Linear algebra. Vectors
Let $ vec{a}, vec{b}, vec{c} $ be non-complanar vectors so that $ vec{a} times vec{b}, vec{b}, vec{c} times vec{a}$ on the same plane. How can I show that $ vec{a}$ is perpendicular to $vec{b}$?
linear-algebra vectors
asked Dec 1 at 17:25
nene123
52
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Let $ vec{a}, vec{b}, vec{c} $ be non-complanar vectors so that $ vec{a} times vec{b}, vec{b}, vec{c} times vec{a}$ on the same plane. How can I show that $ vec{a}$ is perpendicular to $vec{b}$?
linear-algebra vectors
asked Dec 1 at 17:25
nene123
52
add a comment |
Let $ vec{a}, vec{b}, vec{c} $ be non-complanar vectors so that $ vec{a} times vec{b}, vec{b}, vec{c} times vec{a}$ on the same plane. How can I show that $ vec{a}$ is perpendicular to $vec{b}$?
linear-algebra vectors
asked Dec 1 at 17:25
nene123
52
Let $ vec{a}, vec{b}, vec{c} $ be non-complanar vectors so that $ vec{a} times vec{b}, vec{b}, vec{c} times vec{a}$ on the same plane. How can I show that $ vec{a}$ is perpendicular to $vec{b}$?
linear-algebra vectors
linear-algebra vectors
asked Dec 1 at 17:25
nene123
52
asked Dec 1 at 17:25
nene123
52
asked Dec 1 at 17:25
nene123
52
asked Dec 1 at 17:25
nene123
52
asked Dec 1 at 17:25
nene123
52
52
add a comment |
add a comment |
1 Answer
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Idea:
We can (probably) write $$vec{b} = lambda (vec{a} times vec{b}) + mu (vec{c} times vec{a})$$ for some scalars $lambda, mu$ as the three vectors are co-planar. Now the inner product with $vec{a}$ of this expression will be $0$, as the inner product is linear and $vec{a} cdot (vec{a} times vec{b})=0$ and $vec{a} cdot (vec{c} times vec{a}) =0$.
The previous will only fail if $vec{c} times vec{a}$ and $(vec{a} times vec{b})$ are dependent; explore what you can do in that case.
answered Dec 1 at 17:43
Henno Brandsma
104k346113
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Idea:
We can (probably) write $$vec{b} = lambda (vec{a} times vec{b}) + mu (vec{c} times vec{a})$$ for some scalars $lambda, mu$ as the three vectors are co-planar. Now the inner product with $vec{a}$ of this expression will be $0$, as the inner product is linear and $vec{a} cdot (vec{a} times vec{b})=0$ and $vec{a} cdot (vec{c} times vec{a}) =0$.
The previous will only fail if $vec{c} times vec{a}$ and $(vec{a} times vec{b})$ are dependent; explore what you can do in that case.
answered Dec 1 at 17:43
Henno Brandsma
104k346113
add a comment |
Idea:
We can (probably) write $$vec{b} = lambda (vec{a} times vec{b}) + mu (vec{c} times vec{a})$$ for some scalars $lambda, mu$ as the three vectors are co-planar. Now the inner product with $vec{a}$ of this expression will be $0$, as the inner product is linear and $vec{a} cdot (vec{a} times vec{b})=0$ and $vec{a} cdot (vec{c} times vec{a}) =0$.
The previous will only fail if $vec{c} times vec{a}$ and $(vec{a} times vec{b})$ are dependent; explore what you can do in that case.
answered Dec 1 at 17:43
Henno Brandsma
104k346113
add a comment |
Idea:
We can (probably) write $$vec{b} = lambda (vec{a} times vec{b}) + mu (vec{c} times vec{a})$$ for some scalars $lambda, mu$ as the three vectors are co-planar. Now the inner product with $vec{a}$ of this expression will be $0$, as the inner product is linear and $vec{a} cdot (vec{a} times vec{b})=0$ and $vec{a} cdot (vec{c} times vec{a}) =0$.
The previous will only fail if $vec{c} times vec{a}$ and $(vec{a} times vec{b})$ are dependent; explore what you can do in that case.
answered Dec 1 at 17:43
Henno Brandsma
104k346113
Idea:
We can (probably) write $$vec{b} = lambda (vec{a} times vec{b}) + mu (vec{c} times vec{a})$$ for some scalars $lambda, mu$ as the three vectors are co-planar. Now the inner product with $vec{a}$ of this expression will be $0$, as the inner product is linear and $vec{a} cdot (vec{a} times vec{b})=0$ and $vec{a} cdot (vec{c} times vec{a}) =0$.
The previous will only fail if $vec{c} times vec{a}$ and $(vec{a} times vec{b})$ are dependent; explore what you can do in that case.
answered Dec 1 at 17:43
Henno Brandsma
104k346113
answered Dec 1 at 17:43
Henno Brandsma
104k346113
answered Dec 1 at 17:43
Henno Brandsma
104k346113
answered Dec 1 at 17:43
Henno Brandsma
104k346113
104k346113
add a comment |
add a comment |
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