QuestionGiven lines $O A N, O M B$ and $A P B$ , with $A N = 2 O A$ . If $M$ is the midpoint of $O B$ , find $k$ in $\overrightarrow{A P} = k \overrightarrow{A B}$ , given $MPN$ is straight.

Studdy Solution

STEP 1

Assumptions1. $A N, O M B$ and $A B$ are straight lines. . $A N= O A$
3. $M$ is the midpoint of $B$ .
4. $\overrightarrow{ A}=\mathbf{a}$ , $\overrightarrow{ B}=\mathbf{b}$
5. $\overrightarrow{A}=k \overrightarrow{A B}$ where $k$ is a scalar quantity.
6. MP is a straight line.

STEP 2

First, we need to find the vector $\overrightarrow{A N}$ and $\overrightarrow{A M}$ using the given information.
Since $A N=2 O A$ , we have $\overrightarrow{A N} =2 \overrightarrow{ A} =2\mathbf{a}$ .
Since $M$ is the midpoint of $B$ , we have $\overrightarrow{A M} = \overrightarrow{A O} + \overrightarrow{ M} = -\mathbf{a} + \frac{1}{2}\mathbf{b}$ .

STEP 3

Now, we know that $\overrightarrow{M N} = \overrightarrow{M A} + \overrightarrow{A N}$ .
Substitute the values of $\overrightarrow{A N}$ and $\overrightarrow{A M}$ into the equation.

STEP 4

Calculate $\overrightarrow{M N}$ .
$\overrightarrow{M N} = -\overrightarrow{A M} + \overrightarrow{A N} = -(-\mathbf{a} + \frac{1}{2}\mathbf{b}) +2\mathbf{a} = \mathbf{a} + \frac{1}{2}\mathbf{b}$

STEP 5

Since MP is a straight line, $\overrightarrow{M N}$ is a scalar multiple of $\overrightarrow{M}$ .
So, $\overrightarrow{M N} = k \overrightarrow{M}$ .

STEP 6

We know that $\overrightarrow{M} = \overrightarrow{M A} + \overrightarrow{A}$ .
Substitute the value of $\overrightarrow{A}$ into the equation.

STEP 7

Calculate $\overrightarrow{M}$ .
$\overrightarrow{M} = \overrightarrow{M A} + \overrightarrow{A} = -\overrightarrow{A M} + k\overrightarrow{A B} = -(-\mathbf{a} + \frac{1}{2}\mathbf{b}) + k(\mathbf{b} - \mathbf{a}) = (1+k)\mathbf{a} + (k-\frac{1}{2})\mathbf{b}$

STEP 8

Now, equate $\overrightarrow{M N}$ and $k\overrightarrow{M}$ .
$\mathbf{a} + \frac{1}{2}\mathbf{b} = k[(1+k)\mathbf{a} + (k-\frac{1}{2})\mathbf{b}]$

STEP 9

Equating the coefficients of $\mathbf{a}$ and $\mathbf{b}$ , we get two equations. $= k(+k)$ 2. $\frac{}{2} = k(k-\frac{}{2})$

STEP 10

olving the first equation for $k$ , we get $k^2 + k - =0$ .

STEP 11

olving the quadratic equation, we get $k = \frac{- \pm \sqrt{ +4}}{}$ .

STEP 12

Calculate the value of $k$ .
$k = \frac{- \pm \sqrt{5}}{2}$ Since $k$ cannot be negative in this context, we discard the negative solution.
So, $k = \frac{- + \sqrt{5}}{2}$ .
The value of $k$ is $\frac{- + \sqrt{5}}{2}$ .

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QuestionGiven lines $O A N, O M B$ and $A P B$ , with $A N = 2 O A$ . If $M$ is the midpoint of $O B$ , find $k$ in $\overrightarrow{A P} = k \overrightarrow{A B}$ , given $MPN$ is straight.

Studdy Solution

STEP 1

Assumptions1. $A N, O M B$ and $A B$ are straight lines. . $A N= O A$
3. $M$ is the midpoint of $B$ .
4. $\overrightarrow{ A}=\mathbf{a}$ , $\overrightarrow{ B}=\mathbf{b}$
5. $\overrightarrow{A}=k \overrightarrow{A B}$ where $k$ is a scalar quantity.
6. MP is a straight line.

STEP 2

STEP 3

Now, we know that $\overrightarrow{M N} = \overrightarrow{M A} + \overrightarrow{A N}$ .
Substitute the values of $\overrightarrow{A N}$ and $\overrightarrow{A M}$ into the equation.

STEP 4

Calculate $\overrightarrow{M N}$ .
$\overrightarrow{M N} = -\overrightarrow{A M} + \overrightarrow{A N} = -(-\mathbf{a} + \frac{1}{2}\mathbf{b}) +2\mathbf{a} = \mathbf{a} + \frac{1}{2}\mathbf{b}$

STEP 5

Since MP is a straight line, $\overrightarrow{M N}$ is a scalar multiple of $\overrightarrow{M}$ .
So, $\overrightarrow{M N} = k \overrightarrow{M}$ .

STEP 6

We know that $\overrightarrow{M} = \overrightarrow{M A} + \overrightarrow{A}$ .
Substitute the value of $\overrightarrow{A}$ into the equation.

STEP 7

Calculate $\overrightarrow{M}$ .
$\overrightarrow{M} = \overrightarrow{M A} + \overrightarrow{A} = -\overrightarrow{A M} + k\overrightarrow{A B} = -(-\mathbf{a} + \frac{1}{2}\mathbf{b}) + k(\mathbf{b} - \mathbf{a}) = (1+k)\mathbf{a} + (k-\frac{1}{2})\mathbf{b}$

STEP 8

Now, equate $\overrightarrow{M N}$ and $k\overrightarrow{M}$ .
$\mathbf{a} + \frac{1}{2}\mathbf{b} = k[(1+k)\mathbf{a} + (k-\frac{1}{2})\mathbf{b}]$

STEP 9

Equating the coefficients of $\mathbf{a}$ and $\mathbf{b}$ , we get two equations. $= k(+k)$ 2. $\frac{}{2} = k(k-\frac{}{2})$

STEP 10

olving the first equation for $k$ , we get $k^2 + k - =0$ .

STEP 11

olving the quadratic equation, we get $k = \frac{- \pm \sqrt{ +4}}{}$ .

STEP 12

Calculate the value of $k$ .
$k = \frac{- \pm \sqrt{5}}{2}$ Since $k$ cannot be negative in this context, we discard the negative solution.
So, $k = \frac{- + \sqrt{5}}{2}$ .
The value of $k$ is $\frac{- + \sqrt{5}}{2}$ .

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QuestionGiven lines OAN,OMBO A N, O M BOAN,OMB and APBA P BAPB, with AN=2OAA N = 2 O AAN=2OA. If MMM is the midpoint of OBO BOB, find kkk in APundefined=kABundefined\overrightarrow{A P} = k \overrightarrow{A B}AP=kAB, given MPNMPNMPN is straight.

Studdy Solution

STEP 1

STEP 2

STEP 3

Now, we know that MNundefined=MAundefined+ANundefined\overrightarrow{M N} = \overrightarrow{M A} + \overrightarrow{A N}MN=MA+AN.Substitute the values of ANundefined\overrightarrow{A N}AN and AMundefined\overrightarrow{A M}AM into the equation.

STEP 4

STEP 5

Since MP is a straight line, MNundefined\overrightarrow{M N}MN is a scalar multiple of Mundefined\overrightarrow{M}M.So, MNundefined=kMundefined\overrightarrow{M N} = k \overrightarrow{M}MN=kM.

STEP 6

We know that Mundefined=MAundefined+Aundefined\overrightarrow{M} = \overrightarrow{M A} + \overrightarrow{A}M=MA+A.Substitute the value of Aundefined\overrightarrow{A}A into the equation.

STEP 7

STEP 8

Now, equate MNundefined\overrightarrow{M N}MN and kMundefinedk\overrightarrow{M}kM.a+12b=k[(1+k)a+(k−12)b]\mathbf{a} + \frac{1}{2}\mathbf{b} = k[(1+k)\mathbf{a} + (k-\frac{1}{2})\mathbf{b}]a+21​b=k[(1+k)a+(k−21​)b]

STEP 9

Equating the coefficients of a\mathbf{a}a and b\mathbf{b}b, we get two equations. =k(+k) = k(+k)=k(+k)2. 2=k(k−2)\frac{}{2} = k(k-\frac{}{2})2​=k(k−2​)

STEP 10

olving the first equation for kkk, we get k2+k−=0k^2 + k - =0k2+k−=0.

STEP 11

olving the quadratic equation, we get k=−±+4k = \frac{- \pm \sqrt{ +4}}{}k=−±+4​​.

STEP 12

Calculate the value of kkk.k=−±52k = \frac{- \pm \sqrt{5}}{2}k=2−±5​​Since kkk cannot be negative in this context, we discard the negative solution.So, k=−+52k = \frac{- + \sqrt{5}}{2}k=2−+5​​.The value of kkk is −+52\frac{- + \sqrt{5}}{2}2−+5​​.

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QuestionGiven lines $O A N, O M B$ and $A P B$ , with $A N = 2 O A$ . If $M$ is the midpoint of $O B$ , find $k$ in $\overrightarrow{A P} = k \overrightarrow{A B}$ , given $MPN$ is straight.

Now, we know that $\overrightarrow{M N} = \overrightarrow{M A} + \overrightarrow{A N}$ .
Substitute the values of $\overrightarrow{A N}$ and $\overrightarrow{A M}$ into the equation.

Since MP is a straight line, $\overrightarrow{M N}$ is a scalar multiple of $\overrightarrow{M}$ .
So, $\overrightarrow{M N} = k \overrightarrow{M}$ .

We know that $\overrightarrow{M} = \overrightarrow{M A} + \overrightarrow{A}$ .
Substitute the value of $\overrightarrow{A}$ into the equation.

Now, equate $\overrightarrow{M N}$ and $k\overrightarrow{M}$ .
$\mathbf{a} + \frac{1}{2}\mathbf{b} = k[(1+k)\mathbf{a} + (k-\frac{1}{2})\mathbf{b}]$

Equating the coefficients of $\mathbf{a}$ and $\mathbf{b}$ , we get two equations. $= k(+k)$ 2. $\frac{}{2} = k(k-\frac{}{2})$

olving the first equation for $k$ , we get $k^2 + k - =0$ .

olving the quadratic equation, we get $k = \frac{- \pm \sqrt{ +4}}{}$ .

Calculate the value of $k$ .
$k = \frac{- \pm \sqrt{5}}{2}$ Since $k$ cannot be negative in this context, we discard the negative solution.
So, $k = \frac{- + \sqrt{5}}{2}$ .
The value of $k$ is $\frac{- + \sqrt{5}}{2}$ .