Find two unit vectors perpendicular to both \(\vec{a} = 3\vec{j} + 2\vec{k}\) and \(\vec{b} = -\vec{i} - 2\vec{k}\). Select one: a. \( -\frac{6}{7}\vec{i} - \frac{2}{7}\vec{j} + \frac{3}{7}\vec{k} \) and \( \frac{6}{7}\vec{i} + \frac{2}{7}\vec{j} - \frac{3}{7}\vec{k} \) b. \( -\frac{6}{7}\vec{i} + \frac{2}{7}\vec{j} + \frac{3}{7}\vec{k} \) and \( \frac{6}{7}\vec{i} - \frac{2}{7}\vec{j} - \frac{3}{7}\vec{k} \) c. \(\vec{k}\) and \( -\vec{k} \) d. \( -6\vec{i} - 2\vec{j} + 3\vec{k} \) and \( 6\vec{i} + 2\vec{j} - 3\vec{k} \) e. \( -\frac{1}{2}\vec{i} + \frac{3}{2}\vec{j} \) and \( \frac{1}{2}\vec{i} - \frac{3}{2}\vec{j} \)
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The cross product of two vectors a and b is given by the formula: a x b = (a2b3 - a3b2)i - (a1b3 - a3b1)j + (a1b2 - a2b1)k In this case, a = 3 + 2k and b = 7 - 2k. Plugging in the values, we get: a x b = ((2)(-2) - (0)(7))i - ((3)(-2) - (0)(7))j + ((3)(7) - Show more…
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