Just a quick verification

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If [tex] P , \; Q [/tex] and [tex] R [/tex] each represent a point in [tex] \mathbb{R} ^ 3 [/tex], then is it true that

[tex] \left\| {\overrightarrow {PQ} \times \overrightarrow {PR} } \right\| = \left\| {\overrightarrow {PQ} \times \overrightarrow {QR} } \right\| \; {?} [/tex]
 
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  • #2
quasar987
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Almost any arbitrary choice of points is a counter exemple, the most obvious being with P = R = (0,0,0). If you meant "P,Q,R distincts", try it with P=(0,0,0), Q = (1,0,0), R = (0,3,0).
 
  • #3
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quasar987 said:
Almost any arbitrary choice of points is a counter exemple, the most obvious being with P = R = (0,0,0). If you meant "P,Q,R distincts", try it with P=(0,0,0), Q = (1,0,0), R = (0,3,0).
Right, I meant distinct points P,Q, & R. :shy:
 
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  • #4
rachmaninoff
quasar987 said:
Almost any arbitrary choice of points is a counter exemple, the most obvious being with P = R = (0,0,0). If you meant "P,Q,R distincts", try it with P=(0,0,0), Q = (1,0,0), R = (0,3,0).
PQ x PR = (1,0,0) x (0,3,0) = (0,0,3)
PQ x QR = (1,0,0) x (-1,3,0) = (0,0,3)

seem to have the same magnitude...
 
  • #5
rachmaninoff
bomba923 said:
If [tex] P , \; Q [/tex] and [tex] R [/tex] each represent a point in [tex] \mathbb{R} ^ 3 [/tex], then is it true that

[tex] \left\| {\overrightarrow {PQ} \times \overrightarrow {PR} } \right\| = \left\| {\overrightarrow {PQ} \times \overrightarrow {QR} } \right\| \; {?} [/tex]
It is always true. Furthermore, the resulting vectors are the same, not just their magnitude!
Rewrite your vectors as

[tex]\vec{A} \equiv \overrightarrow{PQ}[/tex],
[tex]\vec{B} \equiv \overrightarrow{PR}[/tex], so that

[tex](\vec{B} - \vec{A}) = \overrightarrow{QR}[/tex].

Take the same cross products algebraically, keeping in mind that
[tex]\vec{A} \times \vec{A} = 0[/tex], and that [tex]\vec{A} \times (\vec{B} - \vec{A}) = (\vec{A} \times \vec{B}) - (\vec{A} \times \vec{A})[/tex].
 
  • #6
quasar987
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eeek!

Sorry bomba.
 
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