Find the volume of the parallelopiped with its edges represented by the vectors $\hat{i} + \hat{j}, \hat{i} + 2 \hat{j} and \hat{i} + \hat{j} + π k .$

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Solution

$Let : \vec{a} = \hat{i} + \hat{j} \vec{b} = \hat{i} + 2 \hat{j} \vec{c} = \hat{i} + \hat{j} + π \hat{k} We know that the volume of a parallelopiped whose three adjecent edges are \vec{a}, \vec{b} and \vec{c} is equal to |[\vec{a} \vec{b} \vec{c}]| . We have [\vec{a} \vec{b} \vec{c}] = |\begin{matrix} 1 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & π \end{matrix}| = 1 (2 π - 0) - 1 (π - 0) + 0 (1 - 2) = 2 π - π = π ∴ Volume = |[\vec{a} \vec{b} \vec{c}]| = |π| = π cubic units$

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^i +^j,^i + 2^j

^i +^j + π^k

\vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}, \vec{b} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{c} = 3 \hat{i} - \hat{j} + 2 \hat{k}

\vec{a} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}, \vec{b} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{c} = 3 \hat{i} - \hat{j} - 2 \hat{k}

\vec{a} = 11 \hat{i}, \vec{b} = 2 \hat{j}, \vec{c} = 13 \hat{k}

\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} - \hat{j} + \hat{k}, \vec{c} = \hat{i} + 2 \hat{j} - \hat{k}

a = 2 i - 3 j + 4 k, b = i + 2 j - k, c = 3 i - j + 2 k

\to a = 2^i - 3^j - 4^k, \to b =^i + 2^j -^k,

\to c = 3^i +^j + 2^k

\begin{matrix} \to a =^i - 2^j +^3 \to b = - 2^i + 3^j - 4^k \to c =^i - 3^j + 5^k \end{matrix}

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