# vectorsWatch

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#1
yes i know i'm at uni... but i'm pretty sure this is on the A level course too (which my recollection of is a bit rusty)

any help with this question will be muchos appreciated (rep will be given)

Let A,B,C and D be vertices of a non-regular tetrahedron, and let P, Q, R and S byt the midpoints Of AB, BC, CD and DA respectively. Show that O, Q, R and S are the vertices of a parallelogram. Verify that P,Q,R and S are coplaner. Prove that lines joining PR and QS intersect and bisect each other.

lou xxx
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14 years ago
#2
Oh, I did some kinds of this question before ...but here I haven't studied it ..so I forgot alot. Give me time to refresh my memory.
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#3
(Original post by BCHL85)
Oh, I did some kinds of this question before ...but here I haven't studied it ..so I forgot alot. Give me time to refresh my memory.

cheers! (and i'm not in a rush, was just doing some questions and honestly didn't know where to start with this on)

lou xxx
0
14 years ago
#4
Ok, it's not a vetors problem , I'm sure . First draw a diagram. Then see PS // BD, and QR // BD. And PS = QR = 1/2BD.
So PS //= QR
Similarly PQ //= SR (//=1/2AC).
So it's parrallegram. The others will follow this
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14 years ago
#5
sounds like you are proving Varignon's Theorem...
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#6
okies.. that makes sense... just checking, does // mean parallel?

lou xxx
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#7
(Original post by El Stevo)
sounds like you are proving Varignon's Theorem...
maybe... although i don't think that was the point of the question though, it's just part of a revsion sheet

lou xxx
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14 years ago
#8
(Original post by lou p lou)
okies.. that makes sense... just checking, does // mean parallel?

lou xxx
yeah, right. // means parallel.
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14 years ago
#9
(Original post by lou p lou)
Let A,B,C and D be vertices of a non-regular tetrahedron, and let P, Q, R and S byt the midpoints Of AB, BC, CD and DA respectively. Show that O, Q, R and S are the vertices of a parallelogram. Verify that P,Q,R and S are coplaner. Prove that lines joining PR and QS intersect and bisect each other.
P = (A+B)/2
Q = (B+C)/2
R = (C+D)/2
S = (A+D)/2

PQ = (C-A)/2 = SR
PS = (D-B)/2 = QR

Hence PQRS is a parallelogram. Hence PQRS are coplanar (being vertices of a planar object!) or note PQ, PR, PS are dependent.

(A+B+C+D)/4 = (P+R)/2 = (Q+S)/2

showing that that point is the midpoint of PR and QS both - hence they intersect.
0
14 years ago
#10
(Original post by RichE)
P = (A+B)/2
Q = (B+C)/2
R = (C+D)/2
S = (A+D)/2

PQ = (C-A)/2 = SR
PS = (D-B)/2 = QR

Hence PQRS is a parallelogram. Hence PQRS are coplanar (being vertices of a planar object!) or note PQ, PR, PS are dependent.

(A+B+C+D)/4 = (P+R)/2 = (Q+S)/2

showing that that point is the midpoint of PR and QS both - hence they intersect.
Woa...using vectors here I think more complicated... I don't know if you can use Tallet rule or you must use vectors.
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#11
(Original post by BCHL85)
Woa...using vectors here I think more complicated... I don't know if you can use Tallet rule or you must use vectors.

i think the question was intended for vectors. grrr, i don't liek questiosn like this, they're really obvious but have to be done an unnecessarily complicated way(i'd proved in about 2 seconds by drawing a diagram)

anyway... thanks guys

lou xxx
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