Linear Algebra: orthogonality in $mathbb{R}^{3}$
$begingroup$
$$v_{1} = (1, 1, 1)$$
$$v_{2} = (-1, 1, 0)$$
$$v_{3} = (-1, -1, 2)$$
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
edited Dec 9 '18 at 4:31
dantopa
6,46942243
asked Dec 8 '18 at 23:25
bradpbradp
1
$endgroup$
add a comment |
$begingroup$
$$v_{1} = (1, 1, 1)$$
$$v_{2} = (-1, 1, 0)$$
$$v_{3} = (-1, -1, 2)$$
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
edited Dec 9 '18 at 4:31
dantopa
6,46942243
asked Dec 8 '18 at 23:25
bradpbradp
1
$endgroup$
$begingroup$
Welcome to math.se. Here are some tips on how to ask a good question.
$endgroup$
– André 3000
Dec 9 '18 at 3:43
add a comment |
$begingroup$
$$v_{1} = (1, 1, 1)$$
$$v_{2} = (-1, 1, 0)$$
$$v_{3} = (-1, -1, 2)$$
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
edited Dec 9 '18 at 4:31
dantopa
6,46942243
asked Dec 8 '18 at 23:25
bradpbradp
1
$endgroup$
$$v_{1} = (1, 1, 1)$$
$$v_{2} = (-1, 1, 0)$$
$$v_{3} = (-1, -1, 2)$$
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
edited Dec 9 '18 at 4:31
dantopa
6,46942243
asked Dec 8 '18 at 23:25
bradpbradp
1
edited Dec 9 '18 at 4:31
dantopa
6,46942243
asked Dec 8 '18 at 23:25
bradpbradp
1
edited Dec 9 '18 at 4:31
dantopa
6,46942243
edited Dec 9 '18 at 4:31
dantopa
6,46942243
edited Dec 9 '18 at 4:31
dantopa
6,46942243
6,46942243
asked Dec 8 '18 at 23:25
bradpbradp
1
asked Dec 8 '18 at 23:25
bradpbradp
1
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
add a comment |
$begingroup$
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.
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– André 3000
Dec 9 '18 at 3:43
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1 Answer
1
active
oldest
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$begingroup$
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?
answered Dec 8 '18 at 23:37
DonAntonioDonAntonio
178k1493228
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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?
answered Dec 8 '18 at 23:37
DonAntonioDonAntonio
178k1493228
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?
answered Dec 8 '18 at 23:37
DonAntonioDonAntonio
178k1493228
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?
answered Dec 8 '18 at 23:37
DonAntonioDonAntonio
178k1493228
$endgroup$
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?
answered Dec 8 '18 at 23:37
DonAntonioDonAntonio
178k1493228
answered Dec 8 '18 at 23:37
DonAntonioDonAntonio
178k1493228
answered Dec 8 '18 at 23:37
DonAntonioDonAntonio
178k1493228
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
Welcome to math.se. Here are some tips on how to ask a good question.
$endgroup$
– André 3000
Dec 9 '18 at 3:43