Mixing
Neep help justifying a vector relation given in a question
$begingroup$
i'm trying to do a question for which I was given the following line equations:
$underline r = underline a + lambda underline u$
$underline r' = underline a' + lambda' underline u'$
They then gave me this relationship without any justification, i've been trying to get my head around it but have not had much luck.
$lvert underline r-underline r'rvert^2lvert underline u times underline u'rvert^2=lvert (underline a - underline a') cdot (underline u times underline u')lvert^2+lvert (underline r - underline r') times (underline u times underline u')lvert^2$
I know that
$lvert (underline r - underline r') times (underline u times underline u')lvert = lvert underline r-underline r'rvertlvert underline u times underline u'rvertsin theta $
but cant get any further.
linear-algebra geometry vectors
asked Jan 29 at 12:52
A. PavlenkoA. Pavlenko
123
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add a comment |
$begingroup$
i'm trying to do a question for which I was given the following line equations:
$underline r = underline a + lambda underline u$
$underline r' = underline a' + lambda' underline u'$
They then gave me this relationship without any justification, i've been trying to get my head around it but have not had much luck.
$lvert underline r-underline r'rvert^2lvert underline u times underline u'rvert^2=lvert (underline a - underline a') cdot (underline u times underline u')lvert^2+lvert (underline r - underline r') times (underline u times underline u')lvert^2$
I know that
$lvert (underline r - underline r') times (underline u times underline u')lvert = lvert underline r-underline r'rvertlvert underline u times underline u'rvertsin theta $
but cant get any further.
linear-algebra geometry vectors
asked Jan 29 at 12:52
A. PavlenkoA. Pavlenko
123
$endgroup$
add a comment |
$begingroup$
i'm trying to do a question for which I was given the following line equations:
$underline r = underline a + lambda underline u$
$underline r' = underline a' + lambda' underline u'$
They then gave me this relationship without any justification, i've been trying to get my head around it but have not had much luck.
$lvert underline r-underline r'rvert^2lvert underline u times underline u'rvert^2=lvert (underline a - underline a') cdot (underline u times underline u')lvert^2+lvert (underline r - underline r') times (underline u times underline u')lvert^2$
I know that
$lvert (underline r - underline r') times (underline u times underline u')lvert = lvert underline r-underline r'rvertlvert underline u times underline u'rvertsin theta $
but cant get any further.
linear-algebra geometry vectors
asked Jan 29 at 12:52
A. PavlenkoA. Pavlenko
123
$endgroup$
i'm trying to do a question for which I was given the following line equations:
$underline r = underline a + lambda underline u$
$underline r' = underline a' + lambda' underline u'$
They then gave me this relationship without any justification, i've been trying to get my head around it but have not had much luck.
$lvert underline r-underline r'rvert^2lvert underline u times underline u'rvert^2=lvert (underline a - underline a') cdot (underline u times underline u')lvert^2+lvert (underline r - underline r') times (underline u times underline u')lvert^2$
I know that
$lvert (underline r - underline r') times (underline u times underline u')lvert = lvert underline r-underline r'rvertlvert underline u times underline u'rvertsin theta $
but cant get any further.
linear-algebra geometry vectors
linear-algebra geometry vectors
asked Jan 29 at 12:52
A. PavlenkoA. Pavlenko
123
asked Jan 29 at 12:52
A. PavlenkoA. Pavlenko
123
asked Jan 29 at 12:52
A. PavlenkoA. Pavlenko
123
asked Jan 29 at 12:52
A. PavlenkoA. Pavlenko
123
asked Jan 29 at 12:52
A. PavlenkoA. Pavlenko
123
123
add a comment |
add a comment |
1 Answer
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$begingroup$
You should be familiar with some basic properties of the cross product. Especially the identity
$$ || mathbf{a} times mathbf{b} ||^2 = ||mathbf{a}||^2 , || mathbf{b}||^2 - |mathbf{a} cdot mathbf{b}|^2 $$
As can be seen, for instance from:
$$
begin{aligned}
|| mathbf{a} times mathbf{b} ||^2 & = (mathbf{a} times mathbf{b}) cdot (mathbf{a} times mathbf{b}) \
&= mathbf{b} cdot ((mathbf{a} times mathbf{b}) times mathbf{a}) \
&= -mathbf{b} cdot (mathbf{a} times (mathbf{a} times mathbf{b})) \
&= -mathbf{b} cdot (mathbf{a},(mathbf{a}cdotmathbf{b}) - mathbf{b},(mathbf{a}cdotmathbf{a})) \
&= ||mathbf{a}||^2,||mathbf{b}||^2 - |mathbf{a}cdotmathbf{b}|^2
end{aligned}
$$
Now, with $mathbf{a} = mathbf{r} - mathbf{r'}$ and $mathbf{b} = mathbf{u} times mathbf{u'}$, use the bilinearity of the inner product and
$$
mathbf{r} = mathbf{a} + lambda ,mathbf{u}; ,, mathbf{r'} = mathbf{a'} + lambda' ,mathbf{u'} \
mathbf{u} cdot (mathbf{u} times mathbf{u'}) = mathbf{u'} cdot (mathbf{u} times mathbf{u'}) = 0
$$
answered Jan 29 at 14:06
jgbjgb
14813
$endgroup$
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You should be familiar with some basic properties of the cross product. Especially the identity
$$ || mathbf{a} times mathbf{b} ||^2 = ||mathbf{a}||^2 , || mathbf{b}||^2 - |mathbf{a} cdot mathbf{b}|^2 $$
As can be seen, for instance from:
$$
begin{aligned}
|| mathbf{a} times mathbf{b} ||^2 & = (mathbf{a} times mathbf{b}) cdot (mathbf{a} times mathbf{b}) \
&= mathbf{b} cdot ((mathbf{a} times mathbf{b}) times mathbf{a}) \
&= -mathbf{b} cdot (mathbf{a} times (mathbf{a} times mathbf{b})) \
&= -mathbf{b} cdot (mathbf{a},(mathbf{a}cdotmathbf{b}) - mathbf{b},(mathbf{a}cdotmathbf{a})) \
&= ||mathbf{a}||^2,||mathbf{b}||^2 - |mathbf{a}cdotmathbf{b}|^2
end{aligned}
$$
Now, with $mathbf{a} = mathbf{r} - mathbf{r'}$ and $mathbf{b} = mathbf{u} times mathbf{u'}$, use the bilinearity of the inner product and
$$
mathbf{r} = mathbf{a} + lambda ,mathbf{u}; ,, mathbf{r'} = mathbf{a'} + lambda' ,mathbf{u'} \
mathbf{u} cdot (mathbf{u} times mathbf{u'}) = mathbf{u'} cdot (mathbf{u} times mathbf{u'}) = 0
$$
answered Jan 29 at 14:06
jgbjgb
14813
$endgroup$
add a comment |
$begingroup$
You should be familiar with some basic properties of the cross product. Especially the identity
$$ || mathbf{a} times mathbf{b} ||^2 = ||mathbf{a}||^2 , || mathbf{b}||^2 - |mathbf{a} cdot mathbf{b}|^2 $$
As can be seen, for instance from:
$$
begin{aligned}
|| mathbf{a} times mathbf{b} ||^2 & = (mathbf{a} times mathbf{b}) cdot (mathbf{a} times mathbf{b}) \
&= mathbf{b} cdot ((mathbf{a} times mathbf{b}) times mathbf{a}) \
&= -mathbf{b} cdot (mathbf{a} times (mathbf{a} times mathbf{b})) \
&= -mathbf{b} cdot (mathbf{a},(mathbf{a}cdotmathbf{b}) - mathbf{b},(mathbf{a}cdotmathbf{a})) \
&= ||mathbf{a}||^2,||mathbf{b}||^2 - |mathbf{a}cdotmathbf{b}|^2
end{aligned}
$$
Now, with $mathbf{a} = mathbf{r} - mathbf{r'}$ and $mathbf{b} = mathbf{u} times mathbf{u'}$, use the bilinearity of the inner product and
$$
mathbf{r} = mathbf{a} + lambda ,mathbf{u}; ,, mathbf{r'} = mathbf{a'} + lambda' ,mathbf{u'} \
mathbf{u} cdot (mathbf{u} times mathbf{u'}) = mathbf{u'} cdot (mathbf{u} times mathbf{u'}) = 0
$$
answered Jan 29 at 14:06
jgbjgb
14813
$endgroup$
add a comment |
$begingroup$
You should be familiar with some basic properties of the cross product. Especially the identity
$$ || mathbf{a} times mathbf{b} ||^2 = ||mathbf{a}||^2 , || mathbf{b}||^2 - |mathbf{a} cdot mathbf{b}|^2 $$
As can be seen, for instance from:
$$
begin{aligned}
|| mathbf{a} times mathbf{b} ||^2 & = (mathbf{a} times mathbf{b}) cdot (mathbf{a} times mathbf{b}) \
&= mathbf{b} cdot ((mathbf{a} times mathbf{b}) times mathbf{a}) \
&= -mathbf{b} cdot (mathbf{a} times (mathbf{a} times mathbf{b})) \
&= -mathbf{b} cdot (mathbf{a},(mathbf{a}cdotmathbf{b}) - mathbf{b},(mathbf{a}cdotmathbf{a})) \
&= ||mathbf{a}||^2,||mathbf{b}||^2 - |mathbf{a}cdotmathbf{b}|^2
end{aligned}
$$
Now, with $mathbf{a} = mathbf{r} - mathbf{r'}$ and $mathbf{b} = mathbf{u} times mathbf{u'}$, use the bilinearity of the inner product and
$$
mathbf{r} = mathbf{a} + lambda ,mathbf{u}; ,, mathbf{r'} = mathbf{a'} + lambda' ,mathbf{u'} \
mathbf{u} cdot (mathbf{u} times mathbf{u'}) = mathbf{u'} cdot (mathbf{u} times mathbf{u'}) = 0
$$
answered Jan 29 at 14:06
jgbjgb
14813
$endgroup$
You should be familiar with some basic properties of the cross product. Especially the identity
$$ || mathbf{a} times mathbf{b} ||^2 = ||mathbf{a}||^2 , || mathbf{b}||^2 - |mathbf{a} cdot mathbf{b}|^2 $$
As can be seen, for instance from:
$$
begin{aligned}
|| mathbf{a} times mathbf{b} ||^2 & = (mathbf{a} times mathbf{b}) cdot (mathbf{a} times mathbf{b}) \
&= mathbf{b} cdot ((mathbf{a} times mathbf{b}) times mathbf{a}) \
&= -mathbf{b} cdot (mathbf{a} times (mathbf{a} times mathbf{b})) \
&= -mathbf{b} cdot (mathbf{a},(mathbf{a}cdotmathbf{b}) - mathbf{b},(mathbf{a}cdotmathbf{a})) \
&= ||mathbf{a}||^2,||mathbf{b}||^2 - |mathbf{a}cdotmathbf{b}|^2
end{aligned}
$$
Now, with $mathbf{a} = mathbf{r} - mathbf{r'}$ and $mathbf{b} = mathbf{u} times mathbf{u'}$, use the bilinearity of the inner product and
$$
mathbf{r} = mathbf{a} + lambda ,mathbf{u}; ,, mathbf{r'} = mathbf{a'} + lambda' ,mathbf{u'} \
mathbf{u} cdot (mathbf{u} times mathbf{u'}) = mathbf{u'} cdot (mathbf{u} times mathbf{u'}) = 0
$$
answered Jan 29 at 14:06
jgbjgb
14813
answered Jan 29 at 14:06
jgbjgb
14813
answered Jan 29 at 14:06
jgbjgb
14813
answered Jan 29 at 14:06
jgbjgb
14813
14813
add a comment |
add a comment |
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