# Proof using dot and cross product.

I am looking at a derivation of the parallel axis theorem in a book. The photo is below

I'm not sure why but I just cannot figure out how they got from line 2 to line 3. I have thought through all of the properties I know (I have expanded and used the vector triple product, the cyclic permutations of scalar triple product etc) but I cannot seem to figure out that step.

Would very much appreciate if someone could enlighten me as to what property/rule was used here.

• lagrange's identity – juan arroyo Jan 5 '17 at 19:25

$$(a \times b) \cdot (c \times d) = \varepsilon_{kij} a_i b_j \varepsilon_{klm} c_l d_m = (\delta_{il} \delta_{jm}-\delta_{im}\delta_{kl}) a_i b_j c_l d_m = (a \cdot c)(b \cdot d) - (a \cdot d)(b \cdot c),$$ so in this case, $$\lVert \hat{n} \times (r-a) \rVert^2 = \lVert \hat{n} \rVert^2 \lVert r-a \rVert^2 - (\hat{n} \cdot (r-a))^2.$$

• Thank you for your reply! I thought I would be OK with the rest of the proof but I have also gotten stuck on the penultimate line. I have read the note below the proof several times but I don't see how it applies as you do not have an independent vector r in the integral that is the third from last line. How can the result of integrating the vector r with respect to the mass be used when inyegrating the magnitude of r or when integrating the dot product of a with r? – Meep Jan 5 '17 at 20:16
• If it's the same as this edition of the solution manual to Riley, Hobson and Bence, the definition of $I_0$ becomes $\int (r^2-(n \cdot r)^2) \, dm$, which deals with the terms with two $r$s, and linearity of the integral gives $\int a \cdot r \, dm = a \cdot \int r \, dm$, which is what happens to the linear terms. – Chappers Jan 5 '17 at 20:36
• Thank you for your reply. I still have questions about this so I feel I should make a new post, but thank you for all of your insights! – Meep Jan 6 '17 at 12:50