Solving Vectors: Help for Q, R & S

[theacerf1]

HELP! Vectors...

Been stuck on these 2 questions for ages, PLEASE HELP! :(

(1) Let a, b and c be three vectors such that a + b + c = 0.
(i) Show that a × b = b × c = c × a.
(ii) Explain what this means geometrically. (Your answer should refer to a triangle.)
(2) (i) Find the normal vector to the plane through the points Q = (1, 1, 0), R = (0, 0, 3)
and S = (2, 2, 1).
(ii) Hence find the distance from the point P = (1, 0, 2) to this plane.

 

[Rasalhague]

You can solve 1.i algebraically, using the fact that the cross product distributes over addition, for example, a x c = a x (a + b), and the fact that the cross product of a vector with itself is the zero vector.

For 1.ii, have you tried drawing a diagram? Consider the connection between the cross product and area.

For 2.i, the cross product of two vectors is at right angles to both of them.

For 2.ii, think projection. What happens when you take the dot product of a vector with a unit vector?

 

[theacerf1]

Just looking at the first part, what do u mean when u say,

"You can solve 1.i algebraically, using the fact that the cross product distributes over addition, for example, a x c = a x (a + b), and the fact that the cross product of a vector with itself is the zero vector."

Like this is what i did,

a x b = a x (a + c)
a x b = a x a + a x c
a x b = a x c

but the questions says a x b = c x a...
also how can or why can you just say a x c = a x (a + b)?

im confused

 

[Char. Limit]

Actually, since c is not equal to a+b, that wouldn't work, I think. Rather, c is equal to -(a+b). That might work better for you.

EDIT: Also, the fact that a x -c = c x a is helpful.

 

[theacerf1]

Im still really confused bout wat Rasalhague wrote, i get what you said a x -c = c x a, but don't understand c is equal to -(a+b)...

 

[Char. Limit]

Well, if a+b+c=0, then we can subtract a and b from both sides, to get c = -a - b, and then just factor out the -1 to get c=-(a+b).

 

[theacerf1]

ahhh right ok so...

c=-(a+b)
axb = ax(a+c) <--------- how do u get that??
axb = axa + axc
axb = 0 + ax-(a+b)
axb = ax-a + ax-b <------- i don't feel like I am seeing/going anywhere with this, am i doing it right?

 

[Char. Limit]

I don't see where you get that a x b = a x (a+c). That's still not true. What is true, and what you have to start with, is...

a x c = a x -(a+b)

And of course, remember to use a+b+c=0 to rewrite a, and then b, in terms of the other two functions to solve the other relations.

 

[theacerf1]

Ok so i get this:

c x a = -(a+b) x -(b+c)
a x b = -(b+c) x -(c+a)
b x c = -(c+a) x -(a+b)

im confused...how am i showing a × b = b × c = c × a?
I got no idea how to solve:

-(a+b) x -(b+c) = -(b+c) x -(c+a) = -(c+a) x -(a+b)

 

[Char. Limit]

Start with a x c = a x (-a - b). Then distribute it, like so:

a x c = a x (-a) + a x (-b)

Then just work out the algebra on the two right cross products.

 

[theacerf1]

a x (-a) = 0

a x (-b) is just left as it is...right?

In my question it has c x a:
c x (-c) = 0
c x (-b) <--- left as it is?

after doing this for all, where do i go from there?
Im just going to end up with equations like

c x a = c x (-b)
a x b = a x (-c)
b x c = b x (-a)

Show that a × b = b × c = c × a...so what would be next? Thanks for the quick replys btw :)

 

[Char. Limit]

You're so close. Now just remember that for two vectors, say a and b, then a x (-b) = b x a. That's all you need now.

 

[Char. Limit]

Just wanted to put this in, by the way, this property here:

a x (-b) = b x a

Can easily be proven by just remembering that a x (-b) = - (a x b), and then using the fact that cross products are anti-commutative to finish it.

 

[theacerf1]

Ahhh i get it now!
cross products are anti-commutative ? wat does that mean?
Is this a law: a x (-b) = - (a x b)?
I think illl need to prove that before i can use it

 

[Char. Limit]

The law that a x (-b) = -(a x b) is just scalar multiplication on a cross product. You could prove it if you wanted, by writing out both sides (let a={a₁,a₂,a₃} and b={b₁,b₂,₃} and you can prove it), but you might be able to get away with just stating it as a law. If you're worried, though, I would prove that as well.

 

[theacerf1]

Ah cool, thanks for the help Char. Limit, understand everything now. Just the last part of the question if you got time,

Explain what this means geometrically...(Your answer should refer to a triangle.)

Draw a triangle with sides a, b, c?

 

[Char. Limit]

That would work. Of course, each of the sides of the triangle is a vector, and the three vectors add up to 0. Can you see what it's telling you?

 

[coki2000]

Been stuck on these 2 questions for ages, PLEASE HELP! :(

(1) Let a, b and c be three vectors such that a + b + c = 0.
(i) Show that a × b = b × c = c × a.
(ii) Explain what this means geometrically. (Your answer should refer to a triangle.)
(2) (i) Find the normal vector to the plane through the points Q = (1, 1, 0), R = (0, 0, 3)
and S = (2, 2, 1).
(ii) Hence find the distance from the point P = (1, 0, 2) to this plane.

The area of a triangle can be found by take cross products of ordered sides. This equality means that the area of triangle is always constant.

 

[coki2000]

For second question, i found that answer;
(i)-You can find a normal vector of the plane by taking cross products of two of these vectors.
(ii)-The plane in your question is xy-plane and the vector given in your question in xz-plane so the answer of it is only x component of the vector: (1,0,0).
I hope this is a true answer :)

 

[Rasalhague]

Actually, since c is not equal to a+b, that wouldn't work, I think. Rather, c is equal to -(a+b). That might work better for you.

Sorry about that, theacerf1. Thanks for the correction, Char. Limit.

 

[Rasalhague]

For second question, i found that answer;
(i)-You can find a normal vector of the plane by taking cross products of two of these vectors.
(ii)-The plane in your question is xy-plane and the vector given in your question in xz-plane so the answer of it is only x component of the vector: (1,0,0).
I hope this is a true answer :)

Question (ii) asks for the distance of P from the plane containing Q, R and S. The answer has to be a nonnegative real number. The plane is not the xy-plane; R and S both have nonzero z-component. I didn't get a multiple of (1,0,0) for the normal vector.

 

[Rasalhague]

The area of a triangle can be found by take cross products of ordered sides. This equality means that the area of triangle is always constant.

What do you mean by "the area of [?a,?the] triangle is always constant"? Perhaps you're saying, "If you draw a triangle with directed line segments for its sides, and find the absolute value of the cross product of any pair of vectors corresponsing to those directed line segments, then the result will be the same, no matter which pair of sides / directed line segments / vectors you choose" (and in that sense the result will stay constant, i.e. stay the same, even if you vary your choice of vectors).

 

[Rasalhague]

For second question, i found that answer;
(i)-You can find a normal vector of the plane by taking cross products of two of these vectors.

An interesting question to think about is: why does the method you describe work for this example? Suppose that instead of Q = (1,1,0), we had Q = (1,0,0). What is different about the example with Q = (1,0,0) that forces us to use a different method? What method should we use then?

 

[coki2000]

An interesting question to think about is: why does the method you describe work for this example? Suppose that instead of Q = (1,1,0), we had Q = (1,0,0). What is different about the example with Q = (1,0,0) that forces us to use a different method? What method should we use then?

I don't understand Rasalhague, why doesn't this method work? Is there any problem with this method? And also I don't understand what is the difference when Q=(1,0,0).