Point Vectors in 3D Space: Are Cross Products of Two Vectors Equal?

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In 3D space, the equality of the magnitudes of the cross products of two vectors, specifically \|PQ × PR\| and \|PQ × QR\|, is confirmed under the condition that points P, Q, and R are distinct. Examples demonstrate that with distinct points, such as P=(0,0,0), Q=(1,0,0), and R=(0,3,0), the magnitudes of the cross products yield the same result. The discussion emphasizes that not only are the magnitudes equal, but the resulting vectors are also identical. The algebraic manipulation of the vectors reinforces this conclusion, illustrating the properties of cross products. Thus, the statement holds true for distinct points in 3D space.
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If P , \; Q and R each represent a point in \mathbb{R} ^ 3, then is it true that

\left\| {\overrightarrow {PQ} \times \overrightarrow {PR} } \right\| = \left\| {\overrightarrow {PQ} \times \overrightarrow {QR} } \right\| \; {?}
 
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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).
 
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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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...
 
bomba923 said:
If P , \; Q and R each represent a point in \mathbb{R} ^ 3, then is it true that

\left\| {\overrightarrow {PQ} \times \overrightarrow {PR} } \right\| = \left\| {\overrightarrow {PQ} \times \overrightarrow {QR} } \right\| \; {?}

It is always true. Furthermore, the resulting vectors are the same, not just their magnitude!
Rewrite your vectors as

\vec{A} \equiv \overrightarrow{PQ},
\vec{B} \equiv \overrightarrow{PR}, so that

(\vec{B} - \vec{A}) = \overrightarrow{QR}.

Take the same cross products algebraically, keeping in mind that
\vec{A} \times \vec{A} = 0, and that \vec{A} \times (\vec{B} - \vec{A}) = (\vec{A} \times \vec{B}) - (\vec{A} \times \vec{A}).
 
eeek!

Sorry bomba.
 
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