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Consistency of Linear System of Equations

If i, j, k ...

Question

If $i, j, k$ are the unit vectors and mutually perpendicular, then the value of $[\begin{matrix} i - j & j - k & k - i \end{matrix}] =$

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Solution

The correct option is B $0$
$(i - j) . {(j - k) \times (k - i)} = (i - j) . (j \times k - j \times i - k \times k + k \times i) = (i - j) (i + k - 0 + j) = i . i + i . k + i . j - j . i - j . k - j . j = i^{2} + 0 + 0 - 0 - 0 - j^{2} = i^{2} - j^{2} = 1 - 1 = 0$

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\to A = a \to i + \to j + \to k, \to B = \to i + b \to j + \to k, \to C = \to i + \to j + c \to k

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

\vec{a} + \vec{b} and \vec{b} - \vec{c}

\hat{i}

\hat{j}

\hat{k}

\vec{a} = 3 \hat{i} + x \hat{j} - \hat{k} and \vec{b} = 2 \hat{i} + \hat{j} + y \hat{k}

\hat{i}, \hat{j}, \hat{k}

\hat{i} \cdot \hat{j} = 1

\hat{i} \cdot \hat{i} = 1

\hat{i} \times \hat{j} = 1

\hat{i} \times (\hat{j} \times \hat{k}) = 1

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