Some vector questions

1.
I'll assume you mean fp+gq+hr=0.
f(2i+3j)+g(4i-5j)+h(i-4j)=0
i(2f+4g+h)+j(3f-5g-4h)=0
Thus (1)2f+4g+h=0 and (2)3f-5g-4h=0
That is, 8f+16g+4h=0 and 3f-5g-4h=0
Adding we get: 11f+11g=0 > g=-f.
Putting this into equation 1 we get: 2f-4f+h=0 > h=2f and putting this into equation 2 we get 3f-5(-f)-4h=0 > 8f=4h > h=2f
Thus we know that h=2f=-2g.
Thus the set of numbers (f,g,h) belong to the set $f \in R, (f,g,h)=(f,-f,2f)$
As for the geometric meaning, possibly: if you move along scalar multiplies f,g,h of the vectors p,q,r respectively then the total displacement is zero.
Clearly there will be more than one answer because you could move along each vector twice (giving the scalar multiple double to what it was before) and you'd still arrive back where you started.

2.
2. Four points A, B, C, and D with position vectors a, b, c, and d are vertices of a tetrahedron. The mid-points of BC, CA, AB, AD, BD, CD are denoted by P, Q, R, U, V, W. Fnid the position vector of the mid-points of PU, QV and RW. State your conclusion as a geometrical theorem.
Clearly the midpoints of BC, CA, AB, AD, BD, CD are P=0.5(b+c), Q=0.5(a+c), R=0.5(a+b), U=0.5(a+d), V=0.5(b+d) and W=0.5(c+d) respectively.
Its then easy to find the mid points of PU, QV and RW respectively; for example, the midpoint of PU is 0.5[0.5(b+c)+0.5(a+d)].
I expect that the position vectors of the mid points of PU, QV and RW will satisfy some relation which is where you can get your geometrical meaning from.


3.
A(3,2,4), B(-3,-7,-8), C(0,1,3), D(-2,5,9)
The line AB has equation r=3i+2j+4k+s(2i+3j+4k)=i(3+2s)+j(2+3s)+k(4+4s)
The line CD has equation r=0i+j+3k+t(-i+2j+3k)=i(0-t)+j(1+2t)+k(3+3t)
Hence when they intersect we have:
-t=3+2s > t=-3-2s
2+3s=1+2t > 1+3s=2(-3-2s) > 1+3s=-6-4s > 7s=-7 > s=-1 > t=-1
Checking the k coefficients we see 4+4s=3+3t is satisfied.
Hence the lines intersect at s=-1, t=-1 and the point of intersection is (1,-1,0)

Sorry - after 16 hours I thought it was fair game though
If you had a solution half typed then still post it - sometimes people find it useful to see two slightly different points of view on something.

1.
I'll assume you mean fp+gq+hr=0.
f(2i+3j)+g(4i-5j)+h(i-4j)=0
i(2f+4g+h)+j(3f-5g-4h)=0
Thus (1)2f+4g+h=0 and (2)3f-5g-4h=0
That is, 8f+16g+4h=0 and 3f-5g-4h=0
Adding we get: 11f+11g=0 > g=-f.
Putting this into equation 1 we get: 2f-4f+h=0 > h=2f and putting this into equation 2 we get 3f-5(-f)-4h=0 > 8f=4h > h=2f
Thus we know that h=2f=-2g.
Thus the set of numbers (f,g,h) belong to the set $f \in R, (f,g,h)=(f,-f,2f)$
As for the geometric meaning, possibly: if you move along scalar multiplies f,g,h of the vectors p,q,r respectively then the total displacement is zero.
Clearly there will be more than one answer because you could move along each vector twice (giving the scalar multiple double to what it was before) and you'd still arrive back where you started.

2.
2. Four points A, B, C, and D with position vectors a, b, c, and d are vertices of a tetrahedron. The mid-points of BC, CA, AB, AD, BD, CD are denoted by P, Q, R, U, V, W. Fnid the position vector of the mid-points of PU, QV and RW. State your conclusion as a geometrical theorem.
Clearly the midpoints of BC, CA, AB, AD, BD, CD are P=0.5(b+c), Q=0.5(a+c), R=0.5(a+b), U=0.5(a+d), V=0.5(b+d) and W=0.5(c+d) respectively.
Its then easy to find the mid points of PU, QV and RW respectively; for example, the midpoint of PU is 0.5[0.5(b+c)+0.5(a+d)].
I expect that the position vectors of the mid points of PU, QV and RW will satisfy some relation which is where you can get your geometrical meaning from.


3.
A(3,2,4), B(-3,-7,-8), C(0,1,3), D(-2,5,9)
The line AB has equation r=3i+2j+4k+s(2i+3j+4k)=i(3+2s)+j(2+3s)+k(4+4s)
The line CD has equation r=0i+j+3k+t(-i+2j+3k)=i(0-t)+j(1+2t)+k(3+3t)
Hence when they intersect we have:
-t=3+2s > t=-3-2s
2+3s=1+2t > 1+3s=2(-3-2s) > 1+3s=-6-4s > 7s=-7 > s=-1 > t=-1
Checking the k coefficients we see 4+4s=3+3t is satisfied.
Hence the lines intersect at s=-1, t=-1 and the point of intersection is (1,-1,0)