# Just a quick verification

bomba923
If $$P , \; Q$$ and $$R$$ each represent a point in $$\mathbb{R} ^ 3$$, then is it true that

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

Last edited:

Homework Helper
Gold Member
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).

bomba923
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:

Last edited:
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!

$$\vec{A} \equiv \overrightarrow{PQ}$$,
$$\vec{B} \equiv \overrightarrow{PR}$$, 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 $$\vec{A} \times (\vec{B} - \vec{A}) = (\vec{A} \times \vec{B}) - (\vec{A} \times \vec{A})$$.