Hi i'm trying to prove the following equality where P,Q and R are any 3D vectors:
PxQxR = (P.R)Q - (Q.R)P
I find it easier by proving that the x coordinate of the left side is equal do the right side of the equation.
The x value of left side would lead us to = ((P3*Q1 - P1*Q3)R3 - (P1*Q2 - P2*Q1)*R2)x
The right side I got:
(Q1 * (P.R),Q2 * (P.R), Q3 * (P.R) - (P1 * (Q.R), P2 * (Q.R), P3 * (Q.R))
As I said before we will just calculate the x value, so we got this:
(Q1*P1*R1 + Q1*P2*R2 + Q1*P3*R3) - (P1*Q1*R1 + P1*Q2*R2 + P1*Q2*R2 + P1*Q3*R3)
The problem is that the previous step leading me to (P3*Q1 - P1*Q3)*R3 + (P2*Q1 - P1*Q2)*R2
Which results too:
(P3*Q1 - P1*Q3)*R3 + (P2*Q1 - P1*Q2)*R2 != (P3*Q1 - P1*Q3)R3 - (P1*Q2 - P2*Q1)*R2)
What am I doing wrong?
since the cross product is non-associative your first error is to write the ambiguous expression $P \times Q \times R$. since the scalar product gives a scalar, your second error is to write the incorrect form $(P.Q).R$
however your method of using co-ordinates should lead to the correct answer. the $i$-component of $P \times (Q \times R)$ is: $$ p_2(q_1r_2-q_2r_1)-p_3(q_3r_1-q_1r_3) $$ if we add to this $0=p_1q_1r_1-p_1q_1r_1$ it may be written as: $$ (p_1r_1+p_2r_2+p_3r_3)q_1 - (p_1q_1+p_2q_2+p_3q_3)r_1 $$ you should be able to see your way from here