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If veca and vecb are any two vectors , t...

If `veca` and `vecb` are any two vectors , then prove that `|vecaxxvecb|^(2)=|veca|^(2)|vecb|^(2)-(veca.vecb)^(2)=|{:(veca.veca,veca.vecb),(veca.vecb,vecb.vecb):}|` or `|vecaxxvecb|^(2)+(veca.vecb)^(2)=|veca|^(2)|vecb|^(2)` (This is also known as Lagrange identily)

Text Solution

Verified by Experts

We have,
Here `|vecaxxvecb|=|veca||vecb|sintheta`
`:. " " |vecaxxvecb|^(2)=|veca|^(2)|vecb|^(2) sin^(2)theta`
`rArr " " |vecaxxvecb|^(2)=|veca|^(2)|vecb|^(2)(1-cos^(2)theta)`
`rArr " " |vecaxxvecb|^(2)=|veca^(2)||vecb|^(2)-|veca|^(2)|vecb|^(2)cos^(2)theta`
`rArr " " |vecaxxvecb|^(2)=|veca|^(2)|vecb|^(2)-(|veca||vecb|costheta)^(2)`
`rArr " " |vecaxxvecb|^(2)=|veca|^(2)|vecb|^(2)-(veca.vecb)^(2)`
`|vecaxx vecb|^(2)=|{:(|veca|^(2),veca.vecb),(veca.vecb,|vecb|^(2)):}|`
`=|{:(veca.veca,veca.vecb),(veca.vecb,vecb.vecb):}|`
Hence `|vecaxxvecb|^(2)=|veca|^(2)|vecb|^(2)-(veca.vecb)^(2)`
`rArr |veca.vecb|^(2)+(veca.vecb)^(2)=|veca|^(2)|vecb|^(2)` proved
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