Linear Algebra problems im having

joey12345

LeixlipRed wrote: »

Yeh you prove the point P is not on the plane. But what about the line joining P to Q? It tells you nothing about that.

ah ok i get you now. so just to clear it up here after i got the parametric representation and couldnt get T. i should just write ... because we cant solve for T this proves that theres no point of intersection??

LeixlipRed

Yeh, something to that effect will do.

joey12345

LeixlipRed wrote: »

Yeh, something to that effect will do.

good stuff thanks a million for the help man. im sure theyll be a lot more questions i get stuck on.

Michael Collins

LeixlipRed wrote: »

Just subbing in the point proves nothing other than that point doesn't lie on the plane. A line is a collection of infinitely many points so no that method does not answer the original question.

As for subbing in the parametric equations into the plane, the reason this shows that our line and plane intersect is because we cannot find a t that satisfies this equation as you alluded to.

I meant you could do this after doing the first part the way joey12345 did. Since we know from the first part that the line is either everywhere parallel to - or everywhere in - the plane in question, isn't subbing one point and seeing that it's not in the plane valid?

Obviously without the first part proving the line is parallel to the plane, this way isn't enough to answer the question.

The parametic method is good though, nice and quick.

joey12345

Michael Collins wrote: »

I meant you could do this after doing the first part the way joey12345 did. Since we know from the first part that the line is either everywhere parallel to - or everywhere in - the plane in question, isn't subbing one point and seeing that it's not in the plane valid?

Obviously without the first part proving the line is parallel to the plane, this way isn't enough to answer the question.

The parametic method is good though, nice and quick.

thats what i was trying to say just couldnt word it correct. what do you think is the way around Q3 in the original problems?

LeixlipRed

Apologies, I thought you were asking if originally it's enough to sub P or Q into the plane. Which it's not. But if you know PQ is parallel to the plane then obviously P or Q would satisfy the plane if it lies in it. Plugging either in then is correct.

joey12345

LeixlipRed wrote: »

Apologies, I thought you were asking if originally it's enough to sub P or Q into the plane. Which it's not. But if you know PQ is parallel to the plane then obviously P or Q would satisfy the plane if it lies in it. Plugging either in then is correct.

no probs man its my own fault for not saying it properly.

LeixlipRed

No worries.

joey12345

have another problem here lads.
show that x=-6 is the only real root of the equation

det
x 4 2
1 x 5
3 3 x

=

0



do i just sub in -6 here and get the determinant ?

how do i show that -6 is the ONLY real root?

hope you guys can help

Michael Collins

joey12345 wrote: »

have another problem here lads.
show that x=-6 is the only real root of the equation

det
x 4 2
1 x 5
3 3 x

=

0



do i just sub in -6 here and get the determinant ?

how do i show that -6 is the ONLY real root?

hope you guys can help

Expand the determinant and set it equal to zero. You'll now have a polynomial equation in [latex] x^3 [/latex], i.e. a cubic equation. You should be able to take [latex] x+6 [/latex] out of this as a factor (since -6 is a root, according to the question). Now, does the remaining factor have any real roots?

joey12345

ok so i got the determinant and ended up with

x^3-25x+66=0

i no that x+6 is a factor and then x=-6 so i get -66+66=0

should i do any more checks or would that be all do you think?

Ficheall

What you have shows that -6 is a real root, but you must also show that the other roots are imaginary.

If you divide x^3-25x+66 by x+6 you should be left with a quadratic with imaginary roots. Only then can you say that x+6 is the only real root.

joey12345

ahh yes thank you ill do that now

joey12345

ah yea ok so when i divided in i got
x^2-6x+11

so because this cant be done off the top of your head like more easy quadratics that would mean ive to use the quadratic formula,get them in the form a+bi so -6 IS the only real root?

Michael Collins

joey12345 wrote: »

ah yea ok so when i divided in i got
x^2-6x+11

so because this cant be done off the top of your head like more easy quadratics that would mean ive to use the quadratic formula,get them in the form a+bi so -6 IS the only real root?

Correct. To be extra precise you could show that the quadratic discriminant is negative: [latex] b^2 - 4ac < 0 [/latex] which means there can be no real roots.

joey12345

thanks man. the exams tomorrow and Q1 is built up of easy enough Q's so just getting the seperate questions out so ive a better idea and then hopefully ill be fine.