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"Prove that if vector C is perpendicular to vectors A and B, then vector C is still perpendicular to vector mA + nB where m and n are scalars."

I made a mistake with the wording.

We know so far that A.C = 0 and B.C = 0. Therefore, to prove the above statement....

C.(mA + nB) = C.mA + C.nB

Using scalar properties, C.mA + C.nB = m*(A.C) + n*(B.C)

m*(A.C) + n*(B.C) = m*0 + n*0 = 0

Is it really that easy?

I don't personally see what the fuss is all about, nor do I comprehend the reasonings behind the snide remarks. I know that a scalar only changes the magnitude, but, as MathsIsFun pointed out, in itself that is not a proof.

I was asking a question - I didn't expect to be ridiculed (i.e. Insomnia and Person). Nice to see that internet etiquette is at its best in this forum.

Thank you for your reply, MathsIsFun. As for the other two, keep your bragging for the locker room.

**maths_buff**- Replies: 8

"Prove that if vector C is perpendicular to vectors A and B, then vector C is still perpendicular to vectors A and B when A and B are multiplied by scalars m and n respectively."

My Thoughts....

If we have perpendicular vectors on a 2D plane, then V.W = 0 when the dot product of vectors V and W is found.

It can also be seen from the dot product formula that V.W = ( V ) ( W) cos 90 and cos 90 = 0.

But how to prove they are perpendicular after being multiplied by scalars remains the question.

No update as of yet....

Still working within the problem.

Hopefully other members will have some interesting ideas.

It matters and it doesn't matter; after six years it doesn't matter because on average everyone will have a new car. But it terms of after 8 years, that the questionable part. Perhaps 99% of the population buys their car in that year.

Remember that we're only worrying about current large car owners who'll have a large car in eight years time.

I will try and fiddle with it because as long as the elements in each row add to 1, I can indicate what I did is correct.

I'll wait and see what other forum members say.

Once again, MathsIsFun, thank you!

Here's another theory I have:

Large Small

Large [ 0.60 0.40 ]

Small [ 0.25 0.75 ]

There are two possible ways for a large car owner to own a large car in six years:

A) Someone has a large car now, buys a large car after three years, buys another large car after six years

B) Someone has a large car now, buys a small car after three years, buys another large car after six years

Therefore the probability is P(LL)P(LL) + P(LS)P(SL) = 23/50 like I suspected.

Maybe it's wrong; maybe it's right. ???

Thanks, MathsIsFun!

It's funny because there are two ways of doing it, viz:

http://ceee.rice.edu/Books/LA/markov/

http://ceee.rice.edu/Books/LA/markov/

Which one do you think sounds better?

I think you may have misunderstood me, Milos. I know what x, y and z; I'm looking on how to plot/graph them.

Notice how I said x = 3, y = 2 and z = 1 in my first post?

Cheers,

Kris

I can see your point MathsisFun, I can see it indeed.

But would that limit accuracy? Because I believe the transition matrix refers to three year intervals only. I will type the question word-for-word to clarify things.

"Market analysis in a certain region has established that, on average, a new car is purchased every three years. With respect to those changing cars, the buying patterns are described by the matrix:

Large Small

Large [ 60% 40% ]

small [ 25% 75% ]

a) Rewrite the matrix as a probability matrix

b) Find the probability that a person who now owns a large car will own a large car in eight years' time."

**maths_buff**- Replies: 3

How would I plot linear equations with three variables?

Here's my own problem:

x + y + z = 6

2x - 3y + 4z = 4

x + 2y - z = 6

In this problem I made, x = 3, y = 2 and z = 1

I have tried downloading some programs but they keep saying my syntax is wrong.

Yeah, thank you very much. I had seen that some people did that, whilst other examples didn't.

Cheers,

Kris

After six years I know the respective probabilities are....

[ 23/50 27/50 ]

[ 27/80 53/80 ]

So after six years the probability of someone currently owning a large car and still owning one is 23/50 or 46%.

**maths_buff**- Replies: 13

I am enquiring as to a series of Markov Chains questions I have.

I have come across a question that says market analysis has established that, on average, a new car is purchased every three years. Buying patterns are described by the matrix:

Large Small

Large[ 60% 40% ]

Small[ 25% 75% ]

Am I correct in saying that the probability matrix can be re-written as....

L S

S = L [ 0.6 0.4 ]

S [ 0.25 0.75 ]

In addition, how would I calculate the probability of someone owning a large car still owning a large car in eight years' time, considering that the problem itself deals with car purchases every three years on average?

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