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Mathematics Question on Vectors

Let a⃗,b⃗\vec{a}, \vec{b} & c⃗\vec{c} be non-coplanar unit vectors equally inclined to one another at an acute angle θ\theta. Then ∣[a⃗  b⃗  c⃗]∣| [ \vec{a} \; \vec{b} \; \vec{c}] | in terms of θ\theta is equal to

A

(1+cos⁡θ)cos⁡2θ( 1 + \cos \theta) \sqrt{\cos 2 \theta}

B

(1+cos⁡θ)1−2cos⁡θ( 1 + \cos \theta) \sqrt{1 - 2 \cos \theta}

C

(1+cos⁡θ)1+2cos⁡θ( 1 + \cos \theta) \sqrt{1 + 2 \cos \theta}

D

None of these

Answer

(1+cos⁡θ)1+2cos⁡θ( 1 + \cos \theta) \sqrt{1 + 2 \cos \theta}

Explanation

Solution

we have,
∣a⃗∣=∣b⃗∣=∣c⃗∣=1|\vec{a}|=|\vec{b}|=|\vec{c}|=1
a⃗⋅b⃗=b⃗⋅c⃗=c⃗⋅a⃗=cos⁡θ\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=\cos \theta
Now,
[a⃗b⃗c⃗]2=∣a⃗⋅a⃗a⃗⋅b⃗a⃗⋅c⃗ b⃗⋅a⃗b⃗⋅b⃗b⃗⋅c⃗ c⃗⋅a⃗c⃗⋅b⃗c⃗⋅c⃗∣[\vec{a} \vec{b} \vec{c}]^{2}=\begin{vmatrix}\vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \\\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{c} \\\ \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{c}\end{vmatrix}
=∣1cos⁡θcos⁡θ cos⁡θ1cos⁡θ cos⁡θcos⁡θ1∣=\begin{vmatrix}1 & \cos \theta & \cos \theta \\\ \cos \theta & 1 & \cos \theta \\\ \cos \theta & \cos \theta & 1\end{vmatrix}
=sin⁡2θ−cos⁡(cos⁡θ−cos⁡2θ)+cos⁡θ(cos⁡2θ−cos⁡θ)=\sin ^{2} \theta-\cos \left(\cos \theta-\cos ^{2} \theta\right)+\cos \theta\left(\cos ^{2} \theta-\cos \theta\right)
=sin⁡2θ−cos⁡2θ+cos⁡3θ+cos⁡3θ−cos⁡2θ=\sin ^{2} \theta-\cos ^{2} \theta+\cos ^{3} \theta+\cos ^{3} \theta-\cos ^{2} \theta
=sin⁡2θ−2cos⁡2θ+2cos⁡3θ=\sin ^{2} \theta-2 \cos ^{2} \theta+2 \cos ^{3} \theta
=1−3cos⁡2θ+2cos⁡3θ=1-3 \cos ^{2} \theta+2 \cos ^{3} \theta
=(cos⁡θ−1)(2cos⁡2θ−cos⁡θ−1)=(\cos \theta-1)\left(2 \cos ^{2} \theta-\cos \theta-1\right)
=(1−cos⁡θ)(cos⁡θ+1−2cos⁡2θ)=(1-\cos \theta)\left(\cos \theta+1-2 \cos ^{2} \theta\right)
=(1−cos⁡θ)(1−cos⁡θ)(2cos⁡θ+1)=(1-\cos \theta)(1-\cos \theta)(2 \cos \theta+1)
=(1−cos⁡θ)2(2cos⁡θ+1)=(1-\cos \theta)^{2}(2 \cos \theta+1)
[a⃗b⃗c⃗]=(1−cos⁡θ)2cos⁡θ+1[\vec{ a } \vec{ b } \vec{ c }]=(1-\cos \theta) \sqrt{2 \cos \theta+1}