To prove the vector identity[(A\u00d7B)\u00d7(C\u00d7D)]=[(A\u00d7B)\u22c5D]C\u2212[(A\u00d7B)\u22c5C]D,[(A \\times B) \\times (C \\times D)] = [(A \\times B) \\cdot D] C – [(A \\times B) \\cdot C] D,[(A\u00d7B)\u00d7(C\u00d7D)]=[(A\u00d7B)\u22c5D]C\u2212[(A\u00d7B)\u22c5C]D,<\/p>\n\n\n\n
we will use the Levi-Civita tensor \u03f5ijk\\epsilon_{ijk}\u03f5ijk\u200b, which is a completely antisymmetric tensor used to express cross products and determinants.<\/p>\n\n\n\n
Step 1: Express the Cross Products Using Levi-Civita<\/h3>\n\n\n\n
Recall that the cross product of two vectors can be written in terms of the Levi-Civita symbol as follows:(A\u00d7B)i=\u03f5ijkAjBk,(A \\times B)_i = \\epsilon_{ijk} A_j B_k,(A\u00d7B)i\u200b=\u03f5ijk\u200bAj\u200bBk\u200b,<\/p>\n\n\n\n
where the summation convention is used over repeated indices.<\/p>\n\n\n\n
Step 2: Express the Left-Hand Side<\/h3>\n\n\n\n
Now consider the left-hand side of the equation, (A\u00d7B)\u00d7(C\u00d7D)(A \\times B) \\times (C \\times D)(A\u00d7B)\u00d7(C\u00d7D). Using the Levi-Civita symbol, we write this cross product as:[(A\u00d7B)\u00d7(C\u00d7D)]i=\u03f5ilm(A\u00d7B)l(C\u00d7D)m.[(A \\times B) \\times (C \\times D)]_i = \\epsilon_{ilm} (A \\times B)_l (C \\times D)_m.[(A\u00d7B)\u00d7(C\u00d7D)]i\u200b=\u03f5ilm\u200b(A\u00d7B)l\u200b(C\u00d7D)m\u200b.<\/p>\n\n\n\n
Substitute the expressions for the cross products:[(A\u00d7B)\u00d7(C\u00d7D)]i=\u03f5ilm(\u03f5lqkAqBk)(\u03f5mpnCpDn).[(A \\times B) \\times (C \\times D)]_i = \\epsilon_{ilm} \\left( \\epsilon_{lqk} A_q B_k \\right) \\left( \\epsilon_{mpn} C_p D_n \\right).[(A\u00d7B)\u00d7(C\u00d7D)]i\u200b=\u03f5ilm\u200b(\u03f5lqk\u200bAq\u200bBk\u200b)(\u03f5mpn\u200bCp\u200bDn\u200b).<\/p>\n\n\n\n
Step 3: Simplify Using the Identity for the Product of Two Levi-Civita Symbols<\/h3>\n\n\n\n
We now use the identity for the product of two Levi-Civita symbols:\u03f5lqk\u03f5mpn=\u03b4lm\u03b4qn\u2212\u03b4ln\u03b4qm.\\epsilon_{lqk} \\epsilon_{mpn} = \\delta_{lm} \\delta_{qn} – \\delta_{ln} \\delta_{qm}.\u03f5lqk\u200b\u03f5mpn\u200b=\u03b4lm\u200b\u03b4qn\u200b\u2212\u03b4ln\u200b\u03b4qm\u200b.<\/p>\n\n\n\n
Substitute this identity into the equation:[(A\u00d7B)\u00d7(C\u00d7D)]i=(\u03b4lm\u03b4qn\u2212\u03b4ln\u03b4qm)AqBkCpDn.[(A \\times B) \\times (C \\times D)]_i = \\left( \\delta_{lm} \\delta_{qn} – \\delta_{ln} \\delta_{qm} \\right) A_q B_k C_p D_n.[(A\u00d7B)\u00d7(C\u00d7D)]i\u200b=(\u03b4lm\u200b\u03b4qn\u200b\u2212\u03b4ln\u200b\u03b4qm\u200b)Aq\u200bBk\u200bCp\u200bDn\u200b.<\/p>\n\n\n\n
This expression can be simplified into two terms. Let’s break it down.<\/p>\n\n\n\n
Step 4: Expand and Interpret the Terms<\/h3>\n\n\n\n
The first term is:\u03b4lm\u03b4qnAqBkCpDn=AqBkCpDk,\\delta_{lm} \\delta_{qn} A_q B_k C_p D_n = A_q B_k C_p D_k,\u03b4lm\u200b\u03b4qn\u200bAq\u200bBk\u200bCp\u200bDn\u200b=Aq\u200bBk\u200bCp\u200bDk\u200b,<\/p>\n\n\n\n
which corresponds to (A\u00d7B)\u22c5D(A \\times B) \\cdot D(A\u00d7B)\u22c5D times CCC.<\/p>\n\n\n\n
The second term is:\u2212\u03b4ln\u03b4qmAqBkCpDn=\u2212AqBkCqDk,- \\delta_{ln} \\delta_{qm} A_q B_k C_p D_n = – A_q B_k C_q D_k,\u2212\u03b4ln\u200b\u03b4qm\u200bAq\u200bBk\u200bCp\u200bDn\u200b=\u2212Aq\u200bBk\u200bCq\u200bDk\u200b,<\/p>\n\n\n\n
which corresponds to (A\u00d7B)\u22c5C(A \\times B) \\cdot C(A\u00d7B)\u22c5C times DDD.<\/p>\n\n\n\n
Thus, we have:[(A\u00d7B)\u00d7(C\u00d7D)]i=[(A\u00d7B)\u22c5D]Ci\u2212[(A\u00d7B)\u22c5C]Di.[(A \\times B) \\times (C \\times D)]_i = [(A \\times B) \\cdot D] C_i – [(A \\times B) \\cdot C] D_i.[(A\u00d7B)\u00d7(C\u00d7D)]i\u200b=[(A\u00d7B)\u22c5D]Ci\u200b\u2212[(A\u00d7B)\u22c5C]Di\u200b.<\/p>\n\n\n\n
Step 5: Conclusion<\/h3>\n\n\n\n
We have proven that[(A\u00d7B)\u00d7(C\u00d7D)]=[(A\u00d7B)\u22c5D]C\u2212[(A\u00d7B)\u22c5C]D,[(A \\times B) \\times (C \\times D)] = [(A \\times B) \\cdot D] C – [(A \\times B) \\cdot C] D,[(A\u00d7B)\u00d7(C\u00d7D)]=[(A\u00d7B)\u22c5D]C\u2212[(A\u00d7B)\u22c5C]D,<\/p>\n\n\n\n
using the Levi-Civita tensor and the properties of the cross product and the delta function. This result is consistent with vector identities in classical mechanics and electromagnetism.<\/p>\n\n\n\n Levi-Civita practice; use the Levi-Civita tensor to prove the following vector identities: $[ (A \\times B) \\times (C \\times D) ] = [ (A \\times B) \\cdot D ] C – [ (A \\times B) \\cdot C ] D The Correct Answer and Explanation is: To prove the vector identity[(A\u00d7B)\u00d7(C\u00d7D)]=[(A\u00d7B)\u22c5D]C\u2212[(A\u00d7B)\u22c5C]D,[(A \\times B) \\times (C \\times […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-47885","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47885","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=47885"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47885\/revisions"}],"predecessor-version":[{"id":47887,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47885\/revisions\/47887"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=47885"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=47885"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=47885"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}![\"\"](\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-294.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"