Mixing
Vector geometry involving parallel vectors
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I found this question on vector geometry in an IGCSE text. I got these solutions for the following vectors: $vec {BP}=mathbf a-mathbf b$, $vec {AB}=mathbf b-3mathbf a$, $vec {MB}=frac 12(mathbf b-3mathbf a)$. Then for $vec {MX}$ I got this really long and complicated expression in terms of $mathbf a$, $mathbf b$ and $mathit k$:
$$vec {MX}=frac 12(mathbf b-3mathbf a)+mathit k(mathbf a-mathbf b)=left(mathit k-frac 32right)mathbf a+left(frac 12-mathit kright)mathbf b$$
As I understand, if $vec {MX}$ is parallel to $vec {BO}$, $vec {MX}$ would be something times $vec {BO}$. But how do you use that result to find the value of $mathit k$?
vectors
asked Jan 24 at 14:56
IndulaIndula
506
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add a comment |
$begingroup$
I found this question on vector geometry in an IGCSE text. I got these solutions for the following vectors: $vec {BP}=mathbf a-mathbf b$, $vec {AB}=mathbf b-3mathbf a$, $vec {MB}=frac 12(mathbf b-3mathbf a)$. Then for $vec {MX}$ I got this really long and complicated expression in terms of $mathbf a$, $mathbf b$ and $mathit k$:
$$vec {MX}=frac 12(mathbf b-3mathbf a)+mathit k(mathbf a-mathbf b)=left(mathit k-frac 32right)mathbf a+left(frac 12-mathit kright)mathbf b$$
As I understand, if $vec {MX}$ is parallel to $vec {BO}$, $vec {MX}$ would be something times $vec {BO}$. But how do you use that result to find the value of $mathit k$?
vectors
asked Jan 24 at 14:56
IndulaIndula
506
$endgroup$
add a comment |
$begingroup$
I found this question on vector geometry in an IGCSE text. I got these solutions for the following vectors: $vec {BP}=mathbf a-mathbf b$, $vec {AB}=mathbf b-3mathbf a$, $vec {MB}=frac 12(mathbf b-3mathbf a)$. Then for $vec {MX}$ I got this really long and complicated expression in terms of $mathbf a$, $mathbf b$ and $mathit k$:
$$vec {MX}=frac 12(mathbf b-3mathbf a)+mathit k(mathbf a-mathbf b)=left(mathit k-frac 32right)mathbf a+left(frac 12-mathit kright)mathbf b$$
As I understand, if $vec {MX}$ is parallel to $vec {BO}$, $vec {MX}$ would be something times $vec {BO}$. But how do you use that result to find the value of $mathit k$?
vectors
asked Jan 24 at 14:56
IndulaIndula
506
$endgroup$
I found this question on vector geometry in an IGCSE text. I got these solutions for the following vectors: $vec {BP}=mathbf a-mathbf b$, $vec {AB}=mathbf b-3mathbf a$, $vec {MB}=frac 12(mathbf b-3mathbf a)$. Then for $vec {MX}$ I got this really long and complicated expression in terms of $mathbf a$, $mathbf b$ and $mathit k$:
$$vec {MX}=frac 12(mathbf b-3mathbf a)+mathit k(mathbf a-mathbf b)=left(mathit k-frac 32right)mathbf a+left(frac 12-mathit kright)mathbf b$$
As I understand, if $vec {MX}$ is parallel to $vec {BO}$, $vec {MX}$ would be something times $vec {BO}$. But how do you use that result to find the value of $mathit k$?
vectors
vectors
asked Jan 24 at 14:56
IndulaIndula
506
asked Jan 24 at 14:56
IndulaIndula
506
edited Jan 25 at 1:36
Indula
asked Jan 24 at 14:56
IndulaIndula
506
asked Jan 24 at 14:56
IndulaIndula
506
asked Jan 24 at 14:56
IndulaIndula
506
506
add a comment |
add a comment |
1 Answer
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Two vectors $mathbf u$ and $mathbf v$ are parallel if there is a scalar $c$, so $mathbf u =cmathbf v$. So if $vec{MX}$ is parallel to $vec{OB}$, then there should be $c$:
$$
left(k-frac32right)mathbf a+left(frac12-kright)mathbf b=cmathbf b
$$
But since $mathbf a$ isn't parallel to $mathbf b$, it can't be true until $k-3/2=0$.
answered Jan 24 at 16:14
Vasily MitchVasily Mitch
2,6791312
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$begingroup$
So if $vec {MX}$ is parallel to $vec {OB}$, $vec {MX}=mathit cvec {OB}$ where $mathit c$ is a scalar. But $vec {OB}=mathbf b$ and $vec {MX}=left (mathit k-frac 32 right)mathbf a+left(frac 12-mathit kright)mathbf b$. Unless $mathbf a$ is parallel to $mathbf b$, $vec {MX}$ is parallel to $vec {OB}$ iff $mathit k=frac 32$. Therefore it must be that $mathit k=frac 32$ if $vec {MX}$ is parallel to $vec {OB}$. Is that how the reasoning go?
$endgroup$
– Indula
Jan 25 at 2:15
$begingroup$
Can we reason like that though?
$endgroup$
– Indula
Jan 25 at 2:30
$begingroup$
Yes, that is the right reasoning
$endgroup$
– Vasily Mitch
Jan 25 at 14:22
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
Two vectors $mathbf u$ and $mathbf v$ are parallel if there is a scalar $c$, so $mathbf u =cmathbf v$. So if $vec{MX}$ is parallel to $vec{OB}$, then there should be $c$:
$$
left(k-frac32right)mathbf a+left(frac12-kright)mathbf b=cmathbf b
$$
But since $mathbf a$ isn't parallel to $mathbf b$, it can't be true until $k-3/2=0$.
answered Jan 24 at 16:14
Vasily MitchVasily Mitch
2,6791312
$endgroup$
$begingroup$
So if $vec {MX}$ is parallel to $vec {OB}$, $vec {MX}=mathit cvec {OB}$ where $mathit c$ is a scalar. But $vec {OB}=mathbf b$ and $vec {MX}=left (mathit k-frac 32 right)mathbf a+left(frac 12-mathit kright)mathbf b$. Unless $mathbf a$ is parallel to $mathbf b$, $vec {MX}$ is parallel to $vec {OB}$ iff $mathit k=frac 32$. Therefore it must be that $mathit k=frac 32$ if $vec {MX}$ is parallel to $vec {OB}$. Is that how the reasoning go?
$endgroup$
– Indula
Jan 25 at 2:15
$begingroup$
Can we reason like that though?
$endgroup$
– Indula
Jan 25 at 2:30
$begingroup$
Yes, that is the right reasoning
$endgroup$
– Vasily Mitch
Jan 25 at 14:22
add a comment |
$begingroup$
Two vectors $mathbf u$ and $mathbf v$ are parallel if there is a scalar $c$, so $mathbf u =cmathbf v$. So if $vec{MX}$ is parallel to $vec{OB}$, then there should be $c$:
$$
left(k-frac32right)mathbf a+left(frac12-kright)mathbf b=cmathbf b
$$
But since $mathbf a$ isn't parallel to $mathbf b$, it can't be true until $k-3/2=0$.
answered Jan 24 at 16:14
Vasily MitchVasily Mitch
2,6791312
$endgroup$
$begingroup$
So if $vec {MX}$ is parallel to $vec {OB}$, $vec {MX}=mathit cvec {OB}$ where $mathit c$ is a scalar. But $vec {OB}=mathbf b$ and $vec {MX}=left (mathit k-frac 32 right)mathbf a+left(frac 12-mathit kright)mathbf b$. Unless $mathbf a$ is parallel to $mathbf b$, $vec {MX}$ is parallel to $vec {OB}$ iff $mathit k=frac 32$. Therefore it must be that $mathit k=frac 32$ if $vec {MX}$ is parallel to $vec {OB}$. Is that how the reasoning go?
$endgroup$
– Indula
Jan 25 at 2:15
$begingroup$
Can we reason like that though?
$endgroup$
– Indula
Jan 25 at 2:30
$begingroup$
Yes, that is the right reasoning
$endgroup$
– Vasily Mitch
Jan 25 at 14:22
add a comment |
$begingroup$
Two vectors $mathbf u$ and $mathbf v$ are parallel if there is a scalar $c$, so $mathbf u =cmathbf v$. So if $vec{MX}$ is parallel to $vec{OB}$, then there should be $c$:
$$
left(k-frac32right)mathbf a+left(frac12-kright)mathbf b=cmathbf b
$$
But since $mathbf a$ isn't parallel to $mathbf b$, it can't be true until $k-3/2=0$.
answered Jan 24 at 16:14
Vasily MitchVasily Mitch
2,6791312
$endgroup$
Two vectors $mathbf u$ and $mathbf v$ are parallel if there is a scalar $c$, so $mathbf u =cmathbf v$. So if $vec{MX}$ is parallel to $vec{OB}$, then there should be $c$:
$$
left(k-frac32right)mathbf a+left(frac12-kright)mathbf b=cmathbf b
$$
But since $mathbf a$ isn't parallel to $mathbf b$, it can't be true until $k-3/2=0$.
answered Jan 24 at 16:14
Vasily MitchVasily Mitch
2,6791312
answered Jan 24 at 16:14
Vasily MitchVasily Mitch
2,6791312
answered Jan 24 at 16:14
Vasily MitchVasily Mitch
2,6791312
answered Jan 24 at 16:14
Vasily MitchVasily Mitch
2,6791312
2,6791312
$begingroup$
So if $vec {MX}$ is parallel to $vec {OB}$, $vec {MX}=mathit cvec {OB}$ where $mathit c$ is a scalar. But $vec {OB}=mathbf b$ and $vec {MX}=left (mathit k-frac 32 right)mathbf a+left(frac 12-mathit kright)mathbf b$. Unless $mathbf a$ is parallel to $mathbf b$, $vec {MX}$ is parallel to $vec {OB}$ iff $mathit k=frac 32$. Therefore it must be that $mathit k=frac 32$ if $vec {MX}$ is parallel to $vec {OB}$. Is that how the reasoning go?
$endgroup$
– Indula
Jan 25 at 2:15
$begingroup$
Can we reason like that though?
$endgroup$
– Indula
Jan 25 at 2:30
$begingroup$
Yes, that is the right reasoning
$endgroup$
– Vasily Mitch
Jan 25 at 14:22
add a comment |
$begingroup$
So if $vec {MX}$ is parallel to $vec {OB}$, $vec {MX}=mathit cvec {OB}$ where $mathit c$ is a scalar. But $vec {OB}=mathbf b$ and $vec {MX}=left (mathit k-frac 32 right)mathbf a+left(frac 12-mathit kright)mathbf b$. Unless $mathbf a$ is parallel to $mathbf b$, $vec {MX}$ is parallel to $vec {OB}$ iff $mathit k=frac 32$. Therefore it must be that $mathit k=frac 32$ if $vec {MX}$ is parallel to $vec {OB}$. Is that how the reasoning go?
$endgroup$
– Indula
Jan 25 at 2:15
$begingroup$
Can we reason like that though?
$endgroup$
– Indula
Jan 25 at 2:30
$begingroup$
Yes, that is the right reasoning
$endgroup$
– Vasily Mitch
Jan 25 at 14:22
$begingroup$
So if $vec {MX}$ is parallel to $vec {OB}$, $vec {MX}=mathit cvec {OB}$ where $mathit c$ is a scalar. But $vec {OB}=mathbf b$ and $vec {MX}=left (mathit k-frac 32 right)mathbf a+left(frac 12-mathit kright)mathbf b$. Unless $mathbf a$ is parallel to $mathbf b$, $vec {MX}$ is parallel to $vec {OB}$ iff $mathit k=frac 32$. Therefore it must be that $mathit k=frac 32$ if $vec {MX}$ is parallel to $vec {OB}$. Is that how the reasoning go?
$endgroup$
– Indula
Jan 25 at 2:15
$begingroup$
So if $vec {MX}$ is parallel to $vec {OB}$, $vec {MX}=mathit cvec {OB}$ where $mathit c$ is a scalar. But $vec {OB}=mathbf b$ and $vec {MX}=left (mathit k-frac 32 right)mathbf a+left(frac 12-mathit kright)mathbf b$. Unless $mathbf a$ is parallel to $mathbf b$, $vec {MX}$ is parallel to $vec {OB}$ iff $mathit k=frac 32$. Therefore it must be that $mathit k=frac 32$ if $vec {MX}$ is parallel to $vec {OB}$. Is that how the reasoning go?
$endgroup$
– Indula
Jan 25 at 2:15
$begingroup$
Can we reason like that though?
$endgroup$
– Indula
Jan 25 at 2:30
$begingroup$
Can we reason like that though?
$endgroup$
– Indula
Jan 25 at 2:30
$begingroup$
Yes, that is the right reasoning
$endgroup$
– Vasily Mitch
Jan 25 at 14:22
$begingroup$
Yes, that is the right reasoning
$endgroup$
– Vasily Mitch
Jan 25 at 14:22
add a comment |
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