If p = i - 2j + 3k and q = 3i + j + 2k, then | vedclass

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(C) Let the vector $r$ be a linear combination of $p$ and $q$,so $r = p + \lambda q$.Given $p = i - 2j + 3k$ and $q = 3i + j + 2k$.Since $r$ is perpen

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$-\frac{1}{2}(i + 5j - 4k)$

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(C) Let the vector $r$ be a linear combination of $p$ and $q$,so $r = p + \lambda q$.Given $p = i - 2j + 3k$ and $q = 3i + j + 2k$.Since $r$ is perpendicular to $q$,we have $r \cdot q = 0$.Substituting $r = p + \lambda q$,we get $(p + \lambda q) \cdot q = 0$.$p \cdot q + \lambda (q \cdot q) = 0$.Calculate $p \cdot q = (1)(3) + (-2)(1) + (3)(2) = 3 - 2 + 6 = 7$.Calculate $q \cdot q = |q|^2 = 3^2 + 1^2 + 2^2 = 9 + 1 + 4 = 14$.Substituting these values,$7 + 14\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$.Now,$r = p - \frac{1}{2}q = (i - 2j + 3k) - \frac{1}{2}(3i + j + 2k)$.$r = (1 - \frac{3}{2})i + (-2 - \frac{1}{2})j + (3 - 1)k = -\frac{1}{2}i - \frac{5}{2}j + 2k$.$r = -\frac{1}{2}(i + 5j - 4k)$.

If $p = i - 2j + 3k$ and $q = 3i + j + 2k,$ then a vector along $r$ which is a linear combination of $p$ and $q$ and also perpendicular to $q$ is

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