# How do I make 3 given vectors with an unknown value $t$ in it, into an orthogonal set?

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?

• Use the dot product. – John Douma Dec 22 '18 at 22:16
• Welcome to StackExhange, I have formatted your question using MathJax – lioness99a Dec 22 '18 at 22:36
• 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? – 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:
• $$\mathbf{v_1}=\mathbf{u_1}=(1,t,t)$$
• $$\displaystyle\mathbf{v_2}=\mathbf{u_2}-\frac{\langle\mathbf{u_1},\mathbf{u_2}\rangle}{\langle\mathbf{u_1},\mathbf{u_1}\rangle}\cdot\mathbf{u_1}=(2t,t+1,2t-1)-\Big[\frac{3t^2+2t}{1+2t^2}\Big](1,t,t)\\\displaystyle=\Big(\frac{4t^3-3t^2}{1+2t^2},\frac{-t^3+t+1}{1+2t^2},\frac{t^3-4t^2+2t-1}{1+2t^2}\Big)$$