Let $\to a, \to b$ and $\to c$ be non-coplanar unit vectors equally inclined to one another at an acute angle $θ$ then $[\to a \to b \to c]$ in terms of $θ$ is equal to :

(1 + cos θ) \sqrt{cos 2 θ}

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(1 + cos θ) \sqrt{1 - 2 cos 2 θ}

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(1 - cos θ) \sqrt{1 + 2 cos θ}

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(1 - sin θ) \sqrt{1 + 2 cos 2 θ}

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Solution

The correct option is C $(1 - cos θ) \sqrt{1 + 2 cos θ}$
we have,
$\begin{matrix} | \to a | = ∣ ∣ \to b ∣ ∣ = | \to c | = 1 \to a . \to b = \to b . \to c = \to c . \to a = cos θ N o w, {[\to a \to b \to c]}^{2} = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \to a . \to a & \to a . \to b & \to a . \to c \to b . \to a & \to b . \to b & \to b . \to c \to c . \to a & \to c . \to b & \to c . \to c \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} 1 & cos θ & cos θ cos θ & 1 & cos θ cos θ & cos θ & 1 \end{matrix} ∣ ∣ ∣ ∣ = {sin}^{2} θ - cos (cos θ - {cos}^{2} θ) + cos θ ({cos}^{2} θ - cos θ) = {sin}^{2} θ - {cos}^{2} θ + {cos}^{3} θ + {cos}^{3} θ - {cos}^{2} θ = {sin}^{2} θ - 2 {cos}^{2} θ + 2 {cos}^{3} θ = 1 - 3 {cos}^{2} θ + 2 {cos}^{3} θ = (cos θ - 1) (2 {cos}^{2} θ - cos θ - 1) = (1 - cos θ) (cos θ + 1 - 2 {cos}^{2} θ) = (1 - cos θ) (1 - cos θ) (2 cos θ + 1) = {(1 - cos θ)}^{2} (2 cos θ + 1) [\to a \to b \to c] = (1 - cos θ) \sqrt{2 cos θ + 1} \end{matrix}$

Hence, this is the answer.

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