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












0














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?






linear-algebra vector-spaces vectors gram-schmidt







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  • 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






  • 1




    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


















0














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?






linear-algebra vector-spaces vectors gram-schmidt







share|cite|improve 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






  • 1




    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
















0












0








0







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?






linear-algebra vector-spaces vectors gram-schmidt







share|cite|improve this question















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?






linear-algebra vector-spaces vectors gram-schmidt




linear-algebra vector-spaces vectors gram-schmidt






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 26 '18 at 19:03

























asked Dec 22 '18 at 21:36









ankit vijay

11




11












  • 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






  • 1




    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




















  • 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






  • 1




    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


















Use the dot product.
– John Douma
Dec 22 '18 at 22:16




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




Welcome to StackExhange, I have formatted your question using MathJax
– lioness99a
Dec 22 '18 at 22:36




1




1




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






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












1 Answer
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1














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:




  • $mathbf{v_1}=mathbf{u_1}=(1,t,t)$

  • $displaystylemathbf{v_2}=mathbf{u_2}-frac{langlemathbf{u_1},mathbf{u_2}rangle}{langlemathbf{u_1},mathbf{u_1}rangle}cdotmathbf{u_1}=(2t,t+1,2t-1)-Big[frac{3t^2+2t}{1+2t^2}Big](1,t,t)\displaystyle=Big(frac{4t^3-3t^2}{1+2t^2},frac{-t^3+t+1}{1+2t^2},frac{t^3-4t^2+2t-1}{1+2t^2}Big)$






share|cite|improve this answer





















  • "can easily see" -> put the vectors in a matrix, determine the rank
    – G Cab
    Dec 22 '18 at 23:32











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1 Answer
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1 Answer
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oldest

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active

oldest

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active

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1














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:




  • $mathbf{v_1}=mathbf{u_1}=(1,t,t)$

  • $displaystylemathbf{v_2}=mathbf{u_2}-frac{langlemathbf{u_1},mathbf{u_2}rangle}{langlemathbf{u_1},mathbf{u_1}rangle}cdotmathbf{u_1}=(2t,t+1,2t-1)-Big[frac{3t^2+2t}{1+2t^2}Big](1,t,t)\displaystyle=Big(frac{4t^3-3t^2}{1+2t^2},frac{-t^3+t+1}{1+2t^2},frac{t^3-4t^2+2t-1}{1+2t^2}Big)$






share|cite|improve this answer





















  • "can easily see" -> put the vectors in a matrix, determine the rank
    – G Cab
    Dec 22 '18 at 23:32
















1














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:




  • $mathbf{v_1}=mathbf{u_1}=(1,t,t)$

  • $displaystylemathbf{v_2}=mathbf{u_2}-frac{langlemathbf{u_1},mathbf{u_2}rangle}{langlemathbf{u_1},mathbf{u_1}rangle}cdotmathbf{u_1}=(2t,t+1,2t-1)-Big[frac{3t^2+2t}{1+2t^2}Big](1,t,t)\displaystyle=Big(frac{4t^3-3t^2}{1+2t^2},frac{-t^3+t+1}{1+2t^2},frac{t^3-4t^2+2t-1}{1+2t^2}Big)$






share|cite|improve this answer





















  • "can easily see" -> put the vectors in a matrix, determine the rank
    – G Cab
    Dec 22 '18 at 23:32














1












1








1






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:




  • $mathbf{v_1}=mathbf{u_1}=(1,t,t)$

  • $displaystylemathbf{v_2}=mathbf{u_2}-frac{langlemathbf{u_1},mathbf{u_2}rangle}{langlemathbf{u_1},mathbf{u_1}rangle}cdotmathbf{u_1}=(2t,t+1,2t-1)-Big[frac{3t^2+2t}{1+2t^2}Big](1,t,t)\displaystyle=Big(frac{4t^3-3t^2}{1+2t^2},frac{-t^3+t+1}{1+2t^2},frac{t^3-4t^2+2t-1}{1+2t^2}Big)$






share|cite|improve this answer












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:




  • $mathbf{v_1}=mathbf{u_1}=(1,t,t)$

  • $displaystylemathbf{v_2}=mathbf{u_2}-frac{langlemathbf{u_1},mathbf{u_2}rangle}{langlemathbf{u_1},mathbf{u_1}rangle}cdotmathbf{u_1}=(2t,t+1,2t-1)-Big[frac{3t^2+2t}{1+2t^2}Big](1,t,t)\displaystyle=Big(frac{4t^3-3t^2}{1+2t^2},frac{-t^3+t+1}{1+2t^2},frac{t^3-4t^2+2t-1}{1+2t^2}Big)$







share|cite|improve this answer












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answered Dec 22 '18 at 22:57









Shubham Johri

3,888716




3,888716












  • "can easily see" -> put the vectors in a matrix, determine the rank
    – G Cab
    Dec 22 '18 at 23:32


















  • "can easily see" -> put the vectors in a matrix, determine the rank
    – G Cab
    Dec 22 '18 at 23:32
















"can easily see" -> put the vectors in a matrix, determine the rank
– G Cab
Dec 22 '18 at 23:32




"can easily see" -> put the vectors in a matrix, determine the rank
– G Cab
Dec 22 '18 at 23:32


















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