Vectors Help Urgent (1 Viewer)

csi · Feb 10, 2021

Hi guys,

That attached are some questions that I need help with. The answers printed in small font on the bottom right hand side of each question (the answer for the last question got cut off, it’s meant to be 1/2)

Many thanks

zerdaunal1 · Feb 12, 2021

csi said:
Hi guys,

That attached are some questions that I need help with. The answers printed in small font on the bottom right hand side of each question (the answer for the last question got cut off, it’s meant to be 1/2)

Many thanks

Hii, for the first part, instead of evaluating AB, try using the midpoint formula for vectors OA and OB, this gives you the answer you need.

for ii) vector CM = OM-OC (we just found OM and already are given OC)

for iii, we find the magnitude for both OM and CM, equate it to one another (given) and then expand and solve for k

for iii)
We know that collinearity refers to the direction of the vectors (relative to x-axis) are equal. Consider how to compute the direction of a vector and choose two vectors for example OA and AB, find their direction and equate them, once you’ve done this you can solve for k. (Check attached file)
p.s sorry for the messy handwriting, hope this helps

CM_Tutor · Feb 12, 2021

We are given that:

$\overrightarrow{OA} = 2k\overrightarrow{i} + (k+1)\overrightarrow{j}$
$\overrightarrow{OB} = 4k\overrightarrow{i} + (k+5)\overrightarrow{j}$
$\overrightarrow{OC} = (10+k)\overrightarrow{i}$
$M$ is the midpoint of $AB$

Part (a)

$\begin{align*} \overrightarrow{AB} &= \overrightarrow{AO} + \overrightarrow{OB} \\ &= -\overrightarrow{OA} + \overrightarrow{OB} \\ &= -\left[2k\overrightarrow{i} + (k+1)\overrightarrow{j}\right] + 4k\overrightarrow{i} + (k+5)\overrightarrow{j} \\ &= \left(-2k + 4k\right)\overrightarrow{i} + \left[-(k+1) + (k+5)\right]\overrightarrow{j} \\ &= 2k\overrightarrow{i} + 4\overrightarrow{j} \end{align*}$

Now, as $M$ is the midpoint of $AB$ , $\overrightarrow{AM} = \overrightarrow{MB}$ and $\overrightarrow{AM} + \overrightarrow{MB} = \overrightarrow{AB}$ . Thus,

$\overrightarrow{AM} = \frac{1}{2}\overrightarrow{AB} = \frac{1}{2}\left(2k\overrightarrow{i} + 4\overrightarrow{j}\right) = k\overrightarrow{i} + 2\overrightarrow{j}$

and so

$\begin{align*} \overrightarrow{OM} &= \overrightarrow{OA} + \overrightarrow{AM} \\ &= 2k\overrightarrow{i} + (k+1)\overrightarrow{j} + k\overrightarrow{i} + 2\overrightarrow{j} \\ &= 3k\overrightarrow{i} + (k+3)\overrightarrow{j} \end{align*}$

Part (b)

$\begin{align*} \overrightarrow{CM} &= \overrightarrow{CO} + \overrightarrow{OM} \\ &= -\overrightarrow{OC} + \overrightarrow{OM} \\ &= -(10+k)\overrightarrow{i} + 3k\overrightarrow{i} + (k+3)\overrightarrow{j} \\ &= (2k-10)\overrightarrow{i} + (k+3)\overrightarrow{j} \end{align*}$

Part (c)

$\begin{align*} \left|\overrightarrow{OM}\right| &= \left|\overrightarrow{CM}\right| \\ \left|3k\overrightarrow{i} + (k+3)\overrightarrow{j}\right| &= \left|(2k-10)\overrightarrow{i} + (k+3)\overrightarrow{j}\right| \\ \sqrt{(3k)^2 + (k+3)^2} &= \sqrt{(2k-10)^2 + (k+3)^2} \\ 9k^2 + (k+3)^2 &= (2k-10)^2 + (k+3)^2 \\ 9k^2 &= (2k-10)^2 \\ 3k &= \pm(2k - 10) \\ 3k &= 2k - 10 \qquad \qquad \text{or} \qquad \qquad 3k = -2k + 10 \\ k &= -10 \qquad \qquad \qquad \text{or} \qquad \qquad \qquad k = 2 \end{align*}$

So, since we are given that $k>0$ , the only solution to $\left|\overrightarrow{OM}\right| = \left|\overrightarrow{CM}\right|$ is $k = 2$ .

Part (d)

We already know that $A$ , $M$ , and $B$ , are collinear as $M$ is the midpoint of $AB$ and must lie on $AB$ . If $O$ also lies on this line, then there must be some constant $\lambda \in \mathbb{R}$ such that $\overrightarrow{OA} = \lambda\overrightarrow{AB}$ :

$\begin{align*} \overrightarrow{OA} &= \lambda\overrightarrow{AB} \\ 2k\overrightarrow{i} + (k+1)\overrightarrow{j} &= \lambda\left(2k\overrightarrow{i} + 4\overrightarrow{j}\right) \\ 2k\overrightarrow{i} + (k+1)\overrightarrow{j} &= 2k\lambda\overrightarrow{i} + 4\lambda\overrightarrow{j} \end{align*}$

Equating the $\overrightarrow{i}$ and $\overrightarrow{j}$ components, we see that

$2k = 2k\lambda \implies \lambda = 1$

and then that

$k + 1 = 2k\lambda \implies k=3$

is the required value of $k$ for $O$ to lie on the line that includes $A$ , $M$ , and $B$ , and thus that these four points are collinear.

Note that the same result, $k=3$ , could be obtained with other combinations of vectors (which would then yield different values of $\lambda$ ), including

$\overrightarrow{OA} = \lambda\overrightarrow{OM} \implies \lambda = \frac{2}{3}$
$\overrightarrow{OA} = \lambda\overrightarrow{OB} \implies \lambda = \frac{1}{2}$
$\overrightarrow{OA} = \lambda\overrightarrow{AM} \implies \lambda = 2$
$\overrightarrow{OA} = \lambda\overrightarrow{MB} \implies \lambda = -2$

CM_Tutor · Feb 12, 2021

Question 2

We are given that:

$\overrightarrow{OA} = \overrightarrow{a} + 2\overrightarrow{b}$
$\overrightarrow{OB} = 3\overrightarrow{a} - \overrightarrow{b}$
$\overrightarrow{OC} = 2\overrightarrow{a} + k\overrightarrow{b}$
$A$ , $B$ , and $C$ are collinear

For $C$ to lie on $AB$ , there must be some constant $\lambda \in \mathbb{R}$ such that $\overrightarrow{AC} = \lambda\overrightarrow{AB}$ :

$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = 3\overrightarrow{a} - \overrightarrow{b} - \left(\overrightarrow{a} + 2\overrightarrow{b}\right) = 2\overrightarrow{a} - 3\overrightarrow{b}$

$\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = 2\overrightarrow{a} + k\overrightarrow{b} - \left(\overrightarrow{a} + 2\overrightarrow{b}\right) = \overrightarrow{a} + (k-2)\overrightarrow{b}$

$\begin{align*} \overrightarrow{AC} &= \lambda\overrightarrow{AB} \\ \overrightarrow{a} + (k-2)\overrightarrow{b} &= \lambda\left(2\overrightarrow{a} - 3\overrightarrow{b}\right) \\ \overrightarrow{a} + (k-2)\overrightarrow{b} &= 2\lambda\overrightarrow{a} - 3\lambda\overrightarrow{b} \end{align*}$

Equating the $\overrightarrow{a}$ and $\overrightarrow{b}$ components, we see that

$1 = 2\lambda \implies \lambda = \frac{1}{2}$

and then that

$k - 2 = -3\lambda \implies k=\frac{1}{2}$

is the required value of $k$ for $C$ to lie on the line that includes $A$ and $B$ .

csi · Feb 20, 2021

CM_Tutor said:
Question 2

We are given that:

$\overrightarrow{OA} = \overrightarrow{a} + 2\overrightarrow{b}$

$\overrightarrow{OB} = 3\overrightarrow{a} - \overrightarrow{b}$

$\overrightarrow{OC} = 2\overrightarrow{a} + k\overrightarrow{b}$

$A$ , $B$ , and $C$ are collinear

For $C$ to lie on $AB$ , there must be some constant $\lambda \in \mathbb{R}$ such that $\overrightarrow{AC} = \lambda\overrightarrow{AB}$ :

$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = 3\overrightarrow{a} - \overrightarrow{b} - \left(\overrightarrow{a} + 2\overrightarrow{b}\right) = 2\overrightarrow{a} - 3\overrightarrow{b}$

$\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = 2\overrightarrow{a} + k\overrightarrow{b} - \left(\overrightarrow{a} + 2\overrightarrow{b}\right) = \overrightarrow{a} + (k-2)\overrightarrow{b}$

$\begin{align*} \overrightarrow{AC} &= \lambda\overrightarrow{AB} \\ \overrightarrow{a} + (k-2)\overrightarrow{b} &= \lambda\left(2\overrightarrow{a} - 3\overrightarrow{b}\right) \\ \overrightarrow{a} + (k-2)\overrightarrow{b} &= 2\lambda\overrightarrow{a} - 3\lambda\overrightarrow{b} \end{align*}$

Equating the $\overrightarrow{a}$ and $\overrightarrow{b}$ components, we see that

$1 = 2\lambda \implies \lambda = \frac{1}{2}$

and then that

$k - 2 = -3\lambda \implies k=\frac{1}{2}$

is the required value of $k$ for $C$ to lie on the line that includes $A$ and $B$ .

tysm

csi · Feb 20, 2021

zerdaunal1 said:
Hii, for the first part, instead of evaluating AB, try using the midpoint formula for vectors OA and OB, this gives you the answer you need.

for ii) vector CM = OM-OC (we just found OM and already are given OC)

for iii, we find the magnitude for both OM and CM, equate it to one another (given) and then expand and solve for k

for iii)
We know that collinearity refers to the direction of the vectors (relative to x-axis) are equal. Consider how to compute the direction of a vector and choose two vectors for example OA and AB, find their direction and equate them, once you’ve done this you can solve for k. (Check attached file)
p.s sorry for the messy handwriting, hope this helps

ty!!

zerdaunal1 · Feb 21, 2021

csi said:
ty!!

No worries, hope it was helpful

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