1. ## Question on vectors

I have attached my question

here is my attempt:

I have attached my question

Here's my attempt :
$\displaystyle \overrightarrow{OA}= 2\mathbf{i}+3\mathbf{j}$

$\displaystyle \overrightarrow{OB}= 3\mathbf{i} - \mathbf{j} + k(\mathbf{i}+2\mathbf{j})$

$\displaystyle \overrightarrow{OC}= 4\mathbf{i} + 13\mathbf{j} + k(\mathbf{4i}-\mathbf{j})$

$\displaystyle \overrightarrow{OB}= (3+k)\mathbf{i} + (2k-1)\mathbf{j}$
$\displaystyle \overrightarrow{OC}= (4+4k)\mathbf{i}+(13-k)\mathbf{j}$
$\displaystyle (4+4k)\mathbf{i}+(13-k)\mathbf{j} = (3+k)\mathbf{i} + (2k-1)\mathbf{j}$ since OB is parallel to OC
thus
$\displaystyle (4+4k)\mathbf{i} = (3+k)\mathbf{i}$

$\displaystyle (13-k)\mathbf{j} =(2k-1)\mathbf{j}$

$\displaystyle (4+4k) = (3+k)$

$\displaystyle 4k-k = 3-4$

$\displaystyle k = -\frac{1}{3}$

$\displaystyle (13-k) =(2k-1)$

$\displaystyle 13 +1 = 2k+k$

$\displaystyle k = \frac{14}{3}$

but in the book the answer is 3 or -2/3

please tell me where i did wrong?

thank you for helping

2. ## Re: Question on vectors

positions after motion ...

$A(2,3)$

$B(k+3,2k-1)$

$C(4k+4,13-k)$

for A, B, and C to be collinear, vectors AB and BC need to have the same direction.

slope for AB ... $\dfrac{2k-4}{1+k}$

slope for BC ... $\dfrac{14-3k}{3k+1}$

$\dfrac{2k-4}{1+k} = \dfrac{14-3k}{3k+1}$

solve the resulting quadratic for $k$

3. ## Re: Question on vectors

Originally Posted by bigmansouf
I have attached my question

here is my attempt:

I have attached my question

Here's my attempt :
$\displaystyle \overrightarrow{OA}= 2\mathbf{i}+3\mathbf{j}$

$\displaystyle \overrightarrow{OB}= 3\mathbf{i} - \mathbf{j} + k(\mathbf{i}+2\mathbf{j})$

$\displaystyle \overrightarrow{OC}= 4\mathbf{i} + 13\mathbf{j} + k(\mathbf{4i}-\mathbf{j})$

$\displaystyle \overrightarrow{OB}= (3+k)\mathbf{i} + (2k-1)\mathbf{j}$
$\displaystyle \overrightarrow{OC}= (4+4k)\mathbf{i}+(13-k)\mathbf{j}$
$\displaystyle (4+4k)\mathbf{i}+(13-k)\mathbf{j} = (3+k)\mathbf{i} + (2k-1)\mathbf{j}$ since OB is parallel to OC
No! Saying to vectors are "parallel" does not say they are equal. It says one is a multiple of the other.
4+ 4k= c(3+ k) and 13- k= c(2k- 1), two equations to solve for c and k

thus
$\displaystyle (4+4k)\mathbf{i} = (3+k)\mathbf{i}$

$\displaystyle (13-k)\mathbf{j} =(2k-1)\mathbf{j}$

$\displaystyle (4+4k) = (3+k)$

$\displaystyle 4k-k = 3-4$

$\displaystyle k = -\frac{1}{3}$

$\displaystyle (13-k) =(2k-1)$

$\displaystyle 13 +1 = 2k+k$

$\displaystyle k = \frac{14}{3}$

but in the book the answer is 3 or -2/3

please tell me where i did wrong?

thank you for helping
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