Question

Let `veca, vecb, vecc` be three vectors mutually perpendicular to each other and have same magnitude. If a vector `vecr` satisfies. `veca xx {(vecr - vecb) xx veca} + vecb xx {(vecr - vecc) xx vecb} + vecc xx {(vecr - veca) xx vecc} = vec0`, then `vecr` is equal to ______.

Options

• `1/3(veca + vecb + vecc)`

• `1/3(2veca + vecb - vecc)`

• `1/2(veca + vecb + vecc)`

• `1/2(veca + vecb + 2vecc)`

Answer 1

Let `veca, vecb, vecc` be three vectors mutually perpendicular to each other and have same magnitude. If a vector `vecr` satisfies. `veca xx {(vecr - vecb) xx veca} + vecb xx {(vecr - vecc) xx vecb} + vecc xx {(vecr - veca) xx vecc} = vec0`, then `vecr` is equal to `underlinebb(1/2(veca + vecb + vecc))`.

Explanation:

Let `vecr = xveca + yvecb + 2vecc`

Given that `|veca| = |vecb| = |vecc|` and `veca.vecb = vecb.vecc = vecc.veca` = 0

`veca xx [(vecr - vecb) xx veca] + vecb xx [(vecr - vecc) xx vecb] + vecc xx [(vecr - veca) xx vecc] = vec0`

`\implies (veca.veca)(vecr - vecb) - [veca.(vecr - vecb)]veca + (vecb.vecb)(vecr - vecc) - [vecb.(vecr - vecc)].vecb + (vecc.vecc)(vecc - veca) - [vecc.(vecr - veca)]vecc = vec0`

`\implies |veca|^2(vecr - vecb) - (veca.vecr)veca + |vecb|^2(vecr - vecc) - (vecb.vecr)vecb + |vecc|^2(vecc - veca) - (vecc.vecr)vecc = vec0`

= `|veca|^2[3vecr - (veca + vecb + vecc)] - (vecr.veca)veca + (vecr.vecb)vecb + (vecr.vecc)vecc = vec0`  ...`[∵ |veca|^2 = |vecb|^2 = |vecc|^2]`

= `|veca|^2 [3vecr - (veca + vecb + vecc) - (xveca + yvecb + zvecc)] = vec0`

∴ `3vecr - (veca + vecb + vecc) - vecr = vec0`

`\implies vecr = (veca + vecb + vecc)/2`