YouTube Video: Fibonacci Numbers and the Mysterious Golden Ratio

This is a YouTube Video, based on my earlier post on Fibonacci Numbers and the Mysterious Golden Ratio!

Thanks for watching! Remember to speed up the video when necessary!

Featured Book:

The Fabulous Fibonacci Numbers

The most ubiquitous, and perhaps the most intriguing, number pattern in mathematics is the Fibonacci sequence. In this simple pattern beginning with two ones, each succeeding number is the sum of the two numbers immediately preceding it (1, 1, 2, 3, 5, 8, 13, 21, ad infinitum). Far from being just a curiosity, this sequence recurs in structures found throughout nature – from the arrangement of whorls on a pinecone to the branches of certain plant stems. All of which is astounding evidence for the deep mathematical basis of the natural world.

With admirable clarity, two veteran math educators take us on a fascinating tour of the many ramifications of the Fibonacci numbers. They begin with a brief history of a distinguished Italian discoverer, who, among other accomplishments, was responsible for popularizing the use of Arabic numerals in the West. Turning to botany, the authors demonstrate, through illustrative diagrams, the unbelievable connections between Fibonacci numbers and natural forms (pineapples, sunflowers, and daisies are just a few examples).

Intellectual wealth

An interesting news to share:

Source:http://www.thestandard.com.hk/news_detail.asp?we_cat=21&art_id=152297&sid=43527569&con_type=1&d_str=20141212&fc=8

Billionaire Ronnie Chan rather be mathematician or scientist if he could live life over

Billionaire Ronnie Chan Chi-chung seems to have it all figured out. Were the Hang Lung Properties chairman to live his life over again, it would not be as a businessman — he’d be a mathematician or scientist instead.

Chan, who offered this little gem during a speech at the Hang Lung Mathematics Awards ceremony, said he may have more material wealth than famed mathematician Yau Shing-tung but much less intellectual wealth.

As co-founder of the awards, which were set to encourage secondary school students to pursue maths and sciences, Chan urged youngsters to go the extra mile and become mathematicians or scientists as they can contribute more to society than what a businessman can.

Hopefully this can encourage students currently studying Maths, be it O Level Maths, JC H1 or H2 Maths, or even University Maths!

Featured Book:

The Shape of Inner Space: String Theory and the Geometry of the Universe’s Hidden Dimensions

New Scientist
“It is a testimony to [Yau’s] careful prose (and no doubt to the skills of co-author Steve Nadis) that this book so compellingly captures the essence of what pushes string theorists forward in the face of formidable obstacles. It gives us a rare glimpse into a world as alien as the moons of Jupiter, and just as fascinating…. Yau and Nadis have produced a strangely mesmerizing account of geometry’s role in the universe.”

Nature
“Physicists investigate one cosmos, but mathematicians can explore all possible worlds. So marvels Fields medalist Shing-Tung Yau…. Relating how he solved a major theoretical problem in string theory in the 1970s, Yau explains how the geometries of the vibrating multidimensional strings that may characterize the Universe have implications across physics.”

Mathematicians prove the Umbral Moonshine Conjecture

Source: Science Daily

Mathematicians prove the Umbral Moonshine Conjecture

Date: December 15, 2014

Source: Emory University

Summary: Monstrous moonshine, a quirky pattern of the monster group in theoretical math, has a shadow — umbral moonshine. Mathematicians have now proved this insight, known as the Umbral Moonshine Conjecture, offering a formula with potential applications for everything from number theory to geometry to quantum physics.

“We’ve transformed the statement of the conjecture into something you could test, a finite calculation, and the conjecture proved to be true,” says Ken Ono, a mathematician at Emory University. “Umbral moonshine has created a lot of excitement in the world of math and physics.”

Co-authors of the proof include mathematicians John Duncan from Case Western University and Michael Griffin, an Emory graduate student.

“Sometimes a result is so stunningly beautiful that your mind does get blown a little,” Duncan says. Duncan co-wrote the statement for the Umbral Moonshine Conjecture with Miranda Cheng, a mathematician and physicist at the University of Amsterdam, and Jeff Harvey, a physicist at the University of Chicago.

Ono will present their work on January 11, 2015 at the Joint Mathematics Meetings in San Antonio, the largest mathematics meeting in the world. Ono is delivering one of the highlighted invited addresses.

Review

“An excellent introduction to this area for anyone who is looking for an informal survey… written in a lively and readable style.”
R.E. Boucherds, University of California at Berkeley for the Bulletin of the AMS

“It is written in a breezy, informal style which eschews the familiar Lemma-Theorem-Remark style in favor of a more relaxed and continuous narrative which allows a wide range of material to be included. Gannon has written an attractive and fun introduction to what is an attractive and fun area of research.”
Geoffrey Mason, Mathematical Reviews

“Gannon wants to explain to us “what is really going on.” His book is like a conversation at the blackboard, with ideas being explained in informal terms, proofs being sketched, and unknowns being explored. Given the complexity and breadth of this material, this is exactly the right approach. The result is informal, inviting, and fascinating.”
Fernando Q. Gouvea, MAA Reviews

Integration by Substitution (H2 Maths Tuition)

This is a 1 page article prepared by me for students to learn how to do Integration by Substitution, a very useful technique that can integrate many functions.

This is especially useful for students taking H2 Maths, as it is one of the two tools for integration. The other is Integration by Parts. SMU First Year Students also have to take a calculus course which includes Integration by Substitution too.

This is also my first time trying out embedding Scribd into WordPress, so that users can view the document on the website itself without downloading anything. 🙂

View this document on Scribd

Print version: Integration by Substitution

Featured book:

Calculus: An Intuitive and Physical Approach (Second Edition) (Dover Books on Mathematics)

Application-oriented introduction relates the subject as closely as possible to science. In-depth explorations of the derivative, the differentiation and integration of the powers of x, and theorems on differentiation and antidifferentiation lead to a definition of the chain rule and examinations of trigonometric functions, logarithmic and exponential functions, techniques of integration, polar coordinates, much more. Clear-cut explanations, numerous drills, illustrative examples. 1967 edition. Solution guide available upon request.

Bishan-Ang Mo Kio area to get new JC in 2017

Source: http://news.asiaone.com/news/edvantage/bishan-ang-mo-kio-area-get-new-jc-2017

Bishan-Ang Mo Kio area to get new JC in 2017

The site for the new JC at the junction of Sin Ming Avenue and Marymount Road.

Lee Jian Xuan

The Straits Times

Saturday, Jan 04, 2014

SINGAPORE – A new junior college that will open in 2017 for students from three Integrated Programme (IP) schools will likely be built on the site of the Asian Golf Academy near Bishan.

A statement on the Ministry of Education (MOE) website says the new campus will be at the junction of Sin Ming Avenue and Marymount Road, where the driving range is located.

The area is also zoned for an educational institution, according to the Urban Redevelopment Authority’s Draft Master Plan 2013.

Singapore’s 20th school to offer a JC programme will take in IP students from Catholic High School, CHIJ St Nicholas Girls’ School and the Singapore Chinese Girls’ School. It will also admit more than 100 students from other secondary schools who have completed their O levels.

It will be the newest JC since Innova JC in Woodlands was completed in 2005.

Undergraduate Study in Mathematics (NUS)

Maths Group Tuition to start in 2014!

If you are interested in Mathematics, do consider to study Mathematics at NUS!

Source: http://ww1.math.nus.edu.sg/undergrad.aspx

Quote:

Undergraduate Study in Mathematics (NUS)

Overview

The Department of Mathematics at NUS is the largest department in the Faculty of Science. We offer a wide range of modules catered to specialists contemplating careers in mathematical science research as well as to those interested in applications of advanced mathematics to science, technology and commerce. The curriculum strives to maintain a balance between mathematical rigour and applications to other disciplines.

We offer a variety of major and minor programmes, covering different areas of mathematical sciences, for students pursuing full-time undergraduate studies. Those keen in multidisciplinary studies would also find learning opportunities in special combinations such as double degree, double major and interdisciplinary programmes.

Honours graduates may further their studies with the Graduate Programme in Mathematics by Research leading to M.Sc. or Ph.D. degree, or with the M.Sc. Programme in Mathematics by Course Work.

Prime Minister Lee Hsien Loong Truly Outstanding Mathematics Student

Just to share an inspirational story about studying Mathematics, and our very own Prime Minister Lee Hsien Loong. 🙂

Source: http://www2.ims.nus.edu.sg/imprints/interviews/BelaBollobas.pdf

(page 8/8)

Interview of Professor Béla Bollobás, Professor and teacher of our Prime Minister Lee Hsien Loong

I: Interviewer Y.K. Leong

B: Professor Béla Bollobás

I: I understand that you have taught our present Prime
Minister Lee Hsien Loong.

B: I certainly taught him more than anybody else in
Cambridge. I can truthfully say that he was an exceptionally
good student. I’m not sure that this is really known in
Singapore. “Because he’s now the Prime Minister,” people
may say, “oh, you would say he was good.” No, he was truly
outstanding: he was head and shoulders above the rest of
the students. He was not only the first, but the gap between
him and the man who came second was huge.

I: I believe he did double honors in mathematics and computer science.

B: I think that he did computer science (after mathematics) mostly because his father didn’t want him to stay in pure mathematics. Loong was not only hardworking, conscientious and professional, but he was also very inventive. All the signs indicated that he would have been a world-class research mathematician. I’m sure his father never realized how exceptional Loong was. He thought Loong was very good. No, Loong was much better than that. When I tried to tell Lee Kuan Yew, “Look, your son is phenomenally good: you should encourage him to do mathematics,” then he implied that that was impossible, since as a top-flight professional mathematician Loong would leave Singapore for Princeton, Harvard or Cambridge, and that would send the wrong signal to the people in Singapore. And I have to agree that this was a very good point indeed. Now I am even more impressed by Lee Hsien Loong than I was all those years ago, and I am very proud that I taught him; he seems to be doing very well. I have come round to thinking that it was indeed good for him to go into politics; he can certainly make an awful lot of difference.

H2 Maths 2012 A Level Solution Paper 2 Q6; H2 Maths Group Tuition

6(i)

$H_0: \mu=14.0 cm$

$H_1: \mu\neq 14.0 cm$

(ii)

$\bar{x}\sim N(14,\frac{3.8^2}{20})$

For the null hypothesis not to be rejected,

$Z_{2.5\%}<\frac{\bar{x}-14}{3.8/\sqrt{20}}<Z_{97.5\%}$

$-1.95996<\frac{\bar{x}-14}{3.8/\sqrt{20}}<1.95996$ (use GC invNorm function!)

$12.3<\bar{x}<15.7$ (3 s.f.)

(iii) Since $\bar{x}=15.8$ is out of the set $12.3<\bar{x}<15.7$ , the null hypothesis would be rejected. There is sufficient evidence that the squirrels on the island do not have the same mean tail length as the species known to her.

(technique: put in words what $H_1$ says!)

H2 Maths A Level 2012 Solution, Paper 2 Q5; H2 Maths Tuition

5(i)(a)

$P(\text{patient has the disease and test positive})=0.001(0.995)=9.95\times 10^{-4}$

$P(\text{patient does not have the disease and he tests positive})=(1-0.001)(1-0.995)=4.995\times 10^{-3}$

$P(\text{result of the test is positive})=9.95\times 10^{-4}+4.995\times 10^{-3}=5.99\times 10^{-3}$

(b)

Let A=patient has disease

Let B=result of test is positive

$\displaystyle\begin{array}{rcl}P(A|B)&=&\frac{P(A\cap B)}{P(B)}\\ &=&\frac{(0.001)(0.995)}{5.99\times 10^{-3}}\\ &=&0.166 \end{array}$

Note that the probability is surprisingly quite low! (This is called the False positive paradox, a statistical result where false positive tests are more probable than true positive tests, occurring when the overall population has a low incidence of a condition and the incidence rate is lower than the false positive rate. See http://en.wikipedia.org/wiki/False_positive_paradox)

(ii)

$\displaystyle P(A|B)=\frac{(0.001)p}{(0.001)p+(1-0.001)(1-p)}=0.75$

By GC, $p=0.999666$ (6 d.p.)

H2 Maths 2012 A Level Paper 2 Q4 Solution; H2 Maths Tuition

(i)

1 Jan 2001 –> $100

1 Feb 2001 —> $110

1 Mar 2001 –> $120

Notice that this is an AP with $a=100$ ; $d=10$

$\displaystyle\begin{array}{rcl}S_n&=&\frac{n}{2}(2a+(n-1)d)\\ &=&\frac{n}{2}(200+10(n-1))>5000 \end{array}$

$\frac{n}{2}(200+10(n-1))-5000>0$

From GC, $n>23.5$

$n=24$ (months)

This is inclusive of 1 Jan 2001!!!

Thus, 1 Jan 2001 + 23 months —> 1 Dec 2002

(ii)

1 Jan 2001 –> 100

end of Jan 2001 –> 1.005(100)

1 Feb 2001 –> 1.005(100)+100

end of Feb 2001 –> 1.005[1.005(100)+100]= $1.005^2 (100)+1.005(100)$

From the pattern, we can see that

$\displaystyle\begin{array}{rcl}S_n&=&1.005^n(100)+1.005^{n-1}(100)+\cdots+1.005(100)\\ &=&\frac{a(r^n-1)}{r-1}\\ &=&\frac{1.005(100)[1.005^n-1]}{1.005-1}\\ &=&\frac{100.5(1.005^n-1)}{0.005}\\ &=&20100(1.005^n-1) \end{array}$

$5000-$100=$4900

$20100(1.005^n-1)>4900$

$20100(1.005^n-1)-4900>0$

From GC, $n>43.7$

So $n=44$ months (inclusive of Jan 2001 !!!)

1 Jan 2001+36 months —> 1 Jan 2004

1 Jan 2004+7 months —> 1 Aug 2004

Then on 1 Sep 2004, Mr B will deposit another $100, making the amount greater than $5000.

Hence, answer is 1 Sep 2004.

(iii)

Let the interest rate be x %.

Note that from Jan 2001 to Nov 2003 is 35 months. (Jan 2001 to Dec 2001 is 12 months, Jan 2002 to Dec 2002 is 12 months, Jan 2003 to Nov 2003 is 11 months :))

$5000-$100=$4900

Modifying our formula in part ii, we get

$\displaystyle S_n=\frac{(1+x/100)(100)[(1+x/100)^n-1]}{(1+x/100)-1}=4900$

Setting $n=35$ and using GC, we get

$x=1.80$

Hence, the interest rate is 1.80%.

A Level H2 Maths 2012 Paper 2 Q3 Solution; H2 Maths Tuition

A Level H2 Maths 2012 Paper 2 Q3 Solution

(i)

(The graph above is drawn using the Geogebra software 🙂 )

(ii)

$x^3+x^2-2x-4=4$

$x^3+x^2-2x-8=0$

By GC, $x=2$

By long division, $x^3+x^2-2x-8=(x-2)(x^2+3x+4)$

The discriminant of $x^2+3x+4$ is

$D=b^2-4ac=3^2-4(1)(4)=-7<0$

Hence, there are no other real solutions (proven).

(iii) $x+3=2$

$x=-1$

(iv)

(v)

$|x^3+x^2-2x-4|=4$

$x^3+x^2-2x-4=4$ or $x^3+x^2-2x-4=-4$

$x^3+x^2-2x-8=0$ or $x^3+x^2-2x=0$

$x^3+x^2-2x-8=0 \implies x=2$ (from part ii)

$x^3+x^2-2x=x(x^2+x-2)=x(x-1)(x+2)=0$

$x=0,1,-2$

In summary, the roots are $-2,0,1,2$

List of JCs in Singapore; H2 Maths Tuition

Source: http://en.wikipedia.org/wiki/List_of_schools_in_Singapore#Junior_Colleges_.28JC.29

Junior Colleges (JC)

These offer two-year courses leading to the GCE A-level examination.

Code	Zone	College Name			Established	Address	Type	Special Programmes
Code	Zone	English	Chinese	Abb.	Established	Address	Type	Special Programmes
0705	North	Anderson Junior College	安德逊初级学院	AJC	1984	4500 Ang Mo Kio Avenue 6	Government
7001	West	Anglo-Chinese School (Independent) IB World School	英华中学 (自主)	ACS(I)-IBDP	2004 (IBDP)	121 Dover Road	Independent	IP, MEP
0803	West	Anglo-Chinese Junior College	英华初级学院	ACJC	1977	25 Dover Close East	Government-Aided	MEP, DEP(TSD), LEP (EL)
0802	South	Catholic Junior College	公教初级学院	CJC	1975	129 Whitley Road	Government-Aided	LEP (EL)
3101	East	Dunman High School	德明政府中学	DHS	2005 – IP	10 Tanjong Rhu Road	Autonomous	IP, MEP, BSP, LEP (CL), AEP
0806	Central	Hwa Chong Institution	华侨中学	HCI	1974	661 Bukit Timah Road	Independent	IP, HP, LEP (CL), AEP, BSP
0713	North	Innova Junior College	星烁初级学院	IJC	2005	21 Champions Way	Government	LEP (ML)
0703	West	Jurong Junior College	裕廊初级学院	JJC	1981	800 Corporation Road	Government	LEP (CL)
0712	East	Meridian Junior College	美廉初级学院	MJC	2003	21 Pasir Ris Street 71	Government
0908	West	Millennia Institute	励仁高级中学	MI	2004	60 Bukit Batok West Avenue 8	Government	DTP
0805	North	Nanyang Junior College	南洋初级学院	NYJC	1978	128 Serangoon Avenue 3	Government-Aided	LEP (CL), AEP
0712	Central	National Junior College	国家初级学院	NJC	1969	37 Hillcrest Road	Government	IP, HP, AEP, MEP, STaR
7801	West	NUS High School of Mathematics and Science	新加坡国立大学附属数理中学	NUSHS	2005	20 Clementi Ave 1	Independent	IP, DIP
0711	West	Pioneer Junior College	先驱初级学院	PJC	1999	21 Teck Whye Walk	Government
0704	South	Raffles Institution	莱佛士初级学院	RI	1826	10 Bishan Street 21	Independent	IP, HP, LEP (JL), LEP (EL), MEP, TSD
3103	West	River Valley High School	立化中学	RVHS	1956 2006 – IP	6 Boon Lay Avenue	Autonomous	IP, BSP
0710	North	Serangoon Junior College	实龙岗初级学院	SRJC	1988	1033 Upper Serangoon Road	Government
0804	South	Saint Andrew’s Junior College	圣安德烈初级学院	SAJC	1978	55 Potong Pasir Avenue 1	Government-Aided
0709	East	Tampines Junior College	淡滨尼初级学院	TPJC	1986	2 Tampines Avenue 9	Government	LEP (ML), TSD
0702	East	Temasek Junior College	淡马锡初级学院	TJC	1977	22 Bedok South Road	Government	IP, HP, LEP (CL), MEP
0706	East	Victoria Junior College	维多利亚初级学院	VJC	1984	20 Marine Vista	Government	IP, HP, TSD, NAV
0708	North	Yishun Junior College	义顺初级学院	YJC	1986	3 Yishun Ring Road	Government

Centralised Institutes (CI)

The only centralised institute is Millennia Institute (MI), which offers a three-year course leading to the GCE A-level examination in arts, science, and commerce.^[3]

How to use tables in CASIO FX-9860G Slim (H2 Maths Tuition)

Tables in CASIO FX-9860G Slim

The most popular Graphical Calculator for H2 Maths is currently the TI-84 PLUS series, but some students do use Casio Graphical Calculators.

The manual for CASIO FX-9860G Slim is can be found here:

http://edu.casio.com/products/fx9860g2/data/fx-9860GII_Soft_E.pdf

The information about Tables and how to generate a table is on page 121.

Generating tables is useful to solve some questions in sequences and series, and also probability. It makes guess and check questions much faster to solve.

h2 tuition gc — Picture of the CASIO FX-9860G Slim Calculator

2012 H2 Maths Prelim Solution: SRJC/II/8(iv)

8(iv)

Using the model $\displaystyle y=a+\frac{b}{x-2}$ , estimate the total fertility rate for a particular country Z when its GDP per capita is USD$1000, giving your answer to 1 decimal place and comment on the reliability of the estimate.

First, we need to remove the outlier (40,6.6) as mentioned in part iii.

Then, performing linear regression with GC, (with variables $\frac{1}{x-2}$ , $y$ ), we get:

$a=0.974686$

$b=6.86442$

Substituting $x=1$ , we get $\displaystyle y=a+\frac{b}{1-2}=-5.9$ (1 d.p.)

Since we cannot have a negative fertility rate (the average number of children that would be born to a woman ), the estimate obtained for $y$ is not reliable.

MI H2 Maths Prelim Solutions 2010 Paper 2 Q7 (P&C)

MI H2 Maths Prelim 2010 Paper 2 Q7 (P&C)

Question:
7) Two families are invited to a party. The first family consists of a man and both his parents while the second family consists of a woman and both her parents. The two families sit at a round table with two other men and two other women.
Find the number of possible arrangements if
(i) there is no restriction, [1]
(ii) the men and women are seated alternately, [2]
(iii) members of the same family are seated together and the two other women must be seated separately, [3]
(iv) members of the same family are seated together and the seats are numbered. [2]

Solution:

(i) (10-1)!=9!=362880

(ii) First fix the men’s sitting arrangement: (5-1)!

Then the remaining five women’s total number of arrangements are: 5!

Total=4! x 5!=2880

(iii) Fix the 2 families (as a group) and the 2 men: (4-1)! x 3! x 3!

(3! to permute each family)

By drawing a diagram, the two women have 4 slots to choose from, where order matters: $^4 P_2$

Total = $(4-1)! \times 3! \times 3! \times ^4 P_2 = 2592$

(iv)

We first find the required number of ways by treating the seats as unnumbered: $(6-1)!\times 3!\times 3! =4320$

Since the seats are numbered, there are 10 choices for the point of reference, thus no. of ways = $4320 \times 10 =43200$

H2 JC Maths Tuition Foot of Perpendicular 2007 Paper 1 Q8

One of my students asked me how to solve 2007 Paper 1 Q8 (iii) using Foot of Perpendicular method.

The answer given in the TYS uses a sine method, which is actually shorter in this case, since we have found the angle in part (ii).

Nevertheless, here is how we solve the question using Foot of Perpendicular method.

(Due to copyright issues, I cannot post the whole question here, so please refer to your Ten Year Series.)

Firstly, let F be the foot of the perpendicular.

Then, $\vec{AF}=k\begin{pmatrix}3\\-1\\2\end{pmatrix}$ ——– Eqn (1)

$\vec{OF}\cdot\begin{pmatrix}3\\-1\\2\end{pmatrix}=17$ ——– Eqn (2)

From Eqn (1), $\vec{OF}-\vec{OA}=\begin{pmatrix}3k\\-k\\2k\end{pmatrix}$

$\begin{array}{rcl}\vec{OF}&=&\vec{OA}+\begin{pmatrix}3k\\-k\\2k\end{pmatrix}\\ &=&\begin{pmatrix}1\\2\\4\end{pmatrix}+\begin{pmatrix}3k\\-k\\2k\end{pmatrix}\\ &=&\begin{pmatrix}1+3k\\2-k\\4+2k\end{pmatrix}\end{array}$

Substituting into Eqn (2),

$\begin{pmatrix}1+3k\\2-k\\4+2k\end{pmatrix}\cdot\begin{pmatrix}3\\-1\\2\end{pmatrix}=17$

$14k+9=17$

$k=4/7$

Substituting back into Eqn (1),

$\displaystyle\vec{AF}=\frac{4}{7}\begin{pmatrix}3\\-1\\2\end{pmatrix}$

$\displaystyle|\vec{AF}|=\frac{4}{7}\sqrt{14}$

Maths Tutor Singapore, H2 Maths, A Maths, E Maths

If you or a friend are looking for maths tuition: o level, a level, IB, IP, olympiad, GEP and any other form of mathematics you can think of

Experienced, qualified (Raffles GEP, Deans List, NUS Deans List, Olympiads etc) and most importantly patient even with the most mathematically challenged.

so if you are in need of the solution to your mathematical woes, drop me a message!

Tutor: Mr Wu

Email: mathtuition88@gmail.com

Website: https://mathtuition88.com/

H2 Maths Tuition: Complex Numbers Notes

H2 Maths: Complex Numbers 1 Page Notes

	Modulus	Argument
Cartesian Form		Draw diagram first, then find the appropriate quadrant and use (can use GC to double check)
Polar Form
Exponential Form

☺ When question involvespowers, multiplication or division, it may be helpful toconvert to exponential form.

☺ Please write Ƶ and 2 differently.

☺ De Moivre’s Theorem

Equivalent to

☺

Memory tip: Notice that arg behaves similarly to log.

☺ Locusof z is aset of pointssatisfying certain given conditions.

☺ in English means:The distance between (the point representing)and (the point representing)

☺ Means the distance offromis a constant,.

So this is acircular loci.

Centre:, radius =

☺ means that the distance offromis equal to its distance from

In other words, the locus is theperpendicular bisectorof the line segment joiningand.

☺ represents ahalf-linestarting frommaking an anglewith the positive Re-axis.

(Exclude the point (a,b) )

Common Errors

– Some candidates thought thatis the same asand thatis the same as.

– The “formula”for argumentsdoes not workfor points in the 2^ndand 3^rdquadrant.

– Very many candidates seem unaware that their calculators will work in radians mode and there were many unnecessary “manual” conversions from degrees to radians.

H2 Maths Tuition: Foot of Perpendicular (from point to plane) (Part II)

This is a continuation from H2 Maths Tuition: Foot of Perpendicular (from point to line) (Part I).

Foot of Perpendicular (from point to plane)

From point (B) to Plane ( $p$ )

Equation (I):

Where does F lie?

F lies on the plane $p$ .

$\overrightarrow{\mathit{OF}}\cdot \mathbf{n}=d$

Equation (II):

Perpendicular

$\overrightarrow{\mathit{BF}}=k\mathbf{n}$

$\overrightarrow{\mathit{OF}}-\overrightarrow{\mathit{OB}}=k\mathbf{n}$

$\overrightarrow{\mathit{OF}}=k\mathbf{n}+\overrightarrow{OB}$

Final Step

Substitute Equation (II) into Equation (I) and solve for k.

Example

[VJC 2010 P1Q8i]

The planes $\Pi _{1}$ and $\Pi _{2}$ have equations $\mathbf{r\cdot(i+j-k)}=6$ and $\mathbf{r\cdot(2i-4j+k)}=-12$ respectively. The point $A$ has position vector $\mathbf{{9i-7j+5k}}$ .

(i) Find the position vector of the foot of perpendicular from $A$ to $\Pi _{2}$ .

Solution

Let the foot of perpendicular be F.

Equation (I)

$\overrightarrow{\mathit{OF}}\cdot \left(\begin{matrix}2\\-4\\1\end{matrix}\right)=-12$

Equation (II)

$\overrightarrow{\mathit{OF}}=k\left(\begin{matrix}2\\-4\\1\end{matrix}\right)+\left(\begin{matrix}9\\-7\\5\end{matrix}\right)=\left(\begin{matrix}2k+9\\-4k-7\\k+5\end{matrix}\right)$

Subst. (II) into (I)

$2(2k+9)-4(-4k-7)+(k+5)=-12$

Solve for k, $k=-3$ .

$\overrightarrow{\mathit{OF}}=\left(\begin{matrix}3\\5\\2\end{matrix}\right)$

H2 Maths Tuition

If you are looking for Maths Tuition, contact Mr Wu at:

Email: mathtuition88@gmail.com