Question

Let `veca, vecb` and `vecc` be non-coplanar unit vectors equally inclined to one another at an acute angle θ. Then `[(veca, vecb, vecc)]` in terms of θ is equal to ______.

Options

• `(1 + cos θ) sqrt(cos 2θ)`

• `(1 + cos θ) sqrt(1 - 2 cos 2θ)`

• `(1 - cos θ) sqrt(1 + 2 cos 2θ)`

• None of these

Answer 1

Let `veca, vecb` and `vecc` be non-coplanar unit vectors equally inclined to one another at an acute angle θ. Then `[(veca, vecb, vecc)]` in terms of θ is equal to `underlinebb((1 - cos θ) sqrt(1 + 2 cos 2θ))`.

Explanation:

`|veca| = |vecb| = |vecc|` = 1

`veca.vecb` = (1) (1) cos θ = cos θ and 

`vecc.veca` = cos θ, `vecb.vecc` = cos θ

`[(veca, vecb, vecc)]^2 = |(veca.veca, veca.vecb, veca.vecc),(vecb.veca, vecb.vecb, vecb.vecc),(vecc.veca, vecc.vecb, vecc.vecc)| = |(1, cosθ, cosθ),(cosθ, 1, cosθ),(cosθ, cosθ, 1)|`

Operate `R_1 rightarrow R_1 + R_2 + R_3`

= `(1 + 2 cos θ) |(1, 1, 1),(cosθ, 1, cosθ),(cosθ, cosθ, 1)|`

Operate `C_2 rightarrow C_2 - C_1; C_3 rightarrow C_3 - C_1`

= `(1 + 2 cosθ)|(1, 0, 0),(cosθ, 1 - cosθ, 0),(cosθ, 0, 1 - cosθ)|`

= (1 + 2 cos θ) (1 – cos θ)²

∴ `[(veca, vecb, vecc)] = (1 - cos θ)sqrt(1 + 2 cos θ)`