Proving vector properties 10
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I had the following question:
Which of the following statements are always true for vectors in $R ^3$
If u $cdot($v x w$) = 4$ then w $cdot($v x u$)=-4$
$(5$u + v$)$x$($u x $7$v$) = 36($u x v$)$
If u is orthogonal to v and w then u is also orthogonal to ||w||v+||v||w
I was able to see which ones were false and which were true by simply plugging in vectors and doing the simple calculations, but I feel like that's not the correct approach. Is there properties in from which these are derived? Or was my approach correct in simply just plugging in to find that 1 and 3 are true.
linear-algebra
asked Nov 16 at 18:46
user583753
618
add a comment |
up vote
1
down vote
favorite
I had the following question:
Which of the following statements are always true for vectors in $R ^3$
If u $cdot($v x w$) = 4$ then w $cdot($v x u$)=-4$
$(5$u + v$)$x$($u x $7$v$) = 36($u x v$)$
If u is orthogonal to v and w then u is also orthogonal to ||w||v+||v||w
I was able to see which ones were false and which were true by simply plugging in vectors and doing the simple calculations, but I feel like that's not the correct approach. Is there properties in from which these are derived? Or was my approach correct in simply just plugging in to find that 1 and 3 are true.
linear-algebra
asked Nov 16 at 18:46
user583753
618
To prove any one of the above to be false, you need only supply a counter-example: finding vectors to plug in to find that the statement is not always true. That's legitimate. But to prove any one of them correct, you can not justify correctness merely by finding examples that work.
– amWhy
Nov 16 at 18:53
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I had the following question:
Which of the following statements are always true for vectors in $R ^3$
If u $cdot($v x w$) = 4$ then w $cdot($v x u$)=-4$
$(5$u + v$)$x$($u x $7$v$) = 36($u x v$)$
If u is orthogonal to v and w then u is also orthogonal to ||w||v+||v||w
I was able to see which ones were false and which were true by simply plugging in vectors and doing the simple calculations, but I feel like that's not the correct approach. Is there properties in from which these are derived? Or was my approach correct in simply just plugging in to find that 1 and 3 are true.
linear-algebra
asked Nov 16 at 18:46
user583753
618
I had the following question:
Which of the following statements are always true for vectors in $R ^3$
If u $cdot($v x w$) = 4$ then w $cdot($v x u$)=-4$
$(5$u + v$)$x$($u x $7$v$) = 36($u x v$)$
If u is orthogonal to v and w then u is also orthogonal to ||w||v+||v||w
I was able to see which ones were false and which were true by simply plugging in vectors and doing the simple calculations, but I feel like that's not the correct approach. Is there properties in from which these are derived? Or was my approach correct in simply just plugging in to find that 1 and 3 are true.
linear-algebra
linear-algebra
asked Nov 16 at 18:46
user583753
618
asked Nov 16 at 18:46
user583753
618
asked Nov 16 at 18:46
user583753
618
asked Nov 16 at 18:46
user583753
618
asked Nov 16 at 18:46
user583753
618
618
To prove any one of the above to be false, you need only supply a counter-example: finding vectors to plug in to find that the statement is not always true. That's legitimate. But to prove any one of them correct, you can not justify correctness merely by finding examples that work.
– amWhy
Nov 16 at 18:53
add a comment |
To prove any one of the above to be false, you need only supply a counter-example: finding vectors to plug in to find that the statement is not always true. That's legitimate. But to prove any one of them correct, you can not justify correctness merely by finding examples that work.
– amWhy
Nov 16 at 18:53
To prove any one of the above to be false, you need only supply a counter-example: finding vectors to plug in to find that the statement is not always true. That's legitimate. But to prove any one of them correct, you can not justify correctness merely by finding examples that work.
– amWhy
Nov 16 at 18:53
To prove any one of the above to be false, you need only supply a counter-example: finding vectors to plug in to find that the statement is not always true. That's legitimate. But to prove any one of them correct, you can not justify correctness merely by finding examples that work.
– amWhy
Nov 16 at 18:53
add a comment |
1 Answer
1
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oldest
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up vote
0
down vote
accepted
When you have to prove a result for general vectors $vec{v},vec{w}$ using examples won't do.
On the other hand, when you have to verify that a rule is non-correct, then finding an example is indeed the way to go.
For the first exercise you need to use the property of the triple product which indeed shows that the result is true.
For the second case, just notice that by linearity of the cross product:
begin{align*}(5u+v)times (u+7v)&=5utimes u+5utimes 7v+vtimes u+vtimes 7v\
&=0+35utimes v+vtimes u+0\ &=35utimes-utimes v\& =34utimes vend{align*}
so in general the result is non correct (just take any two vectors whose cross product is non-vanishing).
For the third, orthogonality means:
$$langle u,vrangle=0quad langle u,wrangle=0$$
but by the linearity of the scalar product in the second component
$$langle u,|w|v+|v|wrangle=|w|langle u,vrangle+|v|langle u,wrangle=0$$
which is equivalent to the statement of orthogonality
answered Nov 16 at 18:59
b00n heT
10.2k12134
1
Thank you, that really helps!
– user583753
Nov 16 at 21:15
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
When you have to prove a result for general vectors $vec{v},vec{w}$ using examples won't do.
On the other hand, when you have to verify that a rule is non-correct, then finding an example is indeed the way to go.
For the first exercise you need to use the property of the triple product which indeed shows that the result is true.
For the second case, just notice that by linearity of the cross product:
begin{align*}(5u+v)times (u+7v)&=5utimes u+5utimes 7v+vtimes u+vtimes 7v\
&=0+35utimes v+vtimes u+0\ &=35utimes-utimes v\& =34utimes vend{align*}
so in general the result is non correct (just take any two vectors whose cross product is non-vanishing).
For the third, orthogonality means:
$$langle u,vrangle=0quad langle u,wrangle=0$$
but by the linearity of the scalar product in the second component
$$langle u,|w|v+|v|wrangle=|w|langle u,vrangle+|v|langle u,wrangle=0$$
which is equivalent to the statement of orthogonality
answered Nov 16 at 18:59
b00n heT
10.2k12134
1
Thank you, that really helps!
– user583753
Nov 16 at 21:15
add a comment |
up vote
0
down vote
accepted
When you have to prove a result for general vectors $vec{v},vec{w}$ using examples won't do.
On the other hand, when you have to verify that a rule is non-correct, then finding an example is indeed the way to go.
For the first exercise you need to use the property of the triple product which indeed shows that the result is true.
For the second case, just notice that by linearity of the cross product:
begin{align*}(5u+v)times (u+7v)&=5utimes u+5utimes 7v+vtimes u+vtimes 7v\
&=0+35utimes v+vtimes u+0\ &=35utimes-utimes v\& =34utimes vend{align*}
so in general the result is non correct (just take any two vectors whose cross product is non-vanishing).
For the third, orthogonality means:
$$langle u,vrangle=0quad langle u,wrangle=0$$
but by the linearity of the scalar product in the second component
$$langle u,|w|v+|v|wrangle=|w|langle u,vrangle+|v|langle u,wrangle=0$$
which is equivalent to the statement of orthogonality
answered Nov 16 at 18:59
b00n heT
10.2k12134
1
Thank you, that really helps!
– user583753
Nov 16 at 21:15
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
When you have to prove a result for general vectors $vec{v},vec{w}$ using examples won't do.
On the other hand, when you have to verify that a rule is non-correct, then finding an example is indeed the way to go.
For the first exercise you need to use the property of the triple product which indeed shows that the result is true.
For the second case, just notice that by linearity of the cross product:
begin{align*}(5u+v)times (u+7v)&=5utimes u+5utimes 7v+vtimes u+vtimes 7v\
&=0+35utimes v+vtimes u+0\ &=35utimes-utimes v\& =34utimes vend{align*}
so in general the result is non correct (just take any two vectors whose cross product is non-vanishing).
For the third, orthogonality means:
$$langle u,vrangle=0quad langle u,wrangle=0$$
but by the linearity of the scalar product in the second component
$$langle u,|w|v+|v|wrangle=|w|langle u,vrangle+|v|langle u,wrangle=0$$
which is equivalent to the statement of orthogonality
answered Nov 16 at 18:59
b00n heT
10.2k12134
When you have to prove a result for general vectors $vec{v},vec{w}$ using examples won't do.
On the other hand, when you have to verify that a rule is non-correct, then finding an example is indeed the way to go.
For the first exercise you need to use the property of the triple product which indeed shows that the result is true.
For the second case, just notice that by linearity of the cross product:
begin{align*}(5u+v)times (u+7v)&=5utimes u+5utimes 7v+vtimes u+vtimes 7v\
&=0+35utimes v+vtimes u+0\ &=35utimes-utimes v\& =34utimes vend{align*}
so in general the result is non correct (just take any two vectors whose cross product is non-vanishing).
For the third, orthogonality means:
$$langle u,vrangle=0quad langle u,wrangle=0$$
but by the linearity of the scalar product in the second component
$$langle u,|w|v+|v|wrangle=|w|langle u,vrangle+|v|langle u,wrangle=0$$
which is equivalent to the statement of orthogonality
answered Nov 16 at 18:59
b00n heT
10.2k12134
answered Nov 16 at 18:59
b00n heT
10.2k12134
answered Nov 16 at 18:59
b00n heT
10.2k12134
answered Nov 16 at 18:59
b00n heT
10.2k12134
10.2k12134
1
Thank you, that really helps!
– user583753
Nov 16 at 21:15
add a comment |
1
Thank you, that really helps!
– user583753
Nov 16 at 21:15
1
1
Thank you, that really helps!
– user583753
Nov 16 at 21:15
Thank you, that really helps!
– user583753
Nov 16 at 21:15
add a comment |
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To prove any one of the above to be false, you need only supply a counter-example: finding vectors to plug in to find that the statement is not always true. That's legitimate. But to prove any one of them correct, you can not justify correctness merely by finding examples that work.
– amWhy
Nov 16 at 18:53