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Let u and v be vectors in 3. Prove Lagrange鈥檚 identity, Theorem 10.30:

role="math" localid="1649924794767" uv2=u2v2(uv)2

Short Answer

Expert verified

Hence, prove thatuv2=u2v2(uv)2.

Step by step solution

01

Step 1. Given Information

Let u and v be vectors in 3. Prove Lagrange鈥檚 identity, Theorem 10.30:

uv2=u2v2(uv)2

02

Step 2. We have to prove u×v2=u2v2−(u·v)2

Let u=(u1,u2,u3)andv=(v1,v2,v3)

Firstly finding the value of(uv)2=(u1v1+u2v2+u3v3)2(uv)2=(u1v1)2+(u2v2)2+(u3v3)2+2u1v1u2v2+2u2v2u3v3+2u3v3u1v1(uv)2=u12v12+u22v22+u32v32+2u1v1u2v2+2u2v2u3v3+2u3v3u1v1

03

Step 3. Now finding the value of u×v2

uv=detijku1u2u3v1v2v3uv=iu2v3-u3v2-ju1v3-u3v1+ku1v2-u2v1uv=i(u2v3-u3v2)-j(u1v3-u3v1)+k(u1v2-u2v1)uv=(u2v3-u3v2,u1v3-u3v1,u1v2-u2v1)uv2=(u2v3-u3v2,u1v3-u3v1,u1v2-u2v1)2uv2=(u2v3-u3v2)2+(u1v3-u3v1)2+(u1v2-u2v1)2uv2=u22v32+u32v22-2u2v3u3v2+u12v32+u32v12-2u1v3u3v1+u12v22+u22+v12-2u1v2-u2v1uv2=u12v22+u12v32+u22v12+u22v32+u32v12+u32v22-2u1v1u2v2-2u2v2u3v3-2u3v3u1v1

04

Step 4. Now finding the value of u2v2

u2v2=(u12,u22,u32)(v12,v22,v32)u2v2=u12(v12,v22,v32)+u22(v12,v22,v32)+u32(v12,v22,v32)u2v2=u12v12+u12v22+u12v32+u22v12+u22v22+u22v32+u32v12+u32v22+u32v32

05

Step 5. Now finding the value of u2v2−(u·v)2

u2v2(uv)2=u12v12+u12v22+u12v32+u22v12+u22v22+u22v32+u32v12+u32v22+u32v32-(u12v12+u22v22+u32v32+2u1v1u2v2+2u2v2u3v3+2u3v3u1v1)u2v2(uv)2=u12v12+u12v22+u12v32+u22v12+u22v22+u22v32+u32v12+u32v22+u32v32-u12v12-u22v22-u32v32-2u1v1u2v2-2u2v2u3v3-2u3v3u1v1u2v2(uv)2=u12v22+u12v32+u22v12+u22v32+u32v12+u32v22-2u1v1u2v2-2u2v2u3v3-2u3v3u1v1u2v2(uv)2=uv2

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