Finding possible values in vector.

1

$begingroup$

Just having some difficulty finding the values of '$P$' in this vector.

Relative to an origin $O$, the position vectors of the points $A$ and $B$ are given as follows:

$leftlvert OA rightrvert$ = $2i + 3j - k$
$leftlvert OB rightrvert$ = $4i - 3j + 2k$

The point $C$ is such that $leftlvert OC rightrvert$ = $6j + pk$ where $p$ is a constant. Given the lengths of $leftlvert AB rightrvert$ and $leftlvert AC rightrvert$ are equal, find the possible values of $P$.

--

So what I have done so far is as follows:

$ 2i - 6j + 3k = 9j + pk - 2i - k $

$4i -15j + 3k = pk - k$

Now what I'm doing may be completely stupid or wrong, so apologies in advance, however any pointers or tips would be much appreciated :)

Thanks again!

vectors

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asked Jun 19 '16 at 10:23

M. AndersonM. Anderson

406

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  AB =OB-OA and AC=OC-OA after finding these... equate them and find the value of p.
  $endgroup$
  – Jasser
  Jun 19 '16 at 11:18
  

  
  
  
  

add a comment | 

1

$begingroup$

Just having some difficulty finding the values of '$P$' in this vector.

Relative to an origin $O$, the position vectors of the points $A$ and $B$ are given as follows:

$leftlvert OA rightrvert$ = $2i + 3j - k$
$leftlvert OB rightrvert$ = $4i - 3j + 2k$

The point $C$ is such that $leftlvert OC rightrvert$ = $6j + pk$ where $p$ is a constant. Given the lengths of $leftlvert AB rightrvert$ and $leftlvert AC rightrvert$ are equal, find the possible values of $P$.

--

So what I have done so far is as follows:

$ 2i - 6j + 3k = 9j + pk - 2i - k $

$4i -15j + 3k = pk - k$

Now what I'm doing may be completely stupid or wrong, so apologies in advance, however any pointers or tips would be much appreciated :)

Thanks again!

vectors

share|cite|improve this question

asked Jun 19 '16 at 10:23

M. AndersonM. Anderson

406

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  AB =OB-OA and AC=OC-OA after finding these... equate them and find the value of p.
  $endgroup$
  – Jasser
  Jun 19 '16 at 11:18
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

Just having some difficulty finding the values of '$P$' in this vector.

Relative to an origin $O$, the position vectors of the points $A$ and $B$ are given as follows:

$leftlvert OA rightrvert$ = $2i + 3j - k$
$leftlvert OB rightrvert$ = $4i - 3j + 2k$

The point $C$ is such that $leftlvert OC rightrvert$ = $6j + pk$ where $p$ is a constant. Given the lengths of $leftlvert AB rightrvert$ and $leftlvert AC rightrvert$ are equal, find the possible values of $P$.

--

So what I have done so far is as follows:

$ 2i - 6j + 3k = 9j + pk - 2i - k $

$4i -15j + 3k = pk - k$

Now what I'm doing may be completely stupid or wrong, so apologies in advance, however any pointers or tips would be much appreciated :)

Thanks again!

vectors

share|cite|improve this question

asked Jun 19 '16 at 10:23

M. AndersonM. Anderson

406

$endgroup$

Just having some difficulty finding the values of '$P$' in this vector.

Relative to an origin $O$, the position vectors of the points $A$ and $B$ are given as follows:

$leftlvert OA rightrvert$ = $2i + 3j - k$
$leftlvert OB rightrvert$ = $4i - 3j + 2k$

The point $C$ is such that $leftlvert OC rightrvert$ = $6j + pk$ where $p$ is a constant. Given the lengths of $leftlvert AB rightrvert$ and $leftlvert AC rightrvert$ are equal, find the possible values of $P$.

--

So what I have done so far is as follows:

$ 2i - 6j + 3k = 9j + pk - 2i - k $

$4i -15j + 3k = pk - k$

Now what I'm doing may be completely stupid or wrong, so apologies in advance, however any pointers or tips would be much appreciated :)

Thanks again!

vectors

vectors

share|cite|improve this question

asked Jun 19 '16 at 10:23

M. AndersonM. Anderson

406

share|cite|improve this question

asked Jun 19 '16 at 10:23

M. AndersonM. Anderson

406

share|cite|improve this question

share|cite|improve this question

asked Jun 19 '16 at 10:23

M. AndersonM. Anderson

406

asked Jun 19 '16 at 10:23

M. AndersonM. Anderson

406

asked Jun 19 '16 at 10:23

M. AndersonM. Anderson

406

406

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  AB =OB-OA and AC=OC-OA after finding these... equate them and find the value of p.
  $endgroup$
  – Jasser
  Jun 19 '16 at 11:18
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  AB =OB-OA and AC=OC-OA after finding these... equate them and find the value of p.
  $endgroup$
  – Jasser
  Jun 19 '16 at 11:18
  

  
  
  
  

$begingroup$
AB =OB-OA and AC=OC-OA after finding these... equate them and find the value of p.
$endgroup$
– Jasser
Jun 19 '16 at 11:18

$begingroup$
AB =OB-OA and AC=OC-OA after finding these... equate them and find the value of p.
$endgroup$
– Jasser
Jun 19 '16 at 11:18

add a comment | 

2 Answers
2

active

oldest

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0

$begingroup$

Hint:

It seems that you are confusing the length of a vector with the vector itself. The vector $overrightarrow{AB}=2mathbf{i}-6mathbf{j}+3mathbf{k}$ has lenght $|overrightarrow{AB}|=overline{AB}=sqrt{(2)^2+(-6)^2+(3)^2}$. Do the same for $|overrightarrow{AC}|$ and you find the equation that gives the values of $p$.

share|cite|improve this answer

answered Jun 19 '16 at 13:15

Emilio NovatiEmilio Novati

52.2k43574

$endgroup$

add a comment | 

0

$begingroup$

We know that $OA=(2, 3, -1)$, $OB=(4, -3, 2)$ and $OC=(0, 6, p)$ for $pinBbb K $. Now we use the fact that $AB=OB-OA $ and $AC=OC-OA$ to find the vectors $AB$ and $AC$. We have:
$$AB=(4, -3, 2)-(2, 3, -1)=(2, -6, 3)$$
and:
$$AC=(0, 6, p)-(2, 3, -1)=(-2, 3, p+1)$$
Now we use the formula for the length of a vector in $Bbb K^3$ which yields:
$$||AB||=sqrt {(2)^2+(-6)^2+(3)^2}=7$$
and:
$$||AC||=sqrt {(-2)^2+(3)^2+(p+1)^2}=sqrt {p^2+2p+14}$$
Now you just set $||AB||=||AC||$, since they have the same length, and solve for $p$.

share|cite|improve this answer

answered Jun 19 '16 at 14:45

DaveDave

9,21211033

$endgroup$

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

0

$begingroup$

Hint:

It seems that you are confusing the length of a vector with the vector itself. The vector $overrightarrow{AB}=2mathbf{i}-6mathbf{j}+3mathbf{k}$ has lenght $|overrightarrow{AB}|=overline{AB}=sqrt{(2)^2+(-6)^2+(3)^2}$. Do the same for $|overrightarrow{AC}|$ and you find the equation that gives the values of $p$.

share|cite|improve this answer

answered Jun 19 '16 at 13:15

Emilio NovatiEmilio Novati

52.2k43574

$endgroup$

add a comment | 

0

$begingroup$

Hint:

It seems that you are confusing the length of a vector with the vector itself. The vector $overrightarrow{AB}=2mathbf{i}-6mathbf{j}+3mathbf{k}$ has lenght $|overrightarrow{AB}|=overline{AB}=sqrt{(2)^2+(-6)^2+(3)^2}$. Do the same for $|overrightarrow{AC}|$ and you find the equation that gives the values of $p$.

share|cite|improve this answer

answered Jun 19 '16 at 13:15

Emilio NovatiEmilio Novati

52.2k43574

$endgroup$

add a comment | 

0

0

0

$begingroup$

Hint:

It seems that you are confusing the length of a vector with the vector itself. The vector $overrightarrow{AB}=2mathbf{i}-6mathbf{j}+3mathbf{k}$ has lenght $|overrightarrow{AB}|=overline{AB}=sqrt{(2)^2+(-6)^2+(3)^2}$. Do the same for $|overrightarrow{AC}|$ and you find the equation that gives the values of $p$.

share|cite|improve this answer

answered Jun 19 '16 at 13:15

Emilio NovatiEmilio Novati

52.2k43574

$endgroup$

Hint:

It seems that you are confusing the length of a vector with the vector itself. The vector $overrightarrow{AB}=2mathbf{i}-6mathbf{j}+3mathbf{k}$ has lenght $|overrightarrow{AB}|=overline{AB}=sqrt{(2)^2+(-6)^2+(3)^2}$. Do the same for $|overrightarrow{AC}|$ and you find the equation that gives the values of $p$.

share|cite|improve this answer

answered Jun 19 '16 at 13:15

Emilio NovatiEmilio Novati

52.2k43574

share|cite|improve this answer

share|cite|improve this answer

answered Jun 19 '16 at 13:15

Emilio NovatiEmilio Novati

52.2k43574

answered Jun 19 '16 at 13:15

Emilio NovatiEmilio Novati

52.2k43574

answered Jun 19 '16 at 13:15

Emilio NovatiEmilio Novati

52.2k43574

52.2k43574

add a comment | 

add a comment | 

0

$begingroup$

We know that $OA=(2, 3, -1)$, $OB=(4, -3, 2)$ and $OC=(0, 6, p)$ for $pinBbb K $. Now we use the fact that $AB=OB-OA $ and $AC=OC-OA$ to find the vectors $AB$ and $AC$. We have:
$$AB=(4, -3, 2)-(2, 3, -1)=(2, -6, 3)$$
and:
$$AC=(0, 6, p)-(2, 3, -1)=(-2, 3, p+1)$$
Now we use the formula for the length of a vector in $Bbb K^3$ which yields:
$$||AB||=sqrt {(2)^2+(-6)^2+(3)^2}=7$$
and:
$$||AC||=sqrt {(-2)^2+(3)^2+(p+1)^2}=sqrt {p^2+2p+14}$$
Now you just set $||AB||=||AC||$, since they have the same length, and solve for $p$.

share|cite|improve this answer

answered Jun 19 '16 at 14:45

DaveDave

9,21211033

$endgroup$

add a comment | 

0

$begingroup$

We know that $OA=(2, 3, -1)$, $OB=(4, -3, 2)$ and $OC=(0, 6, p)$ for $pinBbb K $. Now we use the fact that $AB=OB-OA $ and $AC=OC-OA$ to find the vectors $AB$ and $AC$. We have:
$$AB=(4, -3, 2)-(2, 3, -1)=(2, -6, 3)$$
and:
$$AC=(0, 6, p)-(2, 3, -1)=(-2, 3, p+1)$$
Now we use the formula for the length of a vector in $Bbb K^3$ which yields:
$$||AB||=sqrt {(2)^2+(-6)^2+(3)^2}=7$$
and:
$$||AC||=sqrt {(-2)^2+(3)^2+(p+1)^2}=sqrt {p^2+2p+14}$$
Now you just set $||AB||=||AC||$, since they have the same length, and solve for $p$.

share|cite|improve this answer

answered Jun 19 '16 at 14:45

DaveDave

9,21211033

$endgroup$

add a comment | 

0

0

0

$begingroup$

We know that $OA=(2, 3, -1)$, $OB=(4, -3, 2)$ and $OC=(0, 6, p)$ for $pinBbb K $. Now we use the fact that $AB=OB-OA $ and $AC=OC-OA$ to find the vectors $AB$ and $AC$. We have:
$$AB=(4, -3, 2)-(2, 3, -1)=(2, -6, 3)$$
and:
$$AC=(0, 6, p)-(2, 3, -1)=(-2, 3, p+1)$$
Now we use the formula for the length of a vector in $Bbb K^3$ which yields:
$$||AB||=sqrt {(2)^2+(-6)^2+(3)^2}=7$$
and:
$$||AC||=sqrt {(-2)^2+(3)^2+(p+1)^2}=sqrt {p^2+2p+14}$$
Now you just set $||AB||=||AC||$, since they have the same length, and solve for $p$.

share|cite|improve this answer

answered Jun 19 '16 at 14:45

DaveDave

9,21211033

$endgroup$

We know that $OA=(2, 3, -1)$, $OB=(4, -3, 2)$ and $OC=(0, 6, p)$ for $pinBbb K $. Now we use the fact that $AB=OB-OA $ and $AC=OC-OA$ to find the vectors $AB$ and $AC$. We have:
$$AB=(4, -3, 2)-(2, 3, -1)=(2, -6, 3)$$
and:
$$AC=(0, 6, p)-(2, 3, -1)=(-2, 3, p+1)$$
Now we use the formula for the length of a vector in $Bbb K^3$ which yields:
$$||AB||=sqrt {(2)^2+(-6)^2+(3)^2}=7$$
and:
$$||AC||=sqrt {(-2)^2+(3)^2+(p+1)^2}=sqrt {p^2+2p+14}$$
Now you just set $||AB||=||AC||$, since they have the same length, and solve for $p$.

share|cite|improve this answer

answered Jun 19 '16 at 14:45

DaveDave

9,21211033

share|cite|improve this answer

share|cite|improve this answer

answered Jun 19 '16 at 14:45

DaveDave

9,21211033

answered Jun 19 '16 at 14:45

DaveDave

9,21211033

answered Jun 19 '16 at 14:45

DaveDave

9,21211033

9,21211033

add a comment | 

add a comment | 

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