# Thread: Vector

1. ## Vector

In ∆OAB, P is the mid-point of AB and Q is the point on OP such that Q = 3/4P. Given that $\overrightarrow {OA} = \underline{a}$ and $\overrightarrow {OB} = \underline{b}.$

The point R on OB is such that OR = kOB, where 0<k<1.

Find, in terms of a, b and k, the vector $\overrightarrow{AR}.$

I found this: $\overrightarrow{AR} = - \underline{a} + k\underline{b}$

Given that AQR is a straight line:
Find the ratio in which Q divides AR and the value of k.

I did:
$
\overrightarrow{AR} = \overrightarrow{AQ} + \overrightarrow{QR}$

$\overrightarrow{QR} = -\frac{3}{8} \underline{a} + (k - \frac{3}{8})\underline{b}$

I’m confused to find the ratio & the value of k.

Is it like:
$
\overrightarrow{AR} = \overrightarrow{AQ} : \overrightarrow{QR}
=(\frac{3}{8} \underline{b} - \frac{5}{8} \underline{a}) : (k - \frac{3}{8})\underline{b} - \frac{3}{8}\underline{a}$

Then…?

2. Originally Posted by geton
In ∆OAB, P is the mid-point of AB and Q is the point on OP such that Q = 3/4P. Given that $\overrightarrow {OA} = \underline{a}$ and $\overrightarrow {OB} = \underline{b}.$

The point R on OB is such that OR = kOB, where 0<k<1.

Find, in terms of a, b and k, the vector $\overrightarrow{AR}.$

I found this: $\overrightarrow{AR} = - \underline{a} + k\underline{b}$

Given that AQR is a straight line:
Find the ratio in which Q divides AR and the value of k.

I did:
$
\overrightarrow{AR} = \overrightarrow{AQ} + \overrightarrow{QR}$

$\overrightarrow{QR} = -\frac{3}{8} \underline{a} + (k - \frac{3}{8})\underline{b}$

I’m confused to find the ratio & the value of k.

Is it like:
$
\overrightarrow{AR} = \overrightarrow{AQ} : \overrightarrow{QR}
=(\frac{3}{8} \underline{b} - \frac{5}{8} \underline{a}) : (k - \frac{3}{8})\underline{b} - \frac{3}{8}\underline{a}$

Then…?
1. $\overrightarrow{AQ}=-\vec a+\frac34\left(\vec a+\frac12\left(\vec b - \vec a\right) \right) = -\frac58 \vec a + \frac38 \vec b$

2. $\overrightarrow{AR} = r \cdot (\overrightarrow{AQ}) = -\frac58 r \cdot \vec a + \frac38 r \cdot \vec b$

3. $\overrightarrow{AR} = - \vec{a} + k\vec{b}$

4. Compare the coefficients at 2. and 3.:

$-\frac58 r = -1~\implies~r = \frac85$

$\frac38 r = k~\implies~\frac38 \cdot \frac85 = k~\implies~k = \frac35$

5. Calculate the ratio.

3. Originally Posted by earboth
1. $\overrightarrow{AQ}=-\vec a+\frac34\left(\vec a+\frac12\left(\vec b - \vec a\right) \right) = -\frac58 \vec a + \frac38 \vec b$

2. $\overrightarrow{AR} = r \cdot (\overrightarrow{AQ}) = -\frac58 r \cdot \vec a + \frac38 r \cdot \vec b$

3. $\overrightarrow{AR} = - \vec{a} + k\vec{b}$

4. Compare the coefficients at 2. and 3.:

$-\frac58 r = -1~\implies~r = \frac85$

$\frac38 r = k~\implies~\frac38 \cdot \frac85 = k~\implies~k = \frac35$

5. Calculate the ratio.
Thank you so much

,

### divides ar lokkon somoho ki

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