The volume of a parallelepiped whose cotermin | vedclass

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Question

(A) The volume of a parallelepiped with coterminous edges $\hat{a}, \hat{b}, \hat{c}$ is given by the scalar triple product $|[\hat{a} \hat{b} \hat{c}

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(A) The volume of a parallelepiped with coterminous edges $\hat{a}, \hat{b}, \hat{c}$ is given by the scalar triple product $|[\hat{a} \hat{b} \hat{c}]|$.The square of the volume is given by the determinant of the Gram matrix:$V^2 = \begin{vmatrix} \hat{a} \cdot \hat{a} & \hat{a} \cdot \hat{b} & \hat{a} \cdot \hat{c} \ \hat{b} \cdot \hat{a} & \hat{b} \cdot \hat{b} & \hat{b} \cdot \hat{c} \ \hat{c} \cdot \hat{a} & \hat{c} \cdot \hat{b} & \hat{c} \cdot \hat{c} \end{vmatrix}$Since $\hat{a}, \hat{b}, \hat{c}$ are unit vectors,$\hat{a} \cdot \hat{a} = \hat{b} \cdot \hat{b} = \hat{c} \cdot \hat{c} = 1$. Given $\hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{c} = \hat{c} \cdot \hat{a} = \frac{1}{2}$,we have:$V^2 = \begin{vmatrix} 1 & 1/2 & 1/2 \ 1/2 & 1 & 1/2 \ 1/2 & 1/2 & 1 \end{vmatrix}$Calculating the determinant:$V^2 = 1(1 - 1/4) - 1/2(1/2 - 1/4) + 1/2(1/4 - 1/2)$$V^2 = 1(3/4) - 1/2(1/4) + 1/2(-1/4)$$V^2 = 3/4 - 1/8 - 1/8 = 3/4 - 2/8 = 3/4 - 1/4 = 2/4 = 1/2$Therefore,the volume $V = \sqrt{1/2} = \frac{1}{\sqrt{2}}$.

The volume of a parallelepiped whose coterminous edges are represented by unit vectors $\hat{a}, \hat{b}, \hat{c}$ such that $\hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{c} = \hat{c} \cdot \hat{a} = \frac{1}{2}$ is:

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