Linear Algebra problems im having

joey12345

LeixlipRed wrote: »

Probably a million ways of doing that one! If two objects don't intersect each other what would you expect to find if you attempted to find their points of intersection?

Or a vector on the line, what would it's relationship to the normal vector of the plane be if they don't intersect? Think about the relationship the line and the plane must have if they don't intersect....

hi leixlipred thanks for the reply. if the line and the plane dont intersect then they must have different coordinates. and if two objects dont intersect then they must not have any coordinates that are the same?

p.s thanks for the help with the "most general" matrix question. they show up multiple times each year so thanks to you ive hopefully got a little headstart with the exam next week

LeixlipRed

Yeh, that's kind of the general idea but the second way I mentioned is easier to compute. Do you know what the normal vector to a plane is? Try draw a picture, a plane and a line, what is the only type of line that won't touch a plane? One that is parallel to the plane! Otherwise it will pass through it. So if the plane is parallel to the line the normal vector of that plane and a vector on that line will be....?

joey12345

LeixlipRed wrote: »

Yeh, that's kind of the general idea but the second way I mentioned is easier to compute. Do you know what the normal vector to a plane is? Try draw a picture, a plane and a line, what is the only type of line that won't touch a plane? One that is parallel to the plane! Otherwise it will pass through it. So if the plane is parallel to the line the normal vector of that plane and a vector on that line will be....?

when a line and a plane are parallel then they are the same distance from eachother always. a normal vector is perpendicular to any other vector on the plane but i dont know the relationship from the normal vector to the vector on the parallel line? is it that theyll always be perpendicular ?

LeixlipRed

Yeh, if a line and plane are parallel then the normal vector of the plane will be perpendicular to any vector on or parallel to that line. Now, how do you show two vectors are perpendicular?

joey12345

LeixlipRed wrote: »

Yeh, if a line and plane are parallel then the normal vector of the plane will be perpendicular to any vector on or parallel to that line. Now, how do you show two vectors are perpendicular?

two vectors are perpendicular if there dot product is equal to zero.

LeixlipRed

Exactly so do you know what you have to do now?

joey12345

LeixlipRed wrote: »

Exactly so do you know what you have to do now?

just looking on yotube there for a few examples do i get the vector equation of the line?

or is it much more simpler than this and im just not seeing it?

LeixlipRed

The vector equation of a line won't be a vector, it'll be an equation. All you need is to get a vector on that line. You have two points on the line, how do you do that?

joey12345

LeixlipRed wrote: »

The vector equation of a line won't be a vector, it'll be an equation. All you need is to get a vector on that line. You have two points on the line, how do you do that?

i just take away Q from P so ill end up with <1, 6, -4>

and then i get the dot product with the normal vector of the plane.

i take it that if the the plane is this..... 2x+y-z=8

then the normal vector is <2, 1, -1>

i dont think thats right is it?

because when i get there dot products its a non zero answer and that would mean that there not perpindicular?

LeixlipRed

You've calculated QP wrong, check your work.

joey12345

LeixlipRed wrote: »

You've calculated QP wrong, check your work.

ahhhh man sorry didnt see that Q had a MINUS 3.

yea that worked out fine and dandy thanks for the help leixlipred.

say if that question was if they DO intersect?

would it be done a different way?

what you think of this one?

show that the vectors
a= i + 2j - 3k
b=2i - j + 2k
c=3i + j -k

lie in the same plane?(ie coplanar)

so is all i have t do here is find the determinant? and if it equals zero then it they are co planar?

joey12345

k got that one i just got the determinant which was equal to 0 which means the vectors are linearly dependent and therefore coplanar.

Michael Collins

joey12345 wrote: »

ahhhh man sorry didnt see that Q had a MINUS 3.

yea that worked out fine and dandy thanks for the help leixlipred.

say if that question was if they DO intersect?

would it be done a different way?

what you think of this one?

show that the vectors
a= i + 2j - 3k
b=2i - j + 2k
c=3i + j -k

lie in the same plane?(ie coplanar)

so is all i have t do here is find the determinant? and if it equals zero then it they are co planar?

Just thinking for the original problem you should probably also check that the line doesn't lie in the plane, as if it did technically it would intersect it everywhere, but still be perpendicular to the normal to the plane. It's easy enough to do this for a single point.

joey12345

i was actually thinking that myself right ill check that now so.

joey12345

right i plugged in the values and none of them satisfied the equation. thats right isnt it? just plug in one of the points into the equation?

or do i use the parametric representation?

LeixlipRed

Michael Collins wrote: »

Just thinking for the original problem you should probably also check that the line doesn't lie in the plane, as if it did technically it would intersect it everywhere, but still be perpendicular to the normal to the plane. It's easy enough to do this for a single point.

Ah good spot, forgot to mention that. Was trying to think what is the quickest way of approaching this question. I reckon if you find the parametric equation of the line and plug it into the plane to get a contradiction is another quick way of doing it. No matter though, all methods are valid.

joey12345

LeixlipRed wrote: »

Ah good spot, forgot to mention that. Was trying to think what is the quickest way of approaching this question. I reckon if you find the parametric equation of the line and plug it into the plane to get a contradiction is another quick way of doing it. No matter though, all methods are valid.

yea i done the parametric representation there and when i was solving for T it cancelled itself out which would mean that it doesnt intersect the plane right?

is another way to do it just subbing in P or Q?

like for example P(2, -1, 4). could i use 2=x, -1=y, and 4=z and then sub in into the plane to get a contradiction? or is that wrong? well either way the dot product worked out and so did the parametric representation.

LeixlipRed

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.

joey12345

Michael Collins wrote: »

Just thinking for the original problem you should probably also check that the line doesn't lie in the plane, as if it did technically it would intersect it everywhere, but still be perpendicular to the normal to the plane. It's easy enough to do this for a single point.

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.

sorry i wasnt clear there thats what i mean by subbing in P that will prove that its not on the plane if you get a contradiction.

LeixlipRed

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.

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?