Need help, URGENT maths question: (2 Viewers)

1008

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Thanks.


The problem I'm having is that I get cos(DEF) as a number greater than 1, so that the angle is undefined when I take its inverse cos. And for the second part, I plan to use the formula Area = 1/2 ab sin(c) to calculate the area, but can't do this...Maybe I'm doing something wrong???
 

InteGrand

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Thanks.


The problem I'm having is that I get cos(DEF) as a number greater than 1, so that the angle is undefined when I take its inverse cos. And for the second part, I plan to use the formula Area = 1/2 ab sin(c) to calculate the area, but can't do this...Maybe I'm doing something wrong???


 

leehuan

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Thanks.


The problem I'm having is that I get cos(DEF) as a number greater than 1, so that the angle is undefined when I take its inverse cos. And for the second part, I plan to use the formula Area = 1/2 ab sin(c) to calculate the area, but can't do this...Maybe I'm doing something wrong???
MATH1151 ALG Ch4 Q22

a) Vec(DE) . Vec(FE) = (e-d).(e-f) = [1,1,1]T . [-1,-1,2]T = -4

|DE|2=sqrt(1+1+1)=sqrt(3)

|FE|2=sqrt(1+1+4)=sqrt(6)

Hence cos<DEF = (-4)/sqrt(3*6) = -4/(3sqrt2)

But yeah use the method InteGrand gave for part b)
 

1008

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Thanks, it is indeed easier!

I found my mistake, so when I calculate DE = (1,1,1) and EF = (1,1,2), I find cos(DEF) = 4/sqrt(18). Previously, I found 4/(sqrt(12). I feel so dumb....

MATH1151 ALG Ch4 Q22

a) Vec(DE) . Vec(FE) = (e-d).(e-f) = [1,1,1]T . [-1,-1,2]T = -4

|DE|2=sqrt(1+1+1)=sqrt(3)

|FE|2=sqrt(1+1+4)=sqrt(6)

Hence cos<DEF = (-4)/sqrt(3*6) = -4/(3sqrt2)

But yeah use the method InteGrand gave for part b)
Thanks!

Also, there isn't an answer provided at the back of the problems book for this, could you please explain if I am right or not? Here is the question:


The answer I get is:
Equation of altitude through A (l1) = (0,1,2) + t(2,-6,-14)
Equation of altitude through B (l2) = (-1,4,1) + u(0,3,3)
I found u and t by the cross product of BC and CD as well as AD and AC respectively.
Then, to determine if the lines intersect or not:



As evident, b has to be a leading column, and the leading element in b doesn't equal 0 as lines aren't parallel. So altitudes don't intersect as there is no solution to this matrix.
 
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InteGrand

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Thanks, it is indeed easier!



I found my mistake, so when I calculate DE = (1,1,1) and EF = (1,1,2), I find cos(DEF) = 4/sqrt(18). Previously, I found 4/(sqrt(12). I feel so dumb....



Thanks!

Also, there isn't an answer provided at the back of the problems book for this, could you please explain if I am right or not? Here is the question:


The answer I get is:
Equation of altitude through A (l1) = (0,1,2) + t(2,-6,-14)
Equation of altitude through B (l2) = (-1,4,1) + u(0,3,3)
I found u and t by the cross product of BC and CD as well as AD and AC respectively.
Then, to determine if the lines intersect or not:



As evident, b has to be a leading column, and the leading element in b doesn't equal 0 as lines aren't parallel. So altitudes don't intersect as there is no solution to this matrix.
A justification like that for why there's no solution would require us to put the augmented matrix into row-echelon form first, and then inspect it.

Another way to see there's no solution is to note that if there is a solution, the coefficient t needs to be -1/2, in order for the first entries to match. Then looking at second entries, we see that since t = -1/2, u needs to be 0. But with t = -1/2 and u = 0, the third entries don't match. So there can be no solution.

(I've assumed all your prior calculations were correct, didn't check them.)
 

leehuan

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Thanks, it is indeed easier!



I found my mistake, so when I calculate DE = (1,1,1) and EF = (1,1,2), I find cos(DEF) = 4/sqrt(18). Previously, I found 4/(sqrt(12). I feel so dumb....



Thanks!

Also, there isn't an answer provided at the back of the problems book for this, could you please explain if I am right or not? Here is the question:


The answer I get is:
Equation of altitude through A (l1) = (0,1,2) + t(2,-6,-14)
Equation of altitude through B (l2) = (-1,4,1) + u(0,3,3)
I found u and t by the cross product of BC and CD as well as AD and AC respectively.
Then, to determine if the lines intersect or not:



As evident, b has to be a leading column, and the leading element in b doesn't equal 0 as lines aren't parallel. So altitudes don't intersect as there is no solution to this matrix.
Pretty much like InteGrand said, don't forget the methods that Josef Dick explained

1. Use Gaussian elimination to bring out a R-E form.

Via matlab, assuming that your altitude through A is correct (the one through B definitely is correct) the reduced row-echelon form is

1 0 | 0
0 1 | 0
0 0 | 1

(Obviously, there's no need to actually bring out the reduced row echelon form on paper)

2. So since the right hand column is indeed a leading column like you said, there exists no solution. Hence the conclusion is also correct.

Edit: Also, this is me being pedantic now but technically Ax=b is this



Augmented matrix notation is technically [A | b]
 
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1008

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Pretty much like InteGrand said, don't forget the methods that Josef Dick explained

1. Use Gaussian elimination to bring out a R-E form.

Via matlab, assuming that your altitude through A is correct (the one through B definitely is correct) the reduced row-echelon form is

1 0 | 0
0 1 | 0
0 0 | 1

(Obviously, there's no need to actually bring out the reduced row echelon form on paper)

2. So since the right hand column is indeed a leading column like you said, there exists no solution. Hence the conclusion is also correct.

Edit: Also, this is me being pedantic now but technically Ax=b is this



Augmented matrix notation is technically [A | b]
Thanks leehuan! And yeah you're right, my notation is wrong for the augmented matrix, appreciate that.

I have another question:


 
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InteGrand

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1008

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A proof of 1) may be found here: http://community.boredofstudies.org/238/extracurricular-topics/350230/least-squares.html .

For 2), it is known as the Gram-Schmidt orthonormalisation process, and the proof is typically done by induction.
Thanks, I'll leave the latter for later!

I'm pretty frustrated by these...What would be the shortest way of doing them?
Prove the following properties of matrices:
1) If the product AB exists, then A(λB) = λ(AB) = (λA)B
2) Associative law of matrix multiplication. If products AB and BC exist, then A(BC) = (AB)C
3) AI = A and IA = A where I represents identity matrices of the appropriate (possibly different) sizes
4) Left distributive law. If A+B and AC exist, then (A+B)C = AC+BC
5) Right distributive law If B+C and AB exist, then A(B+C) = AB+AC

I know you could do these writing each of them out as an nxn matrix...but it takes way to long to do so test-wise (+ its frustrating).
 
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leehuan

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It's actually possible to prove both distributive laws by considering component wise what goes on and then using some rules over the real numbers.

Associativity is much harder (because you can't assume distribution to prove association obviously). If I remember to I'll go grab my solution when my break ends
 

1008

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It's actually possible to prove both distributive laws by considering component wise what goes on and then using some rules over the real numbers.

Associativity is much harder (because you can't assume distribution to prove association obviously). If I remember to I'll go grab my solution when my break ends
Thanks

But is there a way to prove the distributive laws WITHOUT considering the matrices component wise?
 

InteGrand

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Thanks, I'll leave the latter for later!

I'm pretty frustrated by these...What would be the shortest way of doing them?
Prove the following properties of matrices:
1) If the product AB exists, then A(λB) = λ(AB) = (λA)B
2) Associative law of matrix multiplication. If products AB and BC exist, then A(BC) = (AB)C
3) AI = A and IA = A where I represents identity matrices of the appropriate (possibly different) sizes
4) Left distributive law. If A+B and AC exist, then (A+B)C = AC+BC
5) Right distributive law If B+C and AB exist, then A(B+C) = AB+AC

I know you could do these writing each of them out...but it takes way to long to do so test-wise (+ its frustrating).






 

InteGrand

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Thanks

But is there a way to prove the distributive laws WITHOUT considering the matrices component wise?
You don't need to write out an actual nxn matrix. You just need to use the definition of the i,j entry of a matrix product as a sum (essentially a dot product) and use properties of summations, denoting the i,j entry of a matrix A by say aij.
 

1008

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You don't need to write out an actual nxn matrix. You just need to use the definition of the i,j entry of a matrix product as a sum (essentially a dot product) and use properties of summations, denoting the i,j entry of a matrix A by say aij.
You don't need to write out an actual nxn matrix. You just need to use the definition of the i,j entry of a matrix product as a sum (essentially a dot product) and use properties of summations, denoting the i,j entry of a matrix A by say aij.
Is this right?


This is for the first question, btw
 

InteGrand

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Is this right?


This is for the first question, btw






In other words, the first part of the question follows due to the familiar properties of arithmetic in fields like the real/complex numbers.

(Also, your sums have been referring to the i-j components of matrices rather than the matrices themselves, so it wouldn't be right to say A(lambda B) equals those, for instance.)
 
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