Question

Let `veca, vecb, vecc` be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle θ, with the vector `veca + vecb + vecc`. Then, 36 cos²2θ is equal to ______.

Options

• 1

• 2

• 3

• 4

Answer 1

Let `veca, vecb, vecc` be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle θ, with the vector `veca + vecb + vecc`. Then, 36 cos²2θ is equal to 4.

Explanation:

`|veca + vecb + vecc|^2 = (veca + vecb + vecc).(veca + vecb + vecc)`

= `|veca|^2 + |vecb|^2 + |vecc|^2 + 2(veca.vecb + veca.vecc + vecb.vecc)`

= `|veca|^2 + |vecb|^2 + |vecc|^2 + 0`

∴ `|veca + vecb + vecc|^2 = |veca|^2 + |vecb|^2 + |vecc|^2` = 3k²

∴ `|veca + vecb + vecc| = sqrt(3)k`

Now, `veca.(veca + vecb + vecc) = |veca||veca + vecb + vecc|cosθ`

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

⇒ k² + 0 = `k xx sqrt(3)kcosθ`

⇒ cosθ = `1/sqrt(3)` ⇒ cos2θ = 2cos²θ – 1

⇒ cos2θ = `(-1)/3` ⇒ cos²2θ = `1/9`

⇒ 36 cos²2θ = 4