Let me see if I understand your method... I am doing a dot product of QT and PR to get a vector T, the normal to the line, which I can use with a length formula to get what I need to know? What about the t parameter in PR?
Homework Statement
Find the distance from the point Q to the line passing through the points P and R.
Homework Equations
r=r0+tv
D=|ax0 + by0 + cz0 + d| / sqrt(a2 + b2 + c2)
The Attempt at a Solution
To find the line:
(x,y,z) = (2,3,1) + t(3-2,1-3,1-1)
(x,y,z) = (2,3,1) + t(1,-2,0)...
Ah, right. That makes much sense. Thank you :)
So you're just calculating [-x3- x5, -x3- 2x5, x3, 0, x5], then splitting it up into terms of x3 and x5, then factoring out?
Homework Statement
Consider the matrix A:
1 4 5 0 9
3 -2 1 0 -1
-1 0 -1 0 -1
2 3 5 1 8
(Sorry I don't know how to do TeX matrices on this site)
Find a basis for the row, column, and null space.
Homework Equations
The Attempt at a Solution
I reduced to row echelon form, which...
Both answers were helpful. I do have another question though, since the solution given did not mention the origin. The solution to the question I asked above is:
(x+x') + 2(y+y') + (z+z') = (x+2y+z) + (x'+2y'+z') = 6 + 6 = 12
thus
(x+x') + 2(y+y') + (z+z') is not in P, and so P is not a...
This is a bit of a guess here, since I am a bit confused on the whole thing still, but I believe it must pass through the origin, and since it states x+2y+z=6, that is not satisfied...?
Homework Statement
P={(x,y,z)|x+2y+z=6}, a plane in R3. P is not a subspace of R3. Why?
Homework Equations
See below.
The Attempt at a Solution
I am really quite confused here.
My text says:
"A subset W of a vector space V is called a subspace of V if W is itself a vector space...
Thanks for the tip and clarification :]
I think you are correct, unless I made a mistake. Using this approach I found that all axioms were satisfied except axiom 5. Thoughts on this?
Homework Statement
Homework Equations
The 10 axioms:
1. If u and v are objects in V, then u + v is in V
2. u + v = v + u
3. u + (v+w) = (u+v)+w
4. There is an object 0 in V, called a zero factor for V, such that 0+u = u+0 = u for all u in V
5. For each u in V, there is an object -u in V...
Thanks! Now I have my t and u values, and plugging them into the third equation gives me equality...but where do I go from here? Do I plug them into all of the original equations and use the results as my intersection point?
Homework Statement
Find the point of intersection of two lines
x = -9 + 5t
y = 1 + t
z = 10 - 4t
and
x = -2 -3t
y = 5 + 2t
z = 5 + 3t
Homework Equations
N/A
The Attempt at a Solution
I have read that you should set two of the equations equal to find the value of t, and...