Question

Let `veca, vecb` and `vecc` be three unit vectors such that `|veca - vecb|^2 + |veca - vecc|^2` = 8. Then find the value of `|veca + 2vecb|^2 + |veca + 2vecc|^2`

Options

• 0.00

• 1.00

• 2.00

• 3.00

Answer 1

2.00

Explanation:

`|veca - vecb|^2 + |veca - vecc|^2` = 8

`|veca|^2 + |vecb|^2 - 2veca.vecb + |veca|^2 + |vecc|^2 - 2veca.vecc` = 8

`1 + 1 + 1 + 1 - 2(veca.vecb + veca.vecc)` = 8

`veca.vecb + veca.vecc` = –2  ...(i)

The value of `|veca + 2vecb|^2 + |veca + 2vecc|^2` 

= `|veca|^2 + 4|vecb|^2 + 4veca.vecb + |veca|^2 + 4|vecc|^2 + 4veca.vecc`

= `1 + 4 + 1 + 4 + 4(veca.vecb + veca.vecc)`

= 10 + 4(–2)

= 10 – 8

= 2