Determine whether each pair of vectors is orthogonal.
i + 3√2j, 6i - √2j

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Recall that two vectors are orthogonal if their dot product is zero. The dot product of vectors \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} \) and \( \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} \) is given by the formula: \[ \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 \]

Identify the components of the given vectors. For the first vector \( \mathbf{v_1} = \mathbf{i} + 3\sqrt{2} \mathbf{j} \), the components are \( a_1 = 1 \) and \( a_2 = 3\sqrt{2} \). For the second vector \( \mathbf{v_2} = 6 \mathbf{i} - \sqrt{2} \mathbf{j} \), the components are \( b_1 = 6 \) and \( b_2 = -\sqrt{2} \).

Calculate the dot product using the components: \[ \mathbf{v_1} \cdot \mathbf{v_2} = (1)(6) + (3\sqrt{2})(-\sqrt{2}) \]

Simplify the expression by multiplying the terms: \[ (1)(6) = 6 \] and \[ (3\sqrt{2})(-\sqrt{2}) = 3 \times (-1) \times (\sqrt{2} \times \sqrt{2}) = 3 \times (-1) \times 2 = -6 \]

Add the results from the two products to find the dot product: \[ 6 + (-6) = 0 \] Since the dot product is zero, conclude that the vectors are orthogonal.

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