Dot Products and Projections HELP PLZ (1 Viewer)

csi · Feb 20, 2021

Hi guys,

I'm stuck on the following questions, please give me a hand.
Note: all the letters (vectors) below are meant to have a tilde beneath it, it's just me not knowing how to type it out

1. If a·b=3, a·c=4, b·c=-2 and |b| =1 then evaluate (a-b)·(2b+3c)
ANS: 22

2. Find the possible values of k if the projection of ki+4j onto 12i-5j has length 140/12
ANS: 40/3

3. Points P and A are on the number plane. The vector is OA is (3,1). Point B is chosen so that the area of triangle PAB is 10u^2 and |OB| is 4 root 5. Find all possible vectors for OP.

Thanks!!!

CM_Tutor · Feb 20, 2021

csi said:
Note: all the letters (vectors) below are meant to have a tilde beneath it, it's just me not knowing how to type it out

I've been wondering the same as the methods I am aware of / can find by searching don't work here... so I have posted a question on it.

https://community.boredofstudies.org/threads/vectors-and-tex.394068/

~~I haven't had any answers so far, however.~~ @Trebla has just replied that packages are needed for this

and has also suggested the bold notation that I describe below.

~~While we wait,~~ we can use other conventions. In the following, replace each hyphen (-) with a slash(\):

Display vectors as bold:

$\text{"-mathbf\{v\} = a-mathbf\{i\} + b-mathbf\{j\}"} \qquad \text{produces} \qquad \mathbf{v} = a\mathbf{i} + b\mathbf{j}$

Display vectors using the arrow above notation using the command \overrightarrow:

$\text{"-overrightarrow\{OA\}"} \qquad \text{produces} \qquad \overrightarrow{OA}$

$\text{"-overrightarrow\{v\} = 2-overrightarrow\{i\} + 3-overrightarrow\{j\}"} \\ \\ \text{produces} \\ \\ \overrightarrow{v} = 2\overrightarrow{i} + 3\overrightarrow{j}$

CM_Tutor · Feb 20, 2021

I'll start... hopefully others will jump in too

Question 1

$\text{If} \qquad \overrightarrow{a}\cdot\overrightarrow{b}=3 \qquad \overrightarrow{a}\cdot\overrightarrow{c}=4 \qquad \overrightarrow{b}\cdot\overrightarrow{c}=-2 \qquad \text{and} \qquad |b| =1$

then evaluate

$\left(\overrightarrow{a}-\overrightarrow{b}\right)\cdot\left(2\overrightarrow{b}+3\overrightarrow{c}\right)$

Observations:

Expansions of expressions like the one we are given are just like binomial expansions
Dot products are commutative, and so $\mathbf{a}\cdot\mathbf{b} = \mathbf{b}\cdot\mathbf{a}$
The magnitude of a vector, $\left|\overrightarrow{a}\right|$ , is related to the dot product by $\left|\overrightarrow{a}\right|^2 = \overrightarrow{a}\cdot\overrightarrow{a}$

So, the answer is:

$\begin{align*} \left(\overrightarrow{a}-\overrightarrow{b}\right)\cdot\left(2\overrightarrow{b}+3\overrightarrow{c}\right) &= \overrightarrow{a}\cdot 2\overrightarrow{b} - \overrightarrow{b}\cdot 2\overrightarrow{b} + \overrightarrow{a}\cdot 3 \overrightarrow{c} - \overrightarrow{b}\cdot 3 \overrightarrow{c} \\ &= 2\overrightarrow{a}\cdot\overrightarrow{b} - 2\overrightarrow{b}\cdot\overrightarrow{b} + 3\overrightarrow{a}\cdot\overrightarrow{c} - 3\overrightarrow{b}\cdot\overrightarrow{c} \\ &= 2 \times 3 - 2\left|\overrightarrow{b}\right|^2 + 3 \times 4 - 3 \times -2 \\ &= 6 - 2 \times 1^2 + 12 + 6 \\ &= 22 \end{align*}$

csi · Feb 20, 2021

CM_Tutor said:
I'll start... hopefully others will jump in too

Question 1

$\text{If} \qquad \overrightarrow{a}\cdot\overrightarrow{b}=3 \qquad \overrightarrow{a}\cdot\overrightarrow{c}=4 \qquad \overrightarrow{b}\cdot\overrightarrow{c}=-2 \qquad \text{and} \qquad |b| =1$

then evaluate

$\left(\overrightarrow{a}-\overrightarrow{b}\right)\cdot\left(2\overrightarrow{b}+3\overrightarrow{c}\right)$

Observations:

Expansions of expressions like the one we are given are just like binomial expansions

Dot products are commutative, and so $\mathbf{a}\cdot\mathbf{b} = \mathbf{b}\cdot\mathbf{a}$

The magnitude of a vector, $\left|\overrightarrow{a}\right|$ , is related to the dot product by $\left|\overrightarrow{a}\right|^2 = \overrightarrow{a}\cdot\overrightarrow{a}$

So, the answer is:

$\begin{align*} \left(\overrightarrow{a}-\overrightarrow{b}\right)\cdot\left(2\overrightarrow{b}+3\overrightarrow{c}\right) &= \overrightarrow{a}\cdot 2\overrightarrow{b} - \overrightarrow{b}\cdot 2\overrightarrow{b} + \overrightarrow{a}\cdot 3 \overrightarrow{c} - \overrightarrow{b}\cdot 3 \overrightarrow{c} \\ &= 2\overrightarrow{a}\cdot\overrightarrow{b} - 2\overrightarrow{b}\cdot\overrightarrow{b} + 3\overrightarrow{a}\cdot\overrightarrow{c} - 3\overrightarrow{b}\cdot\overrightarrow{c} \\ &= 2 \times 3 - 2\left|\overrightarrow{b}\right|^2 + 3 \times 4 - 3 \times -2 \\ &= 6 - 2 \times 1^2 + 12 + 6 \\ &= 22 \end{align*}$

Thanks!

CM_Tutor · Feb 21, 2021

Question 2

$\text{Find the possible values of $k$ if the projection of $k\overrightarrow{i}+4\overrightarrow{j}$ onto $12\overrightarrow{i}-5\overrightarrow{j}$ has length $\frac{140}{12}$.$

---

$\text{First, recall that the projection of $\overrightarrow{a}$ onto $\overrightarrow{b}$ is proj}_{\overrightarrow{b}}\overrightarrow{a} = \frac{\overrightarrow{a}\cdot\overrightarrow{b}}{\left|\overrightarrow{b}\right|^2}\overrightarrow{b}$

$\text{So, in this case,} \quad \overrightarrow{a}\cdot\overrightarrow{b} = \left(k\overrightarrow{i}+4\overrightarrow{j}\right)\cdot\left(12\overrightarrow{i}-5\overrightarrow{j}\right) = 12k-20$

$\text{And,} \quad \left|\overrightarrow{b}\right|^2 = 12^2 + (-5)^2 = 169$

$\text{So,} \quad \text{proj}_{\overrightarrow{b}}\overrightarrow{a} = \frac{12k-20}{169}\left(12\overrightarrow{i}-5\overrightarrow{j}\right) = \frac{48(3k-5)}{13^2}\overrightarrow{i}-\frac{20(3k-5)}{13^2}\overrightarrow{j}$

$\begin{align*} \text{And thus,} \quad \left|\text{proj}_{\overrightarrow{b}}\overrightarrow{a}\right|^2 &= \left|\frac{48(3k-5)}{13^2}\overrightarrow{i}-\frac{20(3k-5)}{13^2}\overrightarrow{j}\right|^2 \\ \left(\frac{140}{12}\right)^2 &= \left|\frac{48(3k-5)}{13^2}\overrightarrow{i}-\frac{20(3k-5)}{13^2}\overrightarrow{j}\right|^2 \\ \left(\frac{35}{3}\right)^2 &= \left(\frac{2^4 \times 3(3k-5)}{13^2}\right)^2+\left(-\frac{2^2 \times 5(3k-5)}{13^2}\right)^2 \\ \frac{5^2 \times 7^2}{3^2} &= \frac{2^8 \times 3^2(3k-5)^2}{13^4}+\frac{2^4 \times 5^2(3k-5)^2}{13^4} \\ \frac{5^2 \times 7^2 \times 13^4}{2^4 \times 3^2} &= (3k-5)^2\left(2^4 \times 3^2 + 5^2\right) \\ &= 13^2(3k-5)^2 \\ (3k-5)^2 &= \frac{5^2 \times 7^2 \times 13^4}{2^4 \times 3^2 \times 13^2} \\ &= \left(\frac{5 \times 7 \times 13}{2^2 \times 3}\right)^2 \end{align*}$
$\begin{align*} \text{On taking the square roots} \quad 3k-5 &= \pm\frac{5 \times 7 \times 13}{2^2 \times 3} \\ k &= \frac{1}{3}\left(5\pm\frac{5 \times 7 \times 13}{2^2 \times 3}\right) \\ &= \frac{5}{3}\left(1\pm\frac{7 \times 13}{12}\right) \\ &= \frac{5}{3} \times \frac{12\pm 91}{12} \\ &= \frac{5 \times 103}{36} \quad \text{or} \quad \frac{5 \times -79}{36} \\ &= \frac{515}{36} \quad \text{or} \quad -\frac{395}{36} \end{align*}$

I'm not sure if I have made an arithmetic mistake or not, but I am surprised that there is only supposedly one answer for $k$ and I am also surprised that the length would be given as a fraction with the numerator and denominator having a common factor. @csi, would you please check the question?

CM_Tutor · Feb 21, 2021

CM_Tutor said:
I'm not sure if I have made an arithmetic mistake or not, but I am surprised that there is only supposedly one answer for $k$ and I am also surprised that the length would be given as a fraction with the numerator and denominator having a common factor. @csi, would you please check the question?

For example, if the distance was $\frac{140}{13}$ rather than $\frac{140}{12}$ , we would have

$\begin{align*} \frac{140^2}{13^2} &= \frac{48^2(3k-5)^2}{13^4} + \frac{20^2(3k-5)^2}{13^4} \\ 140^2 \times 13^2 &= (3k-5)^2\left(48^2+20^2\right) = 52^2(3k-5)^2 \\ \pm 140 \times 13 &= 52(3k-5) \\ \pm 35 &= 3k - 5 \\ k &= \frac{40}{3} \quad \text{or} \quad -10 \end{align*}$

So, I strongly suspect that the distance should be $\frac{140}{3}$ and that there are two answers, $k = -10$ and $k = \frac{40}{3}$

Qeru · Feb 21, 2021

CM_Tutor said:
Question 2

$\text{Find the possible values of $k$ if the projection of $k\overrightarrow{i}+4\overrightarrow{j}$ onto $12\overrightarrow{i}-5\overrightarrow{j}$ has length $\frac{140}{12}$.$

---

$\text{First, recall that the projection of $\overrightarrow{a}$ onto $\overrightarrow{b}$ is proj}_{\overrightarrow{b}}\overrightarrow{a} = \frac{\overrightarrow{a}\cdot\overrightarrow{b}}{\left|\overrightarrow{b}\right|^2}\overrightarrow{b}$

$\text{So, in this case,} \quad \overrightarrow{a}\cdot\overrightarrow{b} = \left(k\overrightarrow{i}+4\overrightarrow{j}\right)\cdot\left(12\overrightarrow{i}-5\overrightarrow{j}\right) = 12k-20$

$\text{And,} \quad \left|\overrightarrow{b}\right|^2 = 12^2 + (-5)^2 = 169$

$\text{So,} \quad \text{proj}_{\overrightarrow{b}}\overrightarrow{a} = \frac{12k-20}{169}\left(12\overrightarrow{i}-5\overrightarrow{j}\right) = \frac{48(3k-5)}{13^2}\overrightarrow{i}-\frac{20(3k-5)}{13^2}\overrightarrow{j}$

$\begin{align*} \text{And thus,} \quad \left|\text{proj}_{\overrightarrow{b}}\overrightarrow{a}\right|^2 &= \left|\frac{48(3k-5)}{13^2}\overrightarrow{i}-\frac{20(3k-5)}{13^2}\overrightarrow{j}\right|^2 \\ \left(\frac{140}{12}\right)^2 &= \left|\frac{48(3k-5)}{13^2}\overrightarrow{i}-\frac{20(3k-5)}{13^2}\overrightarrow{j}\right|^2 \\ \left(\frac{35}{3}\right)^2 &= \left(\frac{2^4 \times 3(3k-5)}{13^2}\right)^2+\left(-\frac{2^2 \times 5(3k-5)}{13^2}\right)^2 \\ \frac{5^2 \times 7^2}{3^2} &= \frac{2^8 \times 3^2(3k-5)^2}{13^4}+\frac{2^4 \times 5^2(3k-5)^2}{13^4} \\ \frac{5^2 \times 7^2 \times 13^4}{2^4 \times 3^2} &= (3k-5)^2\left(2^4 \times 3^2 + 5^2\right) \\ &= 13^2(3k-5)^2 \\ (3k-5)^2 &= \frac{5^2 \times 7^2 \times 13^4}{2^4 \times 3^2 \times 13^2} \\ &= \left(\frac{5 \times 7 \times 13}{2^2 \times 3}\right)^2 \end{align*}$
$\begin{align*} \text{On taking the square roots} \quad 3k-5 &= \pm\frac{5 \times 7 \times 13}{2^2 \times 3} \\ k &= \frac{1}{3}\left(5\pm\frac{5 \times 7 \times 13}{2^2 \times 3}\right) \\ &= \frac{5}{3}\left(1\pm\frac{7 \times 13}{12}\right) \\ &= \frac{5}{3} \times \frac{12\pm 91}{12} \\ &= \frac{5 \times 103}{36} \quad \text{or} \quad \frac{5 \times -79}{36} \\ &= \frac{515}{36} \quad \text{or} \quad -\frac{395}{36} \end{align*}$

I'm not sure if I have made an arithmetic mistake or not, but I am surprised that there is only supposedly one answer for $k$ and I am also surprised that the length would be given as a fraction with the numerator and denominator having a common factor. @csi, would you please check the question?

How long did that take you to type out

CM_Tutor · Feb 21, 2021

Qeru said:
How long did that take you to type out

Ummm... a while. Would have been faster if I wasn't trying to figure out why the numbers were getting messy. And if I wasn't an episode of Dr Who in the background at the same time.

csi · Feb 21, 2021

CM_Tutor said:
For example, if the distance was $\frac{140}{13}$ rather than $\frac{140}{12}$ , we would have

$\begin{align*} \frac{140^2}{13^2} &= \frac{48^2(3k-5)^2}{13^4} + \frac{20^2(3k-5)^2}{13^4} \\ 140^2 \times 13^2 &= (3k-5)^2\left(48^2+20^2\right) = 52^2(3k-5)^2 \\ \pm 140 \times 13 &= 52(3k-5) \\ \pm 35 &= 3k - 5 \\ k &= \frac{40}{3} \quad \text{or} \quad -10 \end{align*}$

So, I strongly suspect that the distance should be $\frac{140}{3}$ and that there are two answers, $k = -10$ and $k = \frac{40}{3}$

Yep, you're right. The length is meant to be 140/13 instead of 140/12. That's my bad, sorry about that!!

CM_Tutor · Feb 21, 2021

csi said:
Yep, you're right. The length is meant to be 140/13 instead of 140/12. That's my bad, sorry about that!!

Ok, that's good. Does the source give only one solution, or two?

csi · Feb 21, 2021

CM_Tutor said:
Ok, that's good. Does the source give only one solution, or two?

Just 40/3, it does say all possible solutions though. I wouldn’t be surprised if the the answers were wrong, cause it happens all the time.

CM_Tutor · Feb 21, 2021

csi said:
Just 40/3, it does say all possible solutions though. I wouldn’t be surprised if the the answers were wrong, cause it happens all the time.

Ok, well, I think that's wrong. I think that there are two solutions. If you consider what a projection of one vector onto another is, there can and often will be two vectors that can be projected to give projections of the same length.

Is this from a textbook? If so, which? What's the reference (exercise, page, etc)?

Qeru · Feb 21, 2021

CM_Tutor said:
For example, if the distance was $\frac{140}{13}$ rather than $\frac{140}{12}$ , we would have

$\begin{align*} \frac{140^2}{13^2} &= \frac{48^2(3k-5)^2}{13^4} + \frac{20^2(3k-5)^2}{13^4} \\ 140^2 \times 13^2 &= (3k-5)^2\left(48^2+20^2\right) = 52^2(3k-5)^2 \\ \pm 140 \times 13 &= 52(3k-5) \\ \pm 35 &= 3k - 5 \\ k &= \frac{40}{3} \quad \text{or} \quad -10 \end{align*}$

So, I strongly suspect that the distance should be $\frac{140}{3}$ and that there are two answers, $k = -10$ and $k = \frac{40}{3}$

Wow thats really cool I just realised you did 140/13 since the RHS has 13^4 in the denominator, will that always output a 'nice' answer for k?

CM_Tutor · Feb 21, 2021

Qeru said:
Wow thats really cool I just realised you did 140/13 since the RHS has 13^4 in the denominator

I played a hunch that the 140 / 12 was a typo and, as you said, the 13's in the answer made 140 / 13 a likely guess.

csi · Feb 21, 2021

CM_Tutor said:
Ok, well, I think that's wrong. I think that there are two solutions. If you consider what a projection of one vector onto another is, there can and often will be two vectors that can be projected to give projections of the same length.

Is this from a textbook? If so, which? What's the reference (exercise, page, etc)?

Nope, it’s not from a textbook.

CM_Tutor · Feb 25, 2021

Qeru said:
Wow thats really cool I just realised you did 140/13 since the RHS has 13^4 in the denominator, will that always output a 'nice' answer for k?

Sorry, I just realised that I missed that there was a question here... will it "always" give a nicer $k$ ? I couldn't say always, but you are right that seeing the terms with $13^4$ did make me think that a $13^2$ in the denominator was likely to help get a nicer answer, and the fact that 13 is a one-key typo from 12 made the possibility seem much more likely.

PS: @csi's question 3 above is addressed in another thread and I think the question has infinite answers, you might like to have a look and see if my reasoning is solid... you could even try to figure out all possible positions for $P$ ...

Dot Products and Projections HELP PLZ (1 Viewer)

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