Linear Algebra: orthogonality in $mathbb{R}^{3}$












0












$begingroup$


$$v_{1} = (1, 1, 1)$$



$$v_{2} = (-1, 1, 0)$$



$$v_{3} = (-1, -1, 2)$$



The vectors $v1$, $v2$, and $v3$ form an orthogonal basis for $mathbb{R}^{3}$.



If $w = (4, -2, 4)$, then what is the coordinate vector, $[w]_{B}$, of $w$ with respect to the basis $B = {v1, v2, v3}$






linear-algebra orthogonality







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  • $begingroup$
    Welcome to math.se. Here are some tips on how to ask a good question.
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    Dec 9 '18 at 3:43
















0












$begingroup$


$$v_{1} = (1, 1, 1)$$



$$v_{2} = (-1, 1, 0)$$



$$v_{3} = (-1, -1, 2)$$



The vectors $v1$, $v2$, and $v3$ form an orthogonal basis for $mathbb{R}^{3}$.



If $w = (4, -2, 4)$, then what is the coordinate vector, $[w]_{B}$, of $w$ with respect to the basis $B = {v1, v2, v3}$






linear-algebra orthogonality







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to math.se. Here are some tips on how to ask a good question.
    $endgroup$
    – André 3000
    Dec 9 '18 at 3:43














0












0








0





$begingroup$


$$v_{1} = (1, 1, 1)$$



$$v_{2} = (-1, 1, 0)$$



$$v_{3} = (-1, -1, 2)$$



The vectors $v1$, $v2$, and $v3$ form an orthogonal basis for $mathbb{R}^{3}$.



If $w = (4, -2, 4)$, then what is the coordinate vector, $[w]_{B}$, of $w$ with respect to the basis $B = {v1, v2, v3}$






linear-algebra orthogonality







share|cite|improve this question











$endgroup$




$$v_{1} = (1, 1, 1)$$



$$v_{2} = (-1, 1, 0)$$



$$v_{3} = (-1, -1, 2)$$



The vectors $v1$, $v2$, and $v3$ form an orthogonal basis for $mathbb{R}^{3}$.



If $w = (4, -2, 4)$, then what is the coordinate vector, $[w]_{B}$, of $w$ with respect to the basis $B = {v1, v2, v3}$






linear-algebra orthogonality




linear-algebra orthogonality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 4:31









dantopa

6,46942243




6,46942243










asked Dec 8 '18 at 23:25









bradpbradp

1




1












  • $begingroup$
    Welcome to math.se. Here are some tips on how to ask a good question.
    $endgroup$
    – André 3000
    Dec 9 '18 at 3:43


















  • $begingroup$
    Welcome to math.se. Here are some tips on how to ask a good question.
    $endgroup$
    – André 3000
    Dec 9 '18 at 3:43
















$begingroup$
Welcome to math.se. Here are some tips on how to ask a good question.
$endgroup$
– André 3000
Dec 9 '18 at 3:43




$begingroup$
Welcome to math.se. Here are some tips on how to ask a good question.
$endgroup$
– André 3000
Dec 9 '18 at 3:43










1 Answer
1






active

oldest

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1












$begingroup$

Suppose



$$(4,-2,4)=w=av_1+bv_2+cv_3implies langle v_1,wrangle = 3a_1$$



Can you now complete the deduction? Can you see why having an orthonormal basis makes things even easier than the above?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    what do you mean by this?
    $endgroup$
    – bradp
    Dec 8 '18 at 23:43










  • $begingroup$
    Where did the 3a1 come from?
    $endgroup$
    – bradp
    Dec 8 '18 at 23:46










  • $begingroup$
    ok so I think i know where you got 3a1 from. did you just do the dot product of v1 with itself? Then repeated this to find b and c?
    $endgroup$
    – bradp
    Dec 8 '18 at 23:55










  • $begingroup$
    @bradp Yes, but with $;v_2,v_3;$ the dot product is zero...
    $endgroup$
    – DonAntonio
    Dec 9 '18 at 0:51











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Suppose



$$(4,-2,4)=w=av_1+bv_2+cv_3implies langle v_1,wrangle = 3a_1$$



Can you now complete the deduction? Can you see why having an orthonormal basis makes things even easier than the above?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    what do you mean by this?
    $endgroup$
    – bradp
    Dec 8 '18 at 23:43










  • $begingroup$
    Where did the 3a1 come from?
    $endgroup$
    – bradp
    Dec 8 '18 at 23:46










  • $begingroup$
    ok so I think i know where you got 3a1 from. did you just do the dot product of v1 with itself? Then repeated this to find b and c?
    $endgroup$
    – bradp
    Dec 8 '18 at 23:55










  • $begingroup$
    @bradp Yes, but with $;v_2,v_3;$ the dot product is zero...
    $endgroup$
    – DonAntonio
    Dec 9 '18 at 0:51
















1












$begingroup$

Suppose



$$(4,-2,4)=w=av_1+bv_2+cv_3implies langle v_1,wrangle = 3a_1$$



Can you now complete the deduction? Can you see why having an orthonormal basis makes things even easier than the above?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    what do you mean by this?
    $endgroup$
    – bradp
    Dec 8 '18 at 23:43










  • $begingroup$
    Where did the 3a1 come from?
    $endgroup$
    – bradp
    Dec 8 '18 at 23:46










  • $begingroup$
    ok so I think i know where you got 3a1 from. did you just do the dot product of v1 with itself? Then repeated this to find b and c?
    $endgroup$
    – bradp
    Dec 8 '18 at 23:55










  • $begingroup$
    @bradp Yes, but with $;v_2,v_3;$ the dot product is zero...
    $endgroup$
    – DonAntonio
    Dec 9 '18 at 0:51














1












1








1





$begingroup$

Suppose



$$(4,-2,4)=w=av_1+bv_2+cv_3implies langle v_1,wrangle = 3a_1$$



Can you now complete the deduction? Can you see why having an orthonormal basis makes things even easier than the above?






share|cite|improve this answer









$endgroup$



Suppose



$$(4,-2,4)=w=av_1+bv_2+cv_3implies langle v_1,wrangle = 3a_1$$



Can you now complete the deduction? Can you see why having an orthonormal basis makes things even easier than the above?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 8 '18 at 23:37









DonAntonioDonAntonio

178k1493228




178k1493228












  • $begingroup$
    what do you mean by this?
    $endgroup$
    – bradp
    Dec 8 '18 at 23:43










  • $begingroup$
    Where did the 3a1 come from?
    $endgroup$
    – bradp
    Dec 8 '18 at 23:46










  • $begingroup$
    ok so I think i know where you got 3a1 from. did you just do the dot product of v1 with itself? Then repeated this to find b and c?
    $endgroup$
    – bradp
    Dec 8 '18 at 23:55










  • $begingroup$
    @bradp Yes, but with $;v_2,v_3;$ the dot product is zero...
    $endgroup$
    – DonAntonio
    Dec 9 '18 at 0:51


















  • $begingroup$
    what do you mean by this?
    $endgroup$
    – bradp
    Dec 8 '18 at 23:43










  • $begingroup$
    Where did the 3a1 come from?
    $endgroup$
    – bradp
    Dec 8 '18 at 23:46










  • $begingroup$
    ok so I think i know where you got 3a1 from. did you just do the dot product of v1 with itself? Then repeated this to find b and c?
    $endgroup$
    – bradp
    Dec 8 '18 at 23:55










  • $begingroup$
    @bradp Yes, but with $;v_2,v_3;$ the dot product is zero...
    $endgroup$
    – DonAntonio
    Dec 9 '18 at 0:51
















$begingroup$
what do you mean by this?
$endgroup$
– bradp
Dec 8 '18 at 23:43




$begingroup$
what do you mean by this?
$endgroup$
– bradp
Dec 8 '18 at 23:43












$begingroup$
Where did the 3a1 come from?
$endgroup$
– bradp
Dec 8 '18 at 23:46




$begingroup$
Where did the 3a1 come from?
$endgroup$
– bradp
Dec 8 '18 at 23:46












$begingroup$
ok so I think i know where you got 3a1 from. did you just do the dot product of v1 with itself? Then repeated this to find b and c?
$endgroup$
– bradp
Dec 8 '18 at 23:55




$begingroup$
ok so I think i know where you got 3a1 from. did you just do the dot product of v1 with itself? Then repeated this to find b and c?
$endgroup$
– bradp
Dec 8 '18 at 23:55












$begingroup$
@bradp Yes, but with $;v_2,v_3;$ the dot product is zero...
$endgroup$
– DonAntonio
Dec 9 '18 at 0:51




$begingroup$
@bradp Yes, but with $;v_2,v_3;$ the dot product is zero...
$endgroup$
– DonAntonio
Dec 9 '18 at 0:51


















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