Discussion
If \(\vec{a} = i + 2j - 2k\) and \(\vec{b} = 2i + j + 2k\) then a unit vector perpendicular to \(\vec{a} + \vec{b}\) and \(\vec{a} - \vec{b}\) is
Answer: B
A suitable unit vector is \(\frac{2i - 2j - k}{3}\).
\(\vec{a} + \vec{b} = 3i + 3j\) and \(\vec{a} - \vec{b} = -i + j - 4k\). A vector perpendicular to both is their cross product:
\[ \begin{aligned} (3, 3, 0) \times (-1, 1, -4) &= (-12, 12, 6) \\ &= -6(2, -2, -1) \end{aligned} \]
Its length is \(6 \times 3\), so the unit vectors are \(\pm\frac{2i - 2j - k}{3}\). Check: \((2, -2, -1) \cdot (3, 3, 0) = 0\) and \((2, -2, -1) \cdot (-1, 1, -4) = -2 - 2 + 4 = 0\). \(\frac{2i - 2j + k}{3}\) fails the second check, giving -8.
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