- x_2 $ as i can not able to proceed further...can u help me??? again $\endgroup$
    – user396850
    Apr 19, 2018 at 14:50
  • $\begingroup$ @useruser396850 : Use definition of inner product given in the question and not the standard inner product. $\endgroup$ Nov 15, 2018 at 11:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy