Vectors (1 Viewer)

mathsbrain · Mar 15, 2020

Does anyone have any vectors worksheet they can share, i don't have anything to practice with other than the textbook(which i guess is the only thing most of us has at the moment...), so was wondering if people from selective schools can share some useful harder questions

Pedro123 · Mar 15, 2020

If you would like some hard questions, you should look at a lot of geometric proofs and try to prove it with vectors, i.e. prove in an isosceles triangle, the angle bisector of the angle made by the 2 equal sides is perpendicular to the unequal side, and it bisects it. It sounds easy, but most geometric proofs can be quite tricky only using vectors (so only use dot product and basic vector laws)

mathsbrain · Mar 15, 2020

Pedro123 said:
If you would like some hard questions, you should look at a lot of geometric proofs and try to prove it with vectors, i.e. prove in an isosceles triangle, the angle bisector of the angle made by the 2 equal sides is perpendicular to the unequal side, and it bisects it. It sounds easy, but most geometric proofs can be quite tricky only using vectors (so only use dot product and basic vector laws)

hmm just tried it lol and had no idea how to start...

ultra908 · Mar 15, 2020

Also vectors are a part of the British A-Levels and the Victorian VCE, I've found some A-Levels qs on equations of lines/planes and they're quite similar to 4u textbook qs so they might be handy.

fan96 · Mar 15, 2020

mathsbrain said:
hmm just tried it lol and had no idea how to start...

Recall that for two vectors $\bold x, \bold y$ we have

$\cos \theta = \frac{\bold x \cdot \bold y}{|\bold x||\bold y|},$

where $\theta$ is the angle between $\bold x$ and $\bold y$ .

mathsbrain · Mar 15, 2020

ultra908 said:
Also vectors are a part of the British A-Levels and the Victorian VCE, I've found some A-Levels qs on equations of lines/planes and they're quite similar to 4u textbook qs so they might be handy.

Do we just find them just online?

ultra908 · Mar 16, 2020

Ye i found some here

Exam Questions - Vectors | ExamSolutions

1)View SolutionParts (a) and (b): Part (c): Part (d): Part […]

www.examsolutions.net

mathsbrain · Mar 16, 2020

ultra908 said:
Ye i found some here

Exam Questions - Vectors | ExamSolutions

1)View SolutionParts (a) and (b): Part (c): Part (d): Part […]

www.examsolutions.net

Thanks, but i was looking for more stuff to do with spheres circles (not sure if planes is in syllabus) that is in the last section of 4unit vectors and struggling to find practice...

CM_Tutor · Mar 17, 2020

mathsbrain said:
hmm just tried it lol and had no idea how to start...

One approach is to use vectors on an Argand diagram. For example:

Let $\overrightarrow{OA} = z_1 = x_1 + iy_1 = re^{i\theta_1}$ and $\overrightarrow{OB} = z_2 = x_2 + iy_2 = re^{i\theta_2}$ be two vectors such that $|z_1| = |z_2| = r$ and $\pi > \arg z_1 > \arg z_2 > 0$ .

(i) Show that $\overrightarrow{OZ} = \overrightarrow{OA} + \overrightarrow{OB}$ bisects $\angle AOB$

(ii) Find $\overrightarrow{BA}$ in terms of $z_1$ and $z_2$

(iii) Show that $\frac{z_1 + z_2}{z_1 - z_2}$ is purely imaginary

(iv) Let OZ and AB meet at M. Using the above, show that $\Delta AOB$ is an isosceles triangle where OM bisects angle AOB, $OM \perp AM$ , and M is the midpoint of AB.

A much simpler approach is as follows:

Let $\overrightarrow{OA} = z = x + iy = re^{i\theta}$ be a vector in the Argand diagram such that $0 < \arg z < \frac{\pi}{2}$ . Let $\overrightarrow{OB} = \overline{z}$ . By finding the dot product of $\overrightarrow{OA}$ and $\overrightarrow{OB}$ , prove that AOB is an isosceles triangle and that the bisector of angle AOB is the perpendicular bisector of AB.

HeroWise · Mar 22, 2020

Let the position vectors be $\vec{a}$ and $\vec{b}$ where $|\vec{a}|=| \vec{b}|$ and the ancle bisector as $\vec{c} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$
This stands that, $\vec{a} \cdot \vec{c} = |\vec{a}| |\vec{c}| \cos \theta$
also that: $\vec{b} \cdot \vec{c} = |\vec{c}| |\vec{b}| \cos \theta$
since $|\vec{a}|=| \vec{b}|$ , $\vec{a} \cdot \vec{c} - \vec{b} \cdot \vec{c} = \vec{c} \cdot ( \vec{a} -\vec{b})=0$
Hence perpendicular.

Pedro123 · Mar 22, 2020

HeroWise said:
Let the position vectors be $\vec{a}$ and $\vec{b}$ where $|\vec{a}|=| \vec{b}|$ and the ancle bisector as $\vec{c} \dot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$
This stands that, $\vec{a} \dot \vec{c} = |\vec{a}| |\vec{c}| \cos \theta$
also that: $\vec{b} \dot \vec{c} = |\vec{c}| |\vec{b}| \cos \theta$
since $|\vec{a}|=| \vec{b}|$ , $\vec{a} \dot \vec{c} - \vec{b} \dot \vec{c} = \vec{c} \dot ( \vec{a} -\vec{b})=0$
Hence perpendicular.

How does that work? I think you are missing the step where you say that a and b are not the same, so a-b cannot be 0, and since c= Mag(b)mag(c)*cos(theta)/B, the only possible value that can make that 0 is when cos(theta)=0, meaning it is perpendicular. See video for an answer (How I solved it):

HeroWise · Mar 22, 2020

a-b is the base... That proof i did was fine, I did state they were position vectors. And if u can assume its a median why not just use properties of rhombus and use complex like CM_Tutor, you will just get it out instantly then lol

CM_Tutor · Mar 22, 2020

HeroWise said:
Let the position vectors be $\vec{a}$ and $\vec{b}$ where $|\vec{a}|=| \vec{b}|$ and the ancle bisector as $\vec{c} \dot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$

HeroWise, I don't follow the definition of $\vec{c} \; . \; \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$ . You don't actually need to find $\vec{c}$ , beyond letting letting the angle between $\vec{a}$ and $\vec{b}$ be $2\theta$ so that the angle between $\vec{a}$ and $\vec{c}$ and between $\vec{b}$ and $\vec{c}$ are both $\theta$ . However, if you do want to state it, it is $\vec{c} =k\Big(\frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|}\Big)$ for some constant $k \in \mathbb{R} > 0$ . This is because the unit vectors in the directions of $\vec{a}$ and $\vec{b}$ , when added, represent the diagonal of a rhombus and so must bisect the angles at the vertices through which it passes.

Your method does establish that $\vec{a} \; . \; \vec{c} = \vec{b} \; . \; \vec{c}$ and thus that $(\vec{a} - \vec{b}) \; . \; \vec{c} = 0$ , as you stated, which does make $\vec{a} - \vec{b}$ perpendicular to $\vec{c}$ , proving that the vector bisecting the apex angle of an isosceles triangle is perpendicular to the position vector for its base.

HeroWise · Mar 22, 2020

Oh i just wanted to reach the conclusion with the last paragraph quickly. I constructed that vector as an arbitrary vector that will act as a bisector. Didnt want to use the fact that its half a rhombus and in essence that is the way the complex method I was referring to would've been established. The other reason I did not put it as the form you referred to above was because i just wished to prove that the vector c was perpendicular. Didnt need any other stuff for it so didn't go with the definition you have provided above; so the proof
should be sufficient to say the very least.

CM_Tutor · Mar 22, 2020

HeroWise said:
Oh i just wanted to reach the conclusion with the last paragraph quickly. I constructed that vector as an arbitrary vector that will act as a bisector. Didnt want to use the fact that its half a rhombus and in essence that is the way the complex method I was referring to would've been established. The other reason I did not put it as the form you referred to above was because i just wished to prove that the vector c was perpendicular. Didnt need any other stuff for it so didn't go with the definition you have provided above; so the proof
should be sufficient to say the very least.

It is possible to simply skip defining $\vec{c}$ explicitly, and even if I were to do so, I would not offer an ambiguous definition of $\vec{c}$ as you have done. It simply invites questions as to why that definition was chosen and whether the $\vec{c}$ that you have defined actually has the properties that you claim. In this situation, it is odd to define $\vec{c}$ by means of a dot product as it means that $\vec{c}$ is not a unique vector as there is more than one $\vec{c}$ that satisfies your definition. Under your definition, $\vec{c}$ must satisfy $|\vec{c}| = |\vec{a}|$ and have an angle between it and $\vec{b}$ as $\theta$ - but there are two vectors that fit these criteria and only one of them is the bisector that you seek.

It is sufficient to provide a limited defition: that $\vec{c}$ is a vector that bisects the angles between $\vec{a}$ and $\vec{b}$ , so that the angle between $\vec{a}$ and $\vec{c}$ and the angle between $\vec{b}$ and $\vec{c}$ are the same (i.e. $\theta$ ) and thus that the angle between $\vec{a}$ and $\vec{b}$ is $2\theta$ subject to the requirements that $\vec{a} \neq \vec{b}$ and $\theta \neq 0$ .

In other words, by simply naming $\vec{c}$ and defining it as having the properties of a bisector, your proof that the dot product $\vec{c} \; . \; (\vec{b} - \vec{a}) = 0$ follows quickly and without distraction. As a marker, I would have to stop and consider whether your $\vec{c}$ is actually valid, and the definition itself is not used for the proof in any event so it isn't actually needed.

HeroWise · Mar 22, 2020

HeroWise said:
Let the position vectors be $\vec{a}$ and $\vec{b}$ where $|\vec{a}|=| \vec{b}|$ and the ancle bisector as $\vec{c} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$
This stands that, $\vec{a} \cdot \vec{c} = |\vec{a}| |\vec{c}| \cos \theta$
also that: $\vec{b} \cdot \vec{c} = |\vec{c}| |\vec{b}| \cos \theta$
since $|\vec{a}|=| \vec{b}|$ , $\vec{a} \cdot \vec{c} - \vec{b} \cdot \vec{c} = \vec{c} \cdot ( \vec{a} -\vec{b})=0$
Hence perpendicular.

Im dumb, i can type set as u can see haha. I meant: "and the angle bisector as $\vec{c} \cdot \vec{b} = |\vec{c}| |\vec{b}| \cos \theta$ "

So sorry did it in a hurry, but the maths should be write afterwards

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