For what $t$ will the following vector be an orthogonal basis? \begin{align}u_1&= (1,t,t)\\ u_2&= (2t,t+1,2t-1)\\ u_3&= (2-2t,t-1,1)\end{align}

Till now I have tried using the Gram-Schmidt process but did not really reach anywhere. Can you please provide a hint or some theory that may help me get the solution for this question?

  • $\begingroup$ Use the dot product. $\endgroup$ – John Douma Dec 22 '18 at 22:16
  • $\begingroup$ Welcome to StackExhange, I have formatted your question using MathJax $\endgroup$ – lioness99a Dec 22 '18 at 22:36
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    $\begingroup$ Your title asks a different question than the body of your post. Are you looking for a value of $t$ that make the $u_i$s into an orthogonal set or are you looking for an orthogonal basis that spans the same set as the $u_i$s? $\endgroup$ – John Douma Dec 22 '18 at 23:27

You can easily see that whatever be the value of $t$, we have $\mathbf{u_3}=\mathbf{2u_1}-\mathbf{u_2}$. Therefore, $\text{span}\bf\{u_1,u_2,u_3\}$$=\text{span}\bf\{u_1,u_2\}$ and we only need to perform orthogonalization for $\bf u_1,u_2$.

Using the Gram-Schmidt process, we have $B=\bf\{v_1,v_2\}$, where:

  • $\mathbf{v_1}=\mathbf{u_1}=(1,t,t)$
  • $\displaystyle\mathbf{v_2}=\mathbf{u_2}-\frac{\langle\mathbf{u_1},\mathbf{u_2}\rangle}{\langle\mathbf{u_1},\mathbf{u_1}\rangle}\cdot\mathbf{u_1}=(2t,t+1,2t-1)-\Big[\frac{3t^2+2t}{1+2t^2}\Big](1,t,t)\\\displaystyle=\Big(\frac{4t^3-3t^2}{1+2t^2},\frac{-t^3+t+1}{1+2t^2},\frac{t^3-4t^2+2t-1}{1+2t^2}\Big)$
  • $\begingroup$ "can easily see" -> put the vectors in a matrix, determine the rank $\endgroup$ – G Cab Dec 22 '18 at 23:32

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