The volume V V V a ^ , b ^ , \hat{a}, \hat{b}, a ^ , b ^ , c ^ \hat{c} c ^ V = ∣ [ a ^ b ^ c ^ ] ∣ V = \left| [\hat{a} \ \hat{b} \ \hat{c}] \right| V = [ a ^ b ^ c ^ ]
Using the property of the Gram determinant for vectors, the square of the scalar triple product can be evaluated as:
V 2 = [ a ^ b ^ c ^ ] 2 = ∣ a ^ ⋅ a ^ a ^ ⋅ b ^ a ^ ⋅ c ^ b ^ ⋅ a ^ b ^ ⋅ b ^ b ^ ⋅ c ^ c ^ ⋅ a ^ c ^ ⋅ b ^ c ^ ⋅ c ^ ∣ V^2 = [\hat{a} \ \hat{b} \ \hat{c}]^2 = \begin{vmatrix} \hat{a} \cdot \hat{a} & \hat{a} \cdot \hat{b} & \hat{a} \cdot \hat{c} \\ \hat{b} \cdot \hat{a} & \hat{b} \cdot \hat{b} & \hat{b} \cdot \hat{c} \\ \hat{c} \cdot \hat{a} & \hat{c} \cdot \hat{b} & \hat{c} \cdot \hat{c} \end{vmatrix} V 2 = [ a ^ b ^ c ^ ] 2 = a ^ ⋅ a ^ b ^ ⋅ a ^ c ^ ⋅ a ^ a ^ ⋅ b ^ b ^ ⋅ b ^ c ^ ⋅ b ^ a ^ ⋅ c ^ b ^ ⋅ c ^ c ^ ⋅ c ^ Since a ^ , b ^ , c ^ \hat{a}, \hat{b}, \hat{c} a ^ , b ^ , c ^ a ^ ⋅ a ^ = b ^ ⋅ b ^ = c ^ ⋅ c ^ = 1 \hat{a} \cdot \hat{a} = \hat{b} \cdot \hat{b} = \hat{c} \cdot \hat{c} = 1 a ^ ⋅ a ^ = b ^ ⋅ b ^ = c ^ ⋅ c ^ = 1
Given that:a ^ ⋅ b ^ = b ^ ⋅ c ^ = c ^ ⋅ a ^ = 1 2 \hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{c} = \hat{c} \cdot \hat{a} = \frac{1}{2} a ^ ⋅ b ^ = b ^ ⋅ c ^ = c ^ ⋅ a ^ = 2 1
Substituting these values into the determinant:
V 2 = ∣ 1 1 2 1 2 1 2 1 1 2 1 2 1 2 1 ∣ V^2 = \begin{vmatrix} 1 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 1 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 1 \end{vmatrix} V 2 = 1 2 1 2 1 2 1 1 2 1 2 1 2 1 1 Expanding the determinant along the first row:V 2 = 1 ( 1 ⋅ 1 − 1 2 ⋅ 1 2 ) − 1 2 ( 1 2 ⋅ 1 − 1 2 ⋅ 1 2 ) + 1 2 ( 1 2 ⋅ 1 2 − 1 ⋅ 1 2 ) V^2 = 1 \left(1 \cdot 1 - \frac{1}{2} \cdot \frac{1}{2}\right) - \frac{1}{2} \left(\frac{1}{2} \cdot 1 - \frac{1}{2} \cdot \frac{1}{2}\right) + \frac{1}{2} \left(\frac{1}{2} \cdot \frac{1}{2} - 1 \cdot \frac{1}{2}\right) V 2 = 1 ( 1 ⋅ 1 − 2 1 ⋅ 2 1 ) − 2 1 ( 2 1 ⋅ 1 − 2 1 ⋅ 2 1 ) + 2 1 ( 2 1 ⋅ 2 1 − 1 ⋅ 2 1 )
V 2 = 1 ( 1 − 1 4 ) − 1 2 ( 1 2 − 1 4 ) + 1 2 ( 1 4 − 1 2 ) V^2 = 1 \left(1 - \frac{1}{4}\right) - \frac{1}{2} \left(\frac{1}{2} - \frac{1}{4}\right) + \frac{1}{2} \left(\frac{1}{4} - \frac{1}{2}\right) V 2 = 1 ( 1 − 4 1 ) − 2 1 ( 2 1 − 4 1 ) + 2 1 ( 4 1 − 2 1 )
V 2 = 3 4 − 1 2 ( 1 4 ) + 1 2 ( − 1 4 ) V^2 = \frac{3}{4} - \frac{1}{2} \left(\frac{1}{4}\right) + \frac{1}{2} \left(-\frac{1}{4}\right) V 2 = 4 3 − 2 1 ( 4 1 ) + 2 1 ( − 4 1 )
V 2 = 3 4 − 1 8 − 1 8 = 3 4 − 1 4 = 1 2 V^2 = \frac{3}{4} - \frac{1}{8} - \frac{1}{8} = \frac{3}{4} - \frac{1}{4} = \frac{1}{2} V 2 = 4 3 − 8 1 − 8 1 = 4 3 − 4 1 = 2 1
Taking the positive square root for volume:V = 1 2 V = \frac{1}{\sqrt{2}} V = 2 1
Thus, the volume of the parallelopiped is 1 2 \frac{1}{\sqrt{2}} 2 1
Therefore, the correct option is (A) .